Selina Concise Biology Class 7 ICSE Solutions – Allergy

Selina Concise Biology Class 7 ICSE Solutions – Allergy

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Biology. You can download the Selina Concise Biology ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Biology for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Biology Chapter 7 Allergy

Review Questions

Question 1.
Define the term “allergy”.
Answer:
Allergies, also known as allergic diseases, are a number of conditions caused by hypersensitivity of the immune system to something in the environment that usually causes no adverse effects to most people. These diseases include hay fever, dermatities, asthma, food allergies, etc.

Question 2.
Name the inflammatory chemical released by the basophils causing skin reaction and construction of the bronchioles.
Answer:
The histamine causes inflammation, skin rash and constriction of the bronchioles (as in asthma).

Question 3.
How the allergy causing chemical is naturally removed from the body ?
Answer:
There are no permanent cure for an allergy. Medication for allergies do give relief. Those people prone to allergic reactions should be aware of the allergens that trigger these responses in their bodies and avoid them completely, especially food and medicines. Allergy shots are helpful to those with seasonal allergies like those caused by pollen.

4. Match the items given under column I with those given under column II.
column I                                         column II
Selina Concise Biology Class 7 ICSE Solutions - Allergy 1
Answer:
              column I                                         column II
Selina Concise Biology Class 7 ICSE Solutions - Allergy 2

Question 5.
Name any three routes of which allergens normally enter in our body.
Answer:

  1. Skin — Substances that come into contact with the skin, such as latex, are also common causes of allergic reactions, known as contact dermatitis or eczema.
  2. Mouth — A wide variety of foods can cause allergic reactions, but 90% of allergic responses to foods’are caused by cow’s milk, soy, eggs, wheat, peanuts, tree nuts, fish and shellfish.
  3. Nose — Pollen, spores, dust particles and certain chemicals find entry through the nose.

Question 6.
What are allergens ? Name any three allergens.
Answer:
The substances that cause allergies are called allergens.

  1. Dust particles
  2. Nuts
  3. Pollen

Question 7.
Name the two tests generally used to diagnose allergy and describe any one of them.
Answer:

  1. Prick Test—Skin prick test is also known as “puncture testing” or “prick testing” due to the series of tiny punctures or pricks made into the patients skin. A small plastic or metal device is used to puncture or prick the skin. Sometimes, the allergens are injected intradermally into the patients skin, with a needle and syringe. Small amounts of suspended allergens or their extracts (e.g., pollen grain, insect sting proteins, peanut extracts) are introduced to sites on the skin marked with pen or dye. Common areas for testing include the inside of forearm and the back.
  2. Blood Testing — A sample of the patient’s blood is sent to a laboratory for analysis. Multiple allergence can be detected with a single blood sample.
    The test measures the concentration of specific antibodies in the blood. The quantitative allergy blood result can help determine what a patient is allergic to, help predict and follow the disease development, estimate the risk of a severe reaction, and explain cross-reactivity.

Question 8.
Which test is mostly favoured for testing allergy, prick test or blood test ? Why ?
Answer:
Blood test is mostly favoured for testing allergy.

Blood Testing — A sample of the patient’s blood is sent to a laboratory for analysis. Multiple allergence can be detected with a single blood sample.
The test measures the concentration of specific antibodies in the blood. The quantitative allergy blood result can help determine what a patient is allergic to, help predict and follow the disease development, estimate the risk of a severe reaction, and explain cross-reactivity.

Selina Concise Physics Class 7 ICSE Solutions – Sound

Selina Concise Physics Class 7 ICSE Solutions – Sound

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Physics. You can download the Selina Concise Physics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Physics for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Physics Chapter 6 Sound

  • Points to Remember
  •  Sound is a form of energy which produces the sensation of hearing.
  •  Sound requires a medium to travel. It cannot travel in vacuum.
  •  Sound travels in all directions. Sound can travel through solids, liquids and gases.
  •  Sound travels faster in solids than in liquids and gases.
  •  The loudness of sound is measured on a special scale called the decibel scale.
  •  Sound is classified into two groups. (i) Musical sound, (ii) Noise.
  •  Musical sound is produced by periodic vibrations.
  •  Noise is produced by non-periodic motions.
  •  What the sound gets repeated after reflection from a distant body, is called an echo.
  •  Multiple echoes are called reverberations.
  •  Sonar is method used to measure the depth of a sea and is based on the principle of echo.
  •  Music is pleasent to hear and is caused by periodic vibrations.
  •  Noise is irritating and is produced by an irregular pattern of waves.
  •  The three characteristics of sound are loudness, pitch, quality or timbre.
  •  Loudness depends upon many factors, e.g. amplitude, distance, waves of vibrating body, humidity, pressure, wind velocity etc.
  •  Pitch depends on the frequency of sound. Higher is the pitch, shriller is the sound.
  •  Loudness is merely a sensation.

Test Yourself

A. Objective Questions 

1. Write true or false for each statement

(a) Sound can travel in vacuum.
Answer. False.
Correct — Sound requires medium to travel.

(b) Sound is a form of energy.
Answer. True.

(c) Sound can only be produced by vibrating bodies.
Answer. True.

(d) Larger is the amplitude, feeble is the sound.
Answer.False.
Correct — Larger the amplitude, greater is the sound.

(e) The frequency is measured in hertz.
Answer. True.

(f) Loudness depends on frequency.
Answer. False.
Correct — Loudness depends on the amplitude.

(g) Waveforms of two different stringed instruments can be the same.
Answer. False.
Correct—Waveforms of two different stringed instruments cannot be the same.

(h) Female voice is shriller than the male voice.
Answer. True.

(i) A ticking clock sound is heard late when heard through a metal.
Answer. False.
Correct—A ticking clock sounds is heard early when heard through a metal.

2. Fill in the blanks

(a) Sound is produced when a body vibrates.
(b) The number of times a body vibrates in one second is called its frequency.
(c) The pitch of a sound depends on its frequency.
(d) Sound can travel in a medium solid, liquid or gas.
(e) We can hear sounds of frequency in the range of 20 Hz to 20,000 Hz.
(f) Sound requires a medium for propagation.
(g) Sound travels faster in solids than in liquids.
(h) The sound heard after reflection is echo.
(i) Sound produces sensation in ears.

3. Match the following

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 1

4. Select the correct alternative 

(a) We can distinguish a shrill sound from a flat sound by its

  1.  amplitude
  2.  loudness
  3.  pitch
  4.  none of the above.

(b) We can hear sound of frequency

  1.  10 Hz
  2.  500 Hz
  3.  100,000 Hz
  4.  50,000 Hz

(c) Sound cannot travel in

  1.  gases
  2.  liquids
  3.  solids
  4.  vacuum

(d) The minimum distance required between the source and the reflector so as to hear the echo in air is

  1.  10 m
  2. 17 m
  3.  34 m
  4.  50 m

(e) Wavelength is measured in

  1.  kg
  2.  second
  3.  litre
  4. metre

(f) The speed of sound in water is

  1.  332 m
  2. 1500 m
  3.  5000 m s
  4.  1000 m s

(g) Sound travels the fastest in

  1.  liquids
  2.  solids
  3.  gases
  4.  vacuum

B. Short/Long Answer Questions

Question 1.
What do you mean by a vibratory motion ?
Answer:
The oscillatory motion in which the body assumes a new shape during its motion, is called the vibratory motion.

Question 2.
What is sound ?
Answer:
Sound is a form of energy which produces the sensation of hearing.

Question 3.
How is sound produced ?
Answer:
Sound is produced by vibrating bodies.

Question 4.
Describe an experiment to show that each source of sound is a vibrating body.
Answer:
Sound is produced when a body vibrates. In other words, each source of sound is a vibrating body. This can be demonstrated by the following experiment.
Take a ruler. Press its one end on the table with the left hand as shown in figure. Pull down the other end of the ruler with the right hand and then leave it.
You will notice that the ruler vibrates i.e., the ruler moves to and fro and a humming sound is heard.
After some time, the ruler stops vibrating. No sound is then heard.

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 2
This shows that the humming sound is produced only because of the vibrations of the ruler.

Question 5.
Name two sources of sound.
Answer:
Each vibrating body is a source of sound. We, the human beings, produce sound when our vocal cords vibrate on blowing air through them by our lungs. Some animals like birds, frogs etc., also produce sound due to vibration of their vocal cords. But bees do not have the voice-boxes. They produce sound by moving their wings up and down very fast.

Question 6.
How do we produce sound ?
Answer:
Our throat has a larynx. The voice is produced in the larynx. Larynx is also called the voice box. It is designed to produce voice. It is a box like structure with walls of tough tissues. Inside two folds of the tissue, there is a gap. They are the vocal cords. When we breathe, the vocal cords become loose and the gap between them increases. When we talk, shout or sing, the cords become tight and hence they vibrate, thus produce sound. Given figure shows the part of the body which vibrates to produce sound.

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 3

Question 7.
The bees do not have voice-boxes. How do they produce sound ?
Answer:
The bees do not have the voice-boxes. Still they produce sound.
This happens by the vibrations produced by the quick movement of their wings. Bees buzz while flying and depositing pollen among flowers.

Question 8.
Can sound travel through a vacuum ? Describe an experiment to explain your answer.
Answer:
Experiment — Arrange an electric bell, a glass bell jar, a vacuum pump, a battery and a switch as shown in the figure. When the circuit is closed by pressing the switch, the bell starts ringing and sound can be heard. Now remove the air from the jar with the help of vacuum pump. The loudness of the sound gradually decreases and a stage comes when no sound is heard. Sound requires a medium to travel but cannot travel in vacuum.

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 4
Connect the bell to a battery through a switch. On pressing the switch, the bell starts ringing and a sound is heard. The sound reaches us through the air in the jar.
Now start the vacuum pump. It withdraws the air from the jar. You will notice that as the jar is evacuated, the sound becomes feeble and feeble. After some time when no air is left within the jar, no sound is heard. However, the hammer of the electric bell can be still seen striking the gong. The reason is that when no air is left in the jar, the sound does not reach us, although the bell is still ringing (or vibrating).
Thus, sound cannot travel through a vacuum.

Question 9.
Describe an experiment to show that sound can travel in water.
Answer:
Take a tub filled with water. Hold a bell in one hand and dip it in water. Keep one of your ears gently on the surface of water without letting water into the ear. Now ring the bell inside water. You will be able to hear the sound clearly. This shows that sound can travel through liquids.

Question 10.
Describe an experiment to show that sound can travel in a solid.
Answer:
Take two empty ice-cream cups. Make a small hole at the bottom of each cup and pass a long thread (about 20 m long) through them. Tie a knot or match-stick at each end of the thread so that the thread does not slip out through the holes. This makes a toy – telephone

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 5
Now use the toy-telephone as shown in figure and talk to your friend. You will be able to hear the sound of your friend. This shows that sound travels through the thread and reaches your ear. Thus, sound can travel through a solid.

Question 11.
Can two person hear each other on moon’s surface ? Give reason to support your answer.
Answer:
No, we cannot hear each other since sound requires medium for transmission. It cannot travel through vacuum.

Question 12.
What is a longitudinal wave ?
Answer:
In a longitudinal wave, the particles of air vibrate to and fro about their mean positions in the direction of travel of sound.

Question 13.
Define the following terms :
Amplitude, Time period, Frequency.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 6
(a) Amplitude (A) : The maximum displacement of a wave on either side of its mean position is called Amplitude. A = XY is amplitude.
(b) Time Period (T) : Time taken to complete one vibration is called Time Period, i.e. from A to B

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 7
(c) Frequency (f) or u
Number of oscillations made by a wave in one second is known as its frequency.

Question 14.
Write the audible range of frequency for the normal human ear.
Answer:
The range of frequency from 20 Hz to 20,000 Hz is called the audible range for the normal human ear.

Question 15.
What are ultrasonics ? Can you hear the ultrasonic sound ?
Answer:
Sounds of frequency higher than 20,000 Hz are called the ultrasonics. We cannot hear the ultrasonic sounds.

Question 16.
What are infrasonics ? Can you hear them ?
Answer:
Sounds of frequency lower than 20 Hz are called the infrasonics. We cannot hear the infrasonic sounds.

Question 17.
How does a bat make use of ultrasonics waves to find its way?
Answer:
Use of ultrasonics by bats : Bats have no eyes. But they easily move about without colliding with any object (or obstacle). The reason is that they produce ultrasonic sound as they fly. When this ultrasonic sound comes back after reflection from any object (or obstacle) in their way, they hear it and thus they detect the presence of the object (or obstacle).

Question 18.
Name the two characteristics of sound which differentiate two sounds from each other.
Answer:
A sound wave is characterized by its amplitude and frequency. Depending upon the (amplitude and frequency of the sound wave, the following two characteristics of sound :
(1) Loudness, and (2) Pitch.

Question 19.
On what factor does the loudness of a sound depend ?
Answer:
The loudness of a sound depends on the amplitude of vibration of the vibrating body producing the sound.

Question 20.
How does the loudness of sound produced depend on the vibrating area of the body ?
Answer:
The loudness of sound also depends on the area of the vibrating body. Greater the area of the vibrating body, louder is the sound produced.
If you take two drums, one small and the other big, and beat both of them to produce vibrations in them, We will notice that the sound produced from the big drum is louder than that produced from the small drum. In temples, you must have noticed that the bell with a big case produces a louder sound than that with a small case.

Question 21.
The outer case of the bell in a temple is made big. Give a reason.
Answer:
The outer case of the bell in a temple is made big. So that there is multiple reflection of sound and the sound can be amplified.

Question 22.
State the factors on which the pitch of a sound depends.
Answer:
The pitch of a sound depends on its frequency (i.c., on the frequency of the vibrating body).

Question 23.
Differentiate between a high pitch sound and a low pitch sound.
Answer:
Higher the pitch, the shriller is the sound. Lower the pitch, the flat (or grave) is the sound.

Question 24.
How does a man’s voice differ from a woman’s voice ?
Answer:
A female voice is shriller than a male voice because of higher frequency. Higher is the frequency, shriller is the sound. Female has higher frequency.

Question 25.
Name the characteristic which differentiates two sounds of the same pitch and same loudness.
Answer:
The quality is the characteristic of sound which distinguishes the two sounds of the same pitch and same loudness.

Question 26.
You recognize your friend by hearing his voice on a telephone. Explain.
Answer:
We can recognize our friend by hearing his voice on a telephone due to quality of sound and pitch of sound.

Question 27.
A musician recognizes the musical instrument by hearing the sound produced by it, even without seeing the instrument. Which characteristic of sound makes this possible ?
Answer:
It is the pitch and quality that helps a musician recognize the musical instrument by hearing the sound produced by it, even without seeing the instrument.

Question 28.
Describe an experiment to show the production of sound having low and high pitch.
Answer:
Take few rubber bands some thicker and longer, few thinner and of shorter length. Cut and stretch these rubber bands by holding one end of the string in your mouth under the teeth and the other end in your hand. Now pluck these rubber bands one by one. The rubber bands thicker and longer will produce sound with a lower pitch. The rubber bands thinner and shorter will produce sound with a higher pitch.

Question 29.
How does a musician playing on a flute change the pitch of sound produced by it ?
Answer:
In musical instruments like flute and clarinet, the pitch of sound is changed by changing the length of vibrating air column when different holes in it are closed.

Question 30.
Why are musical instruments provided with more than one string ?
Answer:
The stringed instruments are provided with a number of strings of different thickness and under different tensions so that each string produces sound of a different pitch.

Question 31.
How can the pitch of sound produced in a piano be changed ?
Answer:
In a piano, the string is struck to make the string vibrate and produce sound. The pitch of sound produced can be changed by stretching or loosening the strings of piano.

Question 32.
Explain why you can predict the arrival of a train by placing your ear on the rails without seeing it.
Answer:
The sound produced by the moving wheels of train travels much faster through the track than through the air. Therefore they hear through the track much before it is heard through the – air.

Question 33.
Write the approximate speed of sound in (i) air, (ii) water and (iii) steel.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 8

Question 34.
During a thunderstorm, the sound of a thunder is heard after the lightning is seen. Why ?
Answer:
The velocity of light is 3 × 10m/s whereas velocity of sound is 332 m/s. First we see the flash of light and then we hear the thunder.

Question 35.
Describe an experiment to estimate the speed of sound in air.
Answer:
To estimate the speed of sound in air suppose we choose two  hills A and B about a kilometer apart. A person at the hill A fires  a gun. Another person at the hill B starts a stop watch as he sees the flash of the fire and stops it on hearing the sound. Thus, he measures the time interval between the seeing of flash and hearing of the sound. Let it be t second. Then measure the distance between the hills A and B. Let it be S metre.

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 9
Experimentally, it is found that the speed of sound in air is nearly 330 m s-1

Question 36.
Can sound travel through solids and liquids ? In which of these two does it travel faster ?
Answer:
Sound travels with highest speed in — solids.
and Sound travels with lowest speed in — gases.

Question 37.
What do you mean by reflection of sound ?
Answer:
Reflection of Sound— When a sound wave strikes a rigid surface, it retraces from its path is called reflection of sound.

Question 38.
State one use of reflection of sound.
Answer:
The reflection of sound is used in making the speaking tube (or megaphone), sound board and trumpet.

Question 39.
What is echo ?
Answer:
Echo is the sound heard after reflection from a rigid surface such as a cliff, a hillside, the wall of a building etc.

Question 40.
What minimum distance is required between the source of sound and the reflecting surface to hear an echo ? Give reason.
Answer:
Since sound has to travel an equal distance in going up to the reflecting surface and in coming back from the reflecting surface, therefore it must travel nearly 33/2 = 16.5 m either way. Thus, to hear the echo clearly in air, the reflecting surface should be at a minimum distance of 16.5 m from the source of sound.

Question 41.
List four substances which are good absorbers of sound.
Answer:
When sound falls on sofa, fluffs and light substances such as clothes, papers, thermocol, coating of plaster of paris, carpets, curtains, furniture, wood etc., they absorb the sound to a good extent. These are called good absorbers of sound.

Question 42.
List the measures that you will take when designing a sound-proof room.
Answer:
In order to design such a sound proof room we take the following measures
(1) The roof of the enclosure must be covered by plaster of paris after putting the sheets of thermocol.
(2) The walls of the enclosure should be covered by the wooden strips.
(3) The floor must be laid down by thick carpets.
(4) The machine parts of all the electrical equipments such as fan, air conditioner etc. must be placed outside the enclosure.
(5) Thick curtains should be used to cover the doors and keep them closed.
(6) Thick stripping must be used to cover the openings of doors and windows.

C. Numericals

Question 1.
A boy fires a gun and another boy at a distance of 1020 m hears the sound of firing the gun 3 s after seeing its smoke. Find the speed of sound.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 6 Sound 10

Question 2.
A boy on a hill A fires a gun. The other boy on hill B hears the sound after 4 s. If the speed of sound is 330 ms-1, find the distance between the two hills.
Answer:
Speed, v = 330 ms-1
t = 4s
Distances = v × t
= 330 × 4s = 1320 m Ans.

Selina Concise Physics Class 7 ICSE Solutions – Energy

Selina Concise Physics Class 7 ICSE Solutions – Energy

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Physics. You can download the Selina Concise Physics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Physics for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Physics Chapter 3 Energy

  • Points to Remember
  •  Work is said to be done if the applied force on the body moves it. If no motion takes place, no work is said to be done.
  •  The amount of work done depends on two factors : (i) on the magnitude of the force applied (greater the force applied, greater is the work done), and (ii) on the distance moved in the direction of force (greater the distance moved, greater is the work done).
  •  The work done by a force on a body is equal to the product of the force and the distance moved by the body in the direction of force, i.e. Work done = Force × distance moved in the direction of force Or W = F × d
  •  The S.I. unit of work is joule (J), where 1 joule (J) = 1 newton (N) × 1 metre (m)
  •  The energy of a body is its capacity (or ability) to do work. The energy of a body in a state is equal to the work done on the body to bring it to that state.
  •  The S.I. unit of energy is joule (J).
  •  Kinetic energy of a body is the energy possessed by it due to its motion. It is the energy stored when work is done to bring the body in motion.
  •  Kinetic energy of a moving body depends on two factors :
    (i) on the mass of the body (greater the mass of the body, greater is its kinetic energy), and (iii) on the speed of the body (more the speed of the body, higher is its kinetic energy).
  •  The potential energy changes into the kinetic energy when it is put to use.
  •  In transformation of energy, the total sum of useful and non-useful energy obtained after conversion is equal to the energy converted, i.e. the total energy remains conserved.
  •  According to the conservation of mechanical energy, if friction is neglected, the total sum of potential energy and kinetic energy remains constant. Examples are : motion of roller coaster, free vertical fall of a body etc.
  •  The electricity obtained from the energy possessed by the flowing water is called the hydro-electricity.

Test Yourself

A. Objective Questions 

1. Write true or false for each statement

(a) A man going up has potential energy and kinetic energy both.
Answer. True.

(b) A gum bottle lying on a table has no energy.
Answer. False.
Correct — A gum bottle lying on a table has energy.

(c) In an electric fan, electrical energy changes into the mechanical energy.
Answer. True.

(d) Potential energy changes into kinetic energy when it is put to use.
Answer. True.

(e) One form of energy cannot be converted into another form.
Answer. False.
Correct — One form of energy can be converted into the other form.

(f) There is always some loss of energy in conversion from one form of energy to another form, so the total energy is not conserved.
Answer. False.
Correct — There is always some loss of energy in conversion from one form of energy to the other form, so the total energy is conserved.

(g) The energy of flowing water can be converted into electric energy (electricity).
Answer. True.

2. Fill in the blanks

(a) An electric fan converts electrical energy into mechanical energy.
(b) Cooking gas converts chemical energy into heat energy.
(c) Energy possessed by a compressed spring is potential energy.
(d) The ability to do work is called energy
(e) The energy possessed by a body due to its position is called potential energy.
(f) The energy possessed by a body due to its motion is called kinetic energy.
(g) Green plants convert light energy into chemical energy.
(h) The S.I.unit of energy is joule
(i) An object falling freely from the roof of a multistory building has potential energy and kinetic energy when halfway down the building.

3. Match the following columns

Selina Concise Physics Class 7 ICSE Solutions Chapter 3 Energy 1

Selina Concise Physics Class 7 ICSE Solutions Chapter 3 Energy 2

4. Select the correct alternatives 

(a) When we rub our hands

  1.  kinetic energy changes into potential energy
  2.  mechanical energy changes into heat energy
  3.  potential energy changes into kinetic energy
  4.  heat energy changes into mechanical energy.

(b) A ball rolling on the ground possesses

  1.  kinetic energy
  2.  potential energy
  3.  no energy
  4.  heat energy

(c) The energy stored in an electric cell is

  1.  chemical energy
  2.  electrical energy
  3.  heat energy
  4.  mechanical energy.

(d) When a bulb lights up on passing current, the change of energy is

  1.  from electrical energy to heat energy
  2.  from electrical energy to light energy
  3.  from electrical energy to heat and light energy
  4.  from electrical energy to mechanical energy.

(e) The correct statement is

  1.  Both work and energy have the same units
  2.  Potential energy of a body is due to its motion
  3.  Kinetic energy of a body is due to its position or state
  4.  Kinetic energy can change into potential energy, but potential energy cannot change into kinetic energy.

(f) According to law of conservation of energy, energy changes from one form to another form, but the total energy of that system

  1.  increases
  2.  decreases
  3.  alternates
  4.  remains the same

B. Short/Long Answer Questions

Question 1.
Define the term energy.
Answer:
Energy is the capacity of doing work.

Question 2.
State the unit of energy and define it.
Answer:
The energy is measured in the same unit as work. Therefore the S.I. unit of energy is joule (symbol J).
A body is said to possess an energy of one joule if a force of 1 newton moves the body by a distance of 1 metre in the direction of force.
Another unit of energy is calorie (symbol cal) where 1 cal = 4.2 J. A bigger unit is kilo-calorie (symbol kcal) where 1 kcal = 1000 cal.

Question 3.
Name five different forms of energy.
Answer:
The different forms of energy are :
(i) Mechanical energy
(ii) Heat energy
(iii) Light energy
(iv) Chemical energy
(v) Sound energy
(vi) Magnetic energy
(vii) Electrical energy and
(viii) Atomic energy or nuclear energy.

Question 4.
What are the two kinds of mechanical energy.
Answer:
The mechanical energy is found in two forms namely :
(a) The potential energy, and (b) The kinetic energy.

Question 5.
What is potential energy ? State its unit.
Answer:
The energy of a body at rest is called the potential energy. It is defined as follows :
Potential energy of a body is the energy possessed by it due to its state of rest or position. Actually, it is the work spent in bringing the body to that state of rest or position.
It is written as P.E. or U.
The S.I. unit of potential energy is Joules.

Question 6.
Give one example of a body that has potential energy, in each of the following : (i) due to its position, (ii) due to its state.
Answer:
(i) Potential energy of a body is the energy possessed by it due to its state of rest or position. It is the energy stored when work is done on the body to bring it to that state or position.
(ii) Potential energy of a body in the raised (or lifted) position depends on two factors : (1) the mass of the body greater the mass of the body, greater is the potential energy of the body), and (2) the height of the body above the ground (greater the height of the body, greater is its potential energy.)

Question 7.
State two factors on which the potential energy of a body at a certain height above the ground depends.
Answer:
The potential energy of a body in the raised position depends upon the following two factors :
(a) The mass of the body: Greater the mass of the body, greater is the potential energy of the body.
(b) Its height above the ground : Higher the height of the body, greater is its potential.

Question 8.
Two bodies A and B of masses 10 kg and 20 kg respectively are at the same height above the ground. Which of the two has the greater potential energy ?
Answer:
The body B having mass 20 kg has the greater potential energy. This can be explained as follows :
P.E. = mgh .
For both the bodies gravity and height are same so the body with greater mass possesses greater potential energy.

Question 9.
A bucket full of water .is on the first floor of your house and another identical bucket with same quantity of water is kept on the second floor. Which of the two has greater potential energy ?
Answer:
A bucket full of water kept on second floor has the greater potential energy. This can be explained as follows :
P.E. = mgh
Mass of both bucket and the gravitational force are same, so the body at greater height will possess more potential energy.

Question 10.
Define the term kinetic energy. Give one example of a body which possesses kinetic energy.
Answer:
The energy of a body in motion is called its kinetic energy. It is defined as follows :
Kinetic energy of a body is the energy possessed by it due to its state of motion.
Actually, it is the work done on the body bringing it to the state of motion. In short form it is written as K.E. or K.
Example : In a swinging pendulum moving to and fro, the bob has the kinetic energy.

Selina Concise Physics Class 7 ICSE Solutions Chapter 3 Energy 3

Question 11.
State two factors on which the kinetic energy of a moving body depends.
Answer:
The kinetic energy of a moving body depends on the following two factors :
(a) The mass of the body — Greater the mass of the body, higher is its kinetic energy.
(b) The speed of the body — More the speed of the body, higher is its kinetic energy.

Question 12.
Two toy-cars A and B of masses 500 g and 200 g respectively are moving with the same speed. Which of the two has the greater kinetic energy?
Answer:
The toy car ‘A’ of mass ‘500 gm’ has the greater kinetic energy. This can be explained as :
K.E. = 1 / 2 Mv2
Here, both the cars are moving with same speed. So the car with greater mass will possess greater kinetic energy.

Question 13.
A cyclist doubles his speed. How will his kinetic energy change: increase, decrease or remain same ?
Answer:
When a cyclist doubles his speed. His kinetic energy increases four times.
K.E. = 1 / 2 Mv2
When v = doubles the K.E. quadriples

Question 14.
Name the form of energy which a wound up watch spring possess.
Answer:
A wound up watch spring has the potential energy because of its wound up state. As the spring unwinds itself, the potential energy changes into the kinetic energy. This kinetic energy does work in moving the arms of the watch

Question 15.
Can a body possess energy even when it is not in motion ? Explain your answer with an example.
Answer:
Yes, a body possesses energy even when it is not in motion ;
Consider a body raised to a certain height say h. It its velocity is zero. Kinetic energy will be zero but the body will have.
P.E. = mgh
Thus, a body may possess energy even though it is not in motion.

Question 16.
Name the type of energy (kinetic or potential) possessed by the following :
(i) A moving cricket ball.
(ii) A stone at rest on the top of a building.
(iii) A compressed spring.
(iv) A moving bus.
(v) A bullet fired from a gun.
(vi) Water flowing in a river.
(vii) A stretched rubber band.
Answer:
(i) Kinetic energy.
(ii) Potential energy.
(iii) Potential energy.
(iv) Kinetic energy.
(v) Kinetic energy.
(vi) Potential energy.
(vii) Potential energy.

Question 17.
Give one example to show the conversion of potential energy to kinetic energy when put in use.
Answer:
The example to show the conversion of potential energy to kinetic energy when put in use is :
A stone at a height has the potential energy due to its lifted or raised position. In the figure below when the stone is dropped from that position, it begins to fall. The falling stone has the kinetic energy. Thus, the potential energy stored in the stone in its raised position changes into the kinetic energy when the stone is falling. This kinetic energy does work on the nail as the stone strikes the nail arid makes the nail to move into the wood.

Selina Concise Physics Class 7 ICSE Solutions Chapter 3 Energy 4
Similarly, in the figure below the potential energy possessed by the stone at a height changes into its kinetic energy when it falls, The kinetic energy of the falling stone does work in raising the weight upwards.

Selina Concise Physics Class 7 ICSE Solutions Chapter 3 Energy 5

Question 18.
State the energy changes that occur in the following :
(i) The unwinding of a watch spring.
(ii) Burning coal while operating a steam engine.
(iii) Lighting of a torch bulb.
(iv) An electric generator (or dynamo).
Answer:
(i) Potential energy to kinetic energy.
(ii) Chemical energy of coal changes to heat energy of the steam. Heat energy changes into mechanical energy.
(iii) Chemical energy into light and heat energy.
(iv) Electrical energy change into mechanical energy.

Question 19.
Energy can exist in several forms and may change from one form to another. Give two examples to show the conversion of energy from one form to another.
Answer:
The examples that show the conversion of energy from one form to another are :
(1) In a steam engine, the chemical energy of the coal first changes into the heat energy of the steam. Then heat jenergy of steam changes into the mechanical energy which makes the train to move.
(2) In an electric motor (or in fan), the electrical energy changes into the mechanical energy. This energy rotates the axle of motor (or the blades of the fan).

Question 20.
Give one relevant example for each of the following transformation of energy :
(i) Electrical energy to heat energy.
(ii) Electrical energy to mechanical energy.
(iii) Electrical energy to light energy.
(iv) Chemical energy to heat energy.
(v) Chemical energy to light energy.
Answer:
(i) In an electric heater, oven, geyser, toaster etc., the electrical energy changes into heat energy.
(ii) An electric generator.
(iii) Tube eight of bulbs.
(iv) Burning of wood, coal etc.
(v) Fire crackers burst

Question 21.
What do you mean by conservation of mechanical energy? State the condition when does it hold.
Answer:
This means “The total MECHANICAL ENERGY (P.E + K.E) of an isolated system at any instant is equal to the sum of kinetic ENERGY and the potential ENERGY.”
Condition : Condition under which the mechanical energy is conserved is “WHEN THERE ARE NO FRICTIONAL FORCES.” In other words the mechanical energy is conserved strictly in vacuum where friction due to air is absent.

Selina Concise Physics Class 7 ICSE Solutions Chapter 3 Energy 6

Question 22.
Give one example to show that the sum of potential energy and kinetic energy remains constant if friction is ignored.
Answer:
During the vertical fall of ball, if friction due to air is neglected, the total sum of potential energy and kinetic energy at each point of its path remains same.

Question 23.
A ball is made to fall freely from a height. State the kind/ kinds of energy possessed by the ball when it is
(a) at the highest point
(b) just in the middle
(c) at the ground.
Answer:
(a) Potential energy.
(b) Potential energy + Kinetic energy.
(c) Kinetic energy.

Question 24.
State the changes in form of energy while producing hydro electricity.
Answer:
The water in motion in a river or sea has the kinetic energy. The energy possessed by the flowing water is called the hydro energy. The most important use of hydro energy is to produce electricity from it.

Selina Concise Physics Class 7 ICSE Solutions Chapter 3 Energy 7
Figure shows the principle of a hydroelectric power plant. The flowing water of river is collected in a dam at a high altitude. The water stored in the dam has the potential energy. When water from dam falls on the water turbine, the potential energy of the water stored in dam changes into its kinetic energy and this kinetic energy of water is transferred to the blades of turbine as the kinetic energy which rotates the turbine. As the turbine rotates, it rotates the armature of the generator (or dynamo) to produce electricity.

 

 

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions (Including Problems)

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions (Including Problems)

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POINTS TO REMEMBER

  1. Fraction. A rational number in form of — where a and b are integers is called a fraction.
    ‘a’ is called the numerator and Lb’ is called the denominator of the fraction.
  2. Classification of Fractions :
    Decimal fraction : A fraction whose denominator is 10 or multiple of 10.
    Vulgur fraction : A fraction whose denominator is oilier than 10 or multiple of 10.
    Proper fraction : A fraction whose denominator is greater than its numerator.
    Improper fraction : A fraction whose denominator less than its numerator.
    Mixed fraction : A fraction which consists of an integer and a proper fraction.
    Note. If the numerator of a fraction is equal to its denominator, then the fraction is equal to unity i.e. 1.
  3. Equivalent Fractions
    Fractions having the same value are called the equivalent fractions.
  4. Simple and Complex Fractions
    A fraction whose numerator and denominator both are integers, is called a simple fraction.
    A fraction whose numerator or denominator or both are not integers, is called a complex fraction.
  5. Like and Unlike Fractions
    Fractions having the same denominators are called like fractions.
    The fractions with different denominators are called unlike fractions.
  6. Converting unlike fractions into like fractions
    Find the LCM of the denominators of all the give- fractions.
    For each given fraction, multiply its denominator by a suitable numbers so that the product obtained is equal to the LCM in (i).
    Multiply the numerator also by the same number.
  7. To insert a fraction between two given fractions .
    Add the numerators as well as denominators of the given fractions. Then simplify if required.
  8. Addition and Subtraction of fractions
  9. For like fractions, add or subtract (as required) their numerators, keeping the denominator same.
    For unlike fractions, first change all the fractions into like fractions and then add or subtract as above given in (i).
  10. Multiplication
    To multiply two or more fractions, multiply their numerators as well as their denominators.
  11. Division
    To divide on fraction or integer by some other fractions or integer, multiply the first by the reciprocal of the second as given above in multiplication.
  12. Using ‘BODMAS’
    The word ‘BODMAS’ is the abbreviation formed by taking the initial letters of six operations i.e. ‘Bracket’, ;OF, ‘Division’, ‘Multiplication’, ‘Addition’ and ‘Subtraction’. So, according to the rule of ‘BODMAS’, working must be done in the order corresponding to the letters in the word ‘BODMAS’.
  13. Brackets and their removal
    Brackets are four kinds i.e., bar bracket , circular brackets ( ), curly brackets { } and square brackets [ ] and these can be removed in this order i.e. firstly bar, then circular, then curly and lastly square brackets keeping in considerations of the sign given before them.

Fractions Exercise 3A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Classify, each fraction given below, as decimal or vulgar fraction, proper or improper fraction and mixed fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 1

Solution:
(i) Vulgar and Proper
(ii)Decimal and Improper
(iii) Decimal and Proper
(iv) Vulgar and Improper
(v) Mixed
(vi) Decimal
(vii) Mixed and Decimal
(viii) Vulgar and Proper Ans.

Question 2.
Express the following improper fractions as mixed fractions :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 2

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 3

Question 3.
Express the following mixed fractions as improper fractions :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 4

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 5

Question 4.
Reduce the given fractions to lowest terms
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 6
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 7

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 8

Question 5.
State : true or false
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 9

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 10
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 11

Question 6.
Distinguish each of the following fractions, given below, as a simple fraction or a complex fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 12

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 13

Fractions Exercise 3B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
For each pair, given below, state whether it forms like fractions or unlike fractions :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 18

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 19

Question 2.
Convert given fractions into fractions with equal denominators :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 20

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 21
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 22

Question 3.
Convert given fractions into fractions with equal numerators :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 23

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 24
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 25

Question 4.
Put the given fractions in ascending order by making denominators equal :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 26

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 27
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 28
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 29
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 30

Question 5.
Arrange the given fractions in descending order by making numerators equal :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 31

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 32
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 34

Question 6.
Find the greater fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 35
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 36

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 37
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 38
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 39

Question 7.
Insert one fraction between :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 40

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 41

Question 8.
Insert three fractions between
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 42

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 43
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 44
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 45
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 46
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 47

Question 9.
Insert two fractions between
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 48

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 49

Fractions Exercise 3C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Reduce to a single fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 144

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 51
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 52
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 53
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 54
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 55

Question 2.
Simplify :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 56
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 57

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 58
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 59

Question 3.
Subtract :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 60

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 61

Question 4.
Find the value of
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 62
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 63

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 65

Question 5.
Simplify and reduce to a simple fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 66
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 67
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 68
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 69
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 70
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 71

Question 6.
A bought 3 \(\frac { 3 }{ 4 }\) kg of wheat and 2 \(\frac { 1 }{ 2 }\) kg of rice. Find the total weight of wheat and rice bought.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 72

Question 7.
Which is greater,\(\frac { 3 }{ 5 }\) or \(\frac { 7 }{ 10 }\) and by how much?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 73
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 74

Question 8.
What number should be added to 8 \(\frac { 2 }{ 3 }\) to 12 \(\frac { 5 }{ 6 }\)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 75

Question 9.
What should be subtracted from 8\(\frac { 3 }{ 4 }\) to get 2 \(\frac { 2 }{ 3 }\)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 76

Question 10.
A field is 16 \(\frac { 1 }{ 2 }\) m long and 12 \(\frac { 2 }{ 5 }\) m wide. Find the perimeter of the field.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 77

Question 11.
Sugar costs ₹37 \(\frac { 1 }{ 2 }\)per kg. Find the cost of 8\(\frac { 3 }{ 4 }\) kg sugar.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 78

Question 12.
A motor cycle runs 31\(\frac { 1 }{ 4 }\) km consuming 1 litre of petrol. How much distance will it run consuming 1\(\frac { 3 }{ 5 }\) liter of petrol?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 79

Question 13.
A rectangular park has length = 23 \(\frac { 2 }{ 3 }\) m and breadth = 16 \(\frac { 2 }{ 3 }\) m. Find the area of the park.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 80

Question 14.
Each of 40 identical boxes weighs 4 \(\frac { 4 }{ 5 }\) kg Find the total weight of all the boxes.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 82

Question 15.
Out of 24 kg of wheat, \(\frac { 5 }{ 6 }\) th of wheat is consumed. Find, how much wheat is still left?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 83

Question 16.
A rod of length 2 \(\frac { 2 }{ 5 }\) metre is divided into five equal parts. Find the length of each part so obtained.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 84

Question 17.
IfA = 3\(\frac { 3 }{ 8 }\) and B = 6\(\frac { 5 }{ 8 }\) find :
(i) A+B
(ii) B A
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 85
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 86

Question 18.
Cost of 3 \(\frac { 5 }{ 7 }\) litres of oil is ₹83 \(\frac { 1 }{ 2 }\). Find the
cost of one litre oil.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 87

Question 19.
The product of two numbers is 20 \(\frac { 5 }{ 7 }\). If one of these numbers is 6 \(\frac { 2 }{ 3 }\), find the other.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 88

Question 20.
By what number should 5 \(\frac { 5 }{ 6 }\) be multiplied 1 to get 3\(\frac { 1 }{ 3 }\) ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 89

Fractions Exercise 3D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Simplify
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 90

Question 2.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 91
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 92

Question 3.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 145
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 94

Question 4.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 95
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 96

Question 5.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 97
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 98

Question 6.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 99
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 100

Question 7.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 101
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 102

Question 8.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 103
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 104
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 106

Question 9.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 107
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 108

Question 10.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 109
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 110

Question 11.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 111
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 112

EXERCISE 3 (E)

Question 1.
A line AB is of length 6 cm. Another line CD is of length 15 cm. What fraction is :
(i) The length of AB to that of CD ?
(ii) \(\frac { 1 }{ 2 }\) the length of AB to that of \(\frac { 1 }{ 3 }\) of CD ?
(iii) \(\frac { 1 }{ 5 }\) of CD to that of AB ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 113

Question 2.
Subtract \(\frac { 2 }{ 7 }\) – \(\frac { 5 }{ 21 }\) from the sum of \(\frac { 3 }{ 4 }\) , \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 12 }\)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 114

Question 3.
From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, 5\(\frac { 1 }{ 4 }\) kg, 9\(\frac { 1 }{ 2 }\) kg and 9\(\frac { 3 }{ 4 }\) kg respectively.
(i) How many kg did he sell ?
(ii) How many kg are still left in the sack ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 115

Question 4.
If a boy works for six consecutive days for 8 hours, 7\(\frac { 1 }{ 2 }\) hours, 8\(\frac { 1 }{ 4 }\) hours, 6 \(\frac { 1 }{ 4 }\)4 3hours, 6\(\frac { 3 }{ 4 }\) hours and 7 hours respectively. How much money will he earn at the rate of Rs. 36 per hour ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 116
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 117

Question 5.
A student bought 4 \(\frac { 1 }{ 3 }\) m of yellow ribbon, 6 \(\frac { 1 }{ 6 }\) m of red ribbon and 3\(\frac { 2 }{ 9 }\) m of blue ribbon for decorating a room. How many metres of ribbon did he buy ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 118

Question 6.
In a business, Ram and Deepak invest \(\frac { 3 }{ 5 }\) and \(\frac { 2 }{ 5 }\) of the total investment. IfRs. 40,000 is the total investment, calculate the amount invested by each ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 146
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 120

Question 7.
Geeta had 30 problems for home work. She worked out \(\frac { 2 }{ 5 }\) of them. How many problems were still left to be worked out by her ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 121

Question 8.
A picture was marked at Rs. 90. It was sold at \(\frac { 3 }{ 4 }\) of its marked price. What was the sale price ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 122

Question 9.
Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, if each weighed 10\(\frac { 1 }{ 2 }\) kg ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 123

Question 10.
A rope is 25\(\frac { 1 }{ 2 }\) m long. How many pieces , 1 \(\frac { 1 }{ 2 }\) each of length can be cut out from it?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 124

Question 11.
The heights of two vertical poles, above the earth’s surface, are 14 \(\frac { 1 }{ 4 }\) m and 22 \(\frac { 1 }{ 3 }\) respectively. How much higher is the second pole as compared with the height of the first pole ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 125

Question 12.
Vijay weighed 65\(\frac { 1 }{ 2 }\) kg. He gained 1\(\frac { 2 }{ 5 }\) kg during the first week, 1 \(\frac { 1 }{ 4 }\) kg during the second week, but lost \(\frac { 5 }{ 16 }\) kg during the 16 third week. What was his weight after the third week ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 126
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 127

Question 13.
A man spends \(\frac { 2 }{ 5 }\) of his salary on food and \(\frac { 3 }{ 10 }\) on house rent, electricity, etc. What fraction of his salary is still left with him ?
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 128

Question 14.
A man spends \(\frac { 2 }{ 5 }\) of his salary on food and \(\frac { 3 }{ 10 }\) of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 129

Question 15.
Shyam bought a refrigerator for Rs. 5000. He paid \(\frac { 1 }{ 10 }\) of the price in cash and the rest in 12 equal monthly instalments. How much had he to pay each month ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 130
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 131

Question 16.
A lamp post has half of its length in mud, and \(\frac { 1 }{ 3 }\) of its length in water.
(i) What fraction of its length is above the water ?
(ii) If 3\(\frac { 1 }{ 3 }\) m of the lamp post is above the water, find the whole length of the lamp post.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 132

Question 17.
I spent \(\frac { 3 }{ 5 }\) of my savings and still have Rs. 2,000 left. What were my savings ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 133
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 134

Question 18.
In a school, \(\frac { 4 }{ 5 }\) of the children are boys. If the number of girls is 200, find the number of boys.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 135

Question 19.
If \(\frac { 4 }{ 5 }\) of an estate is worth Rs. 42,000, find the worth of whole estate. Also, find the value of \(\frac { 3 }{ 7 }\) of it.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 136
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 137

Question 20.
After going \(\frac { 3 }{ 4 }\) of my journey, I find that I have covered 16 km. How much Journey is still left ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 138

Question 21.
When Krishna travelled 25 km, he found that \(\frac { 3 }{ 5 }\) of his journey was still left. What was the length of the whole journey.

Solution:Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 139

Question 22.
From a piece of land, one-third is bought by Rajesh and one-third of remaining is bought by Manoj. If 600 m² land is still left unsold, find the total area of the piece of land.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 147
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 140

Question 23.
A boy spent \(\frac { 3 }{ 5 }\) of his money on buying 1 cloth and \(\frac { 1 }{ 4 }\) of the remaining on buying shoes. If initially he has ?2,400; how much did he spend on shoes?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 141

Question 24.
A boy spent \(\frac { 3 }{ 5 }\) of his money on buying cloth and \(\frac { 1 }{ 4 }\) of his money on buying shoes. If initially he has ?2,400; how much did he spend on shoes?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 142
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 143

 

Selina Concise Mathematics class 7 ICSE Solutions – Simple Interest

Selina Concise Mathematics class 7 ICSE Solutions – Simple Interest

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 1

Question 1.
Find the S.I. and amount on :
(i) Rs. 150 for 4 years at 5% per year.
(ii) Rs. 350 for 3\(\frac {1 }{ 2 }\) years at 8% p.a.
(iii) Rs. 620 for 4 months at 8 p. per rupee per month.
(iv) Rs. 3,380 for 30 months at 4 \(\frac { 1 }{ 2 }\) % p.a.
(v) 600 from July 12 to Dec. 5 at 10% p.a.
(vi) Rs. 850 from 10th March to 3rd August at 2 \(\frac { 1 }{ 2 }\) % p.a.
(vii) Rs. 225 for 3 years 9 months at 16% p.a.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 3
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 4
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 5
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 6

Question 2.
On what sum of money does the S.I. for 10 years at 5% become Rs. 1,600 ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 7

Question 3.
Find the time in which Rs. 2,000 will amount to Rs. 2,330 at 11% p.a. ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 8

Question 4.
In what time will a sum of money double it self at 8% p.a ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 9

Question 5.
In how many years will be ₹870 amount to ₹1,044, the rate of interest being 2\(\frac { 1 }{ 2 }\) % p.a ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 10

Question 6.
Find the rate percent if the S.I. on ₹275 is 2 years is ₹22.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 11

Question 7.
Find the sum which will amount to ₹700 in 5 years at 8% rate p.a.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 12
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 13

Question 8.
What is the rate of interest, if ₹3,750 amounts to ₹4,650 in 4 years ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 14

Question 9.
In 4 years, ₹6,000 amount to ₹8,000. In what time will ₹525 amount to ₹700 at the same rate ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 15

Question 10.
The interest on a sum of money at the end of 2\(\frac { 1 }{ 2 }\) years is \(\frac { 4 }{ 5 }\) of the sum. What is the rate percent ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 16

Question 11.
What sum of money lent out at 5% for 3 years will produce the same interest as Rs. 900 lent out at 4% for 5 years ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 17

Question 12.
A sum of Rs. 1,780 become Rs. 2,136 in 4 years,
Find :
(i) the rate of interest.
(ii) the sum that will become Rs. 810 in 7 years at the same rate of interest ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 18

Question 13.
A sum amounts to Rs. 2,652 in 6 years at 5% p.a. simple interest.
Find :
(i) the sum
(ii) the time in which the same sum will double itself at the same rate of interest.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 19

Question 14.
P and Q invest Rs. 36,000 and Rs. 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets Rs. 3,080 more interest than Q; find the rate of interest.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 20

Question 15.
A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one-fourth of the money lent, find the value of R.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 21

Question 16.
The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 22

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Pythagoras Theorem Exercise 16 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Triangle ABC is right-angled at vertex A. Calculate the length of BC, if AB = 18 cm and AC = 24 cm.

Solution:
Given : ∆ABC right angled at A and AB = 18 cm, AC = 24 cm.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 1

To find : Length of BC.
According to Pythagoras Theorem,
BC2 = AB2 + AC2
= 182 + 242 = 324 + 576 = 900
∴BC = \(\sqrt { 900 }\) = \(\sqrt { 30 x 30 }\)= 30 cm

Question 2.
Triangle XYZ is right-angled at vertex Z. Calculate the length of YZ, if XY = 13 cm and XZ = 12 cm.

Solution:
Given : ∆XYZ right angled at Z and XY = 13 cm, XZ = 12 cm.
To find : Length of YZ.
According to Pythagoras Theorem,
XY2 = XZ2 + YZ2
132 = 122 + YZ2
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 2

169= 144 +YZ2
169- 144 = YZ2
25 = YZ2
∴YZ = \(\sqrt { 25 }\)cm \(\sqrt { 5×5 }\) = 5 cm

Question 3.
Triangle PQR is right-angled at vertex R. Calculate the length of PR, if:
PQ = 34 cm and QR = 33.6 cm.

Solution:
Given : ∆PQR right angled at R and PQ = 34 cm, QR = 33.6 cm.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 3

To find : Length of PR.
According to Pythagoras Theorem,
PR2 + QR2 = PQ2
PR2 + 33.62 = 342
PR2+ 1128.96= 1156
PR2 = 1156- 1128.96
∴ PR = \(\sqrt { 27.04 }\) = 5.2 cm

Question 4.
The sides of a certain triangle are given below. Find, which of them is right-triangle
(i) 16 cm, 20 cm and 12 cm
(ii) 6 m, 9 m and 13 m

Solution:
(i) 16 cm, 20 cm and 12 cm
The given triangle will be a right-angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
i.e., If (20)2 = (16)2 = (12)2
(20)2 = (16)2 + (12)2
400 = 256 + 144
400 = 400
So, the given triangle is right angled.
(ii) 6 m, 9 m and 13 m
The given triangle will be a right-angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
i.e., If (13)2 = (9)2 + (6)2
169 = 81+36 169 ≠ 117
So, the given triangle is not right angled.

Question 5.
In the given figure, angle BAC = 90°, AC = 400 m and AB = 300 m. Find the length of BC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 4
Solution:
AC = 400 m
AB = 300 m
BC = ?
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 5
According to Pythagoras Theorem,
BC2 = AB2 + AC2
BC2 = (300)2 + (400)2
BC2 = 90000 + 160000
BC2 = 250000
BC = \(\sqrt { 250000 }\)= 500 m

Question 6.
In the given figure, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA= PB = 15 m. Find:

(i) CP
(ii) PD
(iii) CD
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 6

Solution:
Given : AC = 12 m
BD = 9 m
PA = PB= 15 m
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 7
(i) In right angle triangle ACP
(AP)2 = (AC)2 + (CP)2
152 = 122 + CP2
225 = 144 + CP2
225 – 144 = CP2
81 =CP
\(\sqrt { 81 }\) =CP
∴ CP = 9 m
(ii) In right angle triangle BPD
(PB)2 = (BD)2 + (PD)2
(15)2 = (9)2 + PD2
225 = 81 + PD2
225-81 = PD2
144 = PD2
\(\sqrt { 144 }\)=PD                                               ‘
∴ PD = 12 m
(iii) CP = 9 m
PD = 12 m
∴ CD = CP + PD
= 9+ 12 = 21 m

Question 7.
In triangle PQR, angle Q = 90°, find :
(i) PR, if PQ = 8 cm and QR = 6 cm
(ii) PQ, if PR = 34 cm and QR = 30 cm

Solution:
(i) Given:
PQ = 8 cm
QR = 6 cm
PR = ?
∠PQR = 90°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 8
According to Pythagoras Theorem,
(PR)2 = (PQ)2 + (QR)2
PR2 = 82 + 62
PR2 = 64 + 36
PR2 = 100
∴ PR = \(\sqrt { 100 }\)= 10 cm
(ii) Given :
PR = 34 cm
QR = 30 cm
PQ = ?
∠PQR = 90°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 9
According to Pythagoras Theorem,
(PR)2 = (PQ)2 + (QR)2
(34)2 = PQ2 + (30)2
1156 = PQ2 + 900
1156-900 = PQ2
256 = PQ2
∴ PQ = 16 cm

Question 8.
Show that the triangle ABC is a right-angled triangle; if:
AB = 9 cm, BC = 40 cm and AC = 41 cm

Solution:
AB = 9 cm
CB = 40 cm
AC = 41 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 10
The given triangle will be a right angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
According to Pythagoras Theorem,
(AC)2 = (BC)2 + (AB)2
(41)2 = (40)2 + (9)2
1681 = 1600 + 81
1681 = 1681
Hence, it is a right-angled triangle ABC.

Question 9.
In the given figure, angle ACB = 90° = angle ACD. If AB = 10 m, BC = 6 cm and AD = 17 cm, find :
(i) AC
(ii) CD
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 11

Solution:
Given:
∆ABD
∠ACB = ∠ACD = 90°
and AB = 10 cm, BC = 6 cm and AD = 17 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 12
To find:
(i) Length of AC
(ii) Length of CD
Proof:
(i) In right-angled triangle ABC
BC = 6 cm, AB = 110 cm
According to Pythagoras Theorem,
AB2 = AC2 + BC2
(10)2 = (AC)2 + (6)2
100 = (AC)2 + 36
AC2 = 100-36 = 64 cm
AC2 = 64 cm
∴ AC = \(\sqrt { 8×8 }\) = 8 cm
(ii) In right-angle triangle ACD
AD = 17 cm, AC = 8 cm
According to Pythagoras Theorem,
(AD)2 = (AC)2 + (CD)2
(17)2 = (8)2 + (CD)2
289 – 64 = CD2
225 = CD2
CD =\(\sqrt { 15×15 }\) = 15 cm

Question 10.
In the given figure, angle ADB = 90°, AC = AB = 26 cm and BD = DC. If the length of AD = 24 cm; find the length of BC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 13

Solution:
Given:
∆ABC
∠ADB = 90° and AC = AB = 26 cm
AD = 24 cm
To find : Length of BC In right angled ∆ADC
AB = 26 cm, AD = 24 cm
According to Pythagoras Theorem,
(AC)2 = (AD)2 + (DC)2
(26)2 = (24)2 + (DC)2
676 = 576 + (DC)2
⇒ (DC)2 = 100
⇒ DC = \(\sqrt { 100 }\) = 10 cm
∴ Length of BC = BD + DC
= 10 + 10 = 20 cm

Question 11.
In the given figure, AD = 13 cm, BC = 12 cm, AB = 3 cm and angle ACD = angle ABC = 90°. Find the length of DC.

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 14

Solution:
Given :
∆ACD = ∠ABC = 90°
and AD = 13 cm, BC = 12 cm, AB = 3 cm
To find : Length of DC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 15

(i)In right angled ∆ABC
AB = 3 cm, BC = 12 cm
According to Pythagoras Theorem,
(AC)2 = (AB)2 + (BC)2
(AC)2 = (3)2 + (12)2
(AC) = \(\sqrt { 9+144 }\) = \(\sqrt { 153 }\) cm

 (ii) In right angled triangle ACD
AD = 13 cm, AC =\(\sqrt { 153 }\)
According to Pythagoras Theorem,
DC2 = AB2-AC2
DC2= 169-153
DC = \(\sqrt { 16 }\) = 4 cm
∴ Length of DC is 4 cm

Question 12.
A ladder, 6.5 m long, rests against a vertical wall. Ifthe foot of the ladcler is 2.5 m from the foot of the wall, find upto how much height does the ladder reach?

Solution:
Given :
Length of ladder = 6.5 m
Length of foot of the wall = 2.5 m
To find : Height AC According to Pythagoras Theorem,
(BC)2 = (AB)2 + (AC)2
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 16

(6.5)2 = (2.5)2 + (AC)2
42.25 = 6.25 + AC2
AC2 = 42.25 – 6.25 = 36 m
AC = \(\sqrt { 6×6 }\) = 6 m
∴ Height of wall = 6 m

Question 13.
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.

Solution:
Given : Direction of north = 5 m i.e. AC Direction of east = 12 m i.e. AB
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 17

To find: BC
According to Pythagoras Theorem,
In right angled AABC
(BC)2 = (AC)2 + (AB)2
(BC)2 = (5)2 + (12)2
(BC)2 = 25 + 144
(BC)2 = 25 + 144
(BC)2= 169
∴ BC = \(\sqrt { 169 }\) = \(\sqrt { 13×13 }\) = 13 m

Question 14.
Use the information given in the figure to find the length AD.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 18

Solution:
Given :
AB = 20 cm
∴AO =\(\frac { AB }{ 2 }\) = \(\frac { 20 }{ 2 }\) =10cm
BC = OD = 24 cm
To find : Length of AD
In right angled triangle
AOD (AD)2 = (AO)2 + (OD)2
(AD)2 = (10)2 + (24)2
(AD)2 = 100 + 576
(AD)2 = 676
∴ AD =  \(\sqrt { 26×26 }\)
AD = 26 cm

Selina Concise Mathematics class 7 ICSE Solutions – Profit, Loss and Discount

Selina Concise Mathematics class 7 ICSE Solutions – Profit, Loss and Discount

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER

  1. The Cost Price (C.P.) of an article is the price at which the article is bought.
  2. The Selling Price (S.P.) of an article is the price at which the article is sold.
  3. If Selling Price of an article is more than its cost price ; it is sold at a profit (gain)
    Profit = Selling Price-Cost Price
    i.e., Profit (gain) = S.P. – C.P. and S.P. = C.P. + Gain
  4. If Selling Price of an article is less than its cost price ; it is sold at a loss.
    Loss = Cost Price – Selling Price
    i.e., Loss = C.P. – S.P. and S.P. = C.P. — Loss
  5. Profit percent and loss percent are always calculated on cost price (C.P.) only.
    i.e., (i) Profit % = \(\frac { Profit }{ C.P. }\) x 100% and
    (ii)
    Loss % = \(\frac {Loss }{ C.P. }\) x 100%
  6. Selling Price = Marked price – Discount
    i.e., S.P. = M.P.—(piscount
    Note : (i) Discount is calculated on marked price (M.P.)
    (ii) Marked price is also written as List price.

EXERCISE 9 (A)

Question 1.
Find the gain or loss percent, if
(i) C.P. = Rs. 200 and S.P.: = Rs. 224
(ii) C.P. = Rs. 450 and S.P. = Rs. 400
(iii) C.P. = Rs. 550 and gain = Rs . 22
(iv) CP. = Rs. 216 and loss = Rs. 72
(v) S.P. = Rs. 500 and loss : = Rs. 100
(vi) S.P. = Rs. 12 and profit = Rs. 4
(vii) C.P. = Rs. 5 and gain = 60 P

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 1
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 3

Question 2.
Find the selling price, if:
(i) C.P. = Rs. 500 and gain = 25%
(ii) C.P. = Rs. 60 and loss = 12 1/2%
(iii) C.P. = Rs. 150 and loss = 20%
(iv) C.P. = Rs. 80 and gain = 2.5%

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 4
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 5

Question 3.
Rohit bought a tape-recorder for Rs. 1,500 and sold it for Rs. 1,800. Calculate his profit or loss percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 6

Question 4.
An article bought for Rs. 350 is sold at a profit of 20%. Find its selling price.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 7

Question 5.
An old machine is bought for Rs. 1,400 and is sold at a loss of 15%. Find its selling price.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 8

Question 6.
Oranges are bought at 5 for Rs. 10 and sold at 6 for Rs. 15. Find profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 53

Question 7.
A certain number of articles are bought at 3 for Rs. 150 and all of them are sold at 4 for Rs. 180. Find the loss or gain as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 10

Question 8.
A vendor bought 120 sweets at 20 p each. In his house, 18 were consumed and he sold the remaining at 30 p each. Find his profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 11

Question 9.
The cost price of an article is Rs. 1,200 and selling price is \(\frac { 5 }{ 4 }\) times of its cost price. Find:
(i) selling price of the article
(ii) profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 12

Question 10.
The selling price of an article is Rs. 1,200 and cost price is \(\frac { 5 }{ 4 }\) times of its selling price,
find :
(i) cost price of the article ;
(ii) profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 13

EXERCISE 9 (B)

Question 1.
Find the cost price, if:
(i) S.P. = Rs. 21 and gain = 5%
(ii) S.P. = Rs. 22 and loss = 12%
(iii) S.P. = Rs. 340 and gain = Rs. 20
(iv) S.P. = Rs. 200 and loss = Rs. 50
(v) S.P. = Re. 1 and loss = 5 p.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 14

Question 2.
By selling an article for Rs. 810, a loss of 10 percent is suffered. Find its cost price.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 15

Question 3.
By selling a scooter for Rs. 9,200, a man gains 15%. Find the cost price of the scooter.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 16

Question 4.
On selling an article for Rs. 2,640, a profit of 10 percent is made. Find
(i) cost price of the article
(ii) new selling price of it, in order to gain 15%

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 17

Question 5.
A T.V. set is sold for Rs. 6800 at a loss of 15%. Find
(i)cost price of the T.V. set.
(ii)new selling price of it, in order to gain 12%

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 18

Question 6.
A fruit seller bought mangoes at Rs. 90 per dozen and sold them at a loss of 8 percent. How much will a customer pay for.
(i) one mango
(ii) 40 mangoes

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 19

Question 7.
By selling two transistors for Rs. 00 each, a shopkeeper gains 20 percent on one transistor and loses 20 percent on the other.
Find :
(i) C.P. of each transistor

(ii) total C.P. and total S.P. of both the transistors
(iii) profit or loss percent on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 20
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 21

Question 8.
Mangoes are bought at 20 for Rs. 60. If
they are sold at 33\(\frac { 1 }{ 3 }\) percent profit.
Find:

(i) selling price of each mango.
(ii) S.P. of 8 mangoes.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 22

Question 9.
Find the cost price of an article, which is sold for Rs. 4050 at a loss of 10%. Also, find the new selling price of the article which must give a profit of 8%.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 23
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 24

Question 10.
By selling an article for ₹825, a man loses \(\frac { 1 }{ 3 }\) equal to j of its selling price.
Find :

(i) the cost price of the article,
(ii) the profit percent or the loss percent made, if the same article is sold for ₹1265.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 25

Question 11.
Find the loss or gain as percent, if the C.P. of 10 articles, all of the same kind, is equal to S.P. of 8 articles.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 26
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 27

Question 12.
Find the loss or gain as percent, if the C.P. of 8 articles, all of the same kind, is equal to S.P. of 10 articles.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 28

Question 13.
The cost price of an article is 96% of its selling price. Find the loss or the gain as percent on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 29
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 30

Question 14.
The selling price of an article is 96% of its cost price. Find the loss or the gain as percent on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 31

Question 15.
Hundred oranges are bought for ₹350 and all of them are sold at the rate of ₹48 per dozen. Find the profit percent or loss percent made.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 32

Question 16.
Oranges are bought at 100 for ?80 and all of them are sold at ₹80 for ₹100. Find the loss or gain as percent in this transaction.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 33

Question 17.
An article is bought for ₹5,700 and ₹1,300 is spent on its repairing, transportion, etc. For how much should this article be sold in order to gain 20% on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 34

EXERCISE 9 (C)

Question 1.
A machine is marked at ₹5000 and is sold at a discount of 10%. Find the selling price of the machine.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 35

Question 2.
shopkeeper marked a dinner set for ₹1000. He sold it at ₹900, what percent discount did he give ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 36

Question 3.
A pair of shoes marked at ₹320, are sold at a discount of 15 percent.
Find :
(i) discount
(ii) selling price of the shoes.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 37

Question 4.
The list price of an article is ₹450 and it is sold for ₹360.
Find :
(i) discount
(ii) discount percent

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 38

Question 5.
A shopkeeper buys an article for₹300. He increases its price by 20% and then gives 10% discount on the new price. Find:
(i) the new price (marked price) of the article.
(ii) the discount given by the shopkeeper.
(iii) the selling price.
(iv) profit percent made by the shopkeeper.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 39
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 40

Question 6.
A car is marked at Rs. 50,000. The dealer gives 5% discount on first Rs. 20,000 and 2% discount on the remaining Rs. 30,000.
Find :
(i) the total discount.
(ii) the price charged by the dealer.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 41

Question 7.
A dealer buys a T.V. set for Rs. 2500. He marks it at Rs. 3,200 and then gives a discount of 10% on it.
Find :
(i) the selling price of the T.V. set
(ii) the profit percent made by the dealer.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 42

Question 8.
A sells his goods at 15% discount. Find the price of an article which is sold for Rs. 680.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 43
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 44

Question 9.
A shopkeeper allows 20% discount on the marked price of his articles. Find the marked price of an article for which he charges Rs. 560.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 45

Question 10.
An article is bought for Rs. 1,200 and Rs. 100 is spent on its transportation, etc.
Find :
(i) the total C.P. of the article.
(ii) the selling price of it in order to gain 20% on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 46

Question 11.
40 pens are bought at 4 for Rs. 50 and all of them are sold at 5 for Rs. 80
Find :
(i) C.P. of one pen.
(ii) S/P. of one pen.
(iii) Profit made by selling one pen.
(iv) Profit percent made by selling one pen.
(v) C.P. of 40 pens
(vi) S.P. of 40 pens.
(vii) Profit made by selling 40 pens.
(viii) Profit percent made by selling 40 pens. Are the results of parts (iv) and (viii) same? What conclusion do you draw from the above result ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 47
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 48

Question 12.
The C.P. of 5 identical articles is equal to S.P. of 4 articles. Calculate the profit percent or loss percent made if all the articles bought are sold.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 49

Question 13.
The C.P. of 8 pens is same as S.P. of 10 pens. Calculate the profit or loss percent made, if all the pens bought are considered to be sold

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 50

Question 14.
A certain number of articles are bought at Rs. 450 per dozen and all of them are sold at a profit of 20%. Find the S.P. of:
(i) one article
(ii)seven articles.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 51

Question 15.
An article is marked 60% above the cost price and sold at 20% discount. Find the profit percent made.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 52

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids (Representing 3-D in 2-D)

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids (Representing 3-D in 2-D)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Recognition of solids Exercise 18 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Identify the nets which can be used to form cubes
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 1
Solution:
Nets for a cube are (ii) , (iii) and (v).

Question 2.
Draw at least three different nets for making cube.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 2

Question 3.
The figure, given below, shows shadows of some 3D objects, when seen under the lamp of an overhead projector :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 3
In each case, name the object.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 4

Question 4.
Using Euler’s formula, find the values of a, b, c and d.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 5
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 6
(i) a + 6 – 12 = 2 ⇒ a = 2 – 6 + 12 = 14 – 6 = 8
(ii) b + 5- 9 = 2 ⇒6 = 2 + 9- 5 = 6
(iii) 20+ 12 — c = 2 ⇒32 – c = 2 ⇒ c = 32-2 ⇒ c = 30
(iv) 6 + d-12=2 ⇒ d – 6 = 2 ⇒ d = 2 + 6 = 8

Question 5.
Dice are cubes with dot or dots on each face. Opposite faces of a die always have a total of seven on them.
Below are given two nets to make dice (cube), the numbers inserted in each square indicate the number of dots in it.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 7
Insert suitable numbers in each blank so that numbers in opposite faces of the die have a total of seven dots.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 8

Question 6.
The following figures represent nets of some solids. Name the solids
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 9
Solution:
The given nets are of the solid as given below :
(i) Cube
(ii) Cuboid

Question 7.
Draw a map of your class room using proper scale and symbols for different objects.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 10

Question 8.
Draw a map of your school compound using proper scale and symbols for various features like play ground, main building, garden, etc.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 11

Question 9.
In the map of India, the distance between two cities is 13.8 cm.
Taking scale : 1 cm = 12 km, find the actual distance between these two cities.
Solution:
The scale for a map is given to be 1 cm = 12 km
The distance between these two cities = 13.8 cm on the map
∴ Actual distance between these two cities
= 12 x 13.8 km = 165.6 km

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry (Including Reflection and Rotation)

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 17 Symmetry (Including Reflection and Rotation)

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER
1. Symmetry : A geometrical figure is said to be symmetric about a line if on folding about that line, the two parts of the figure exactly concide each other. The given figure is symmetric about the line PQ. The line is said to be a line of symmetry or an axis of symmetry
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 1

2. Lines of symmetry of given geometrical figures :
It is not necessary that every figure under consideration will definitely have a line symmetry. If we consider different types of triangle ; we find :
1. A scalene triangle has no line of Symmetry : i.e. we can not have a line in a scalene triangle about which if the figure (triangle) is folded, the two parts of the figure will coincide.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 2
2. An isosceles triangle has only one line of symmetry. The bisector of angle of vertex which is also the perpendicular bisector of its base.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 3
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 4

The bisectors of the angle of vertices which are also the perpendicular bisectors of its sides.
4. Line/lincs of symmetry of differed types of quadrilaterals are shown below by dotted lines :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 5
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 6

5. In each of the following, the dotted line/lines are the line/lines of symmetry of the given figure:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 7

6. As shown below:
(i) a circle has infinite lines of symmetry ; every line through its centre is line of symmetry’.
(ii) a semi-circle has one line of symmetry.
(iii) a quadrant (one-fourth) of a circle has one line of symmetry’ and so on.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 8
Note : It is clear from the question numbers 4 and 5, given above that:
1. The largest number of lines of symmetry’ of a triangle is three (3).
2. The largest number of lines of symmetry of a quadrilateral is four (4).
i. e. as the number of sides in a triangle is 3 ; the largest number of lines of symmetry in it is 3 and as the number of sides in a quadrilateral is 4 ; the largest number of lines of symmetry is 4.
In the same way :
1. The largest number of lines of symmetry of a pentagon is 5, as a pentagon has 5 sides.
2. The hexagon has 6 sides and so the largest number of lines of symmetry’ of a hexagon is 6.
In general, we can say, that if a polygon has n sides ; M
the largest number of lines of symmetry, it can have, is n.
3. Reflection (Image): Tire given figure shows a candle place ‘d’ distance before a plane mirror MM’, the image of the candle is obtained in the mirror at the same distance ‘d’ behind the mirror. Geometrically, the line joining the candle (c) and its reflection c’ is (Candle) perpendicular bisector of the mirror line MM’.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 9
4. Reflection in x-axis : Reflection if x-axis means the x-axis is considered as the plane mirror, the given point as the object and then to find its image.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 10
Let P (x, y) be a point and. as shown in the figure, when it is reflected in x-axis to point P’; the co-ordinates of image point P’ are (x, – y).
i. e. reflection of P (x, y) in x-axis = P’ (x, – y)
In other words :
Image of P (x, y) in x-axis = P’ (x, – y)
We can say, when a point (x, y) is reflected in x-axis, the sign of its second component (ordinate) changes i.e. tire sign of y changes and so the image of (x, y) in x-axis is (x, – y).
5. Reflection in y-axis : As is clear from the figure, given alongside, the reflection P (x, y) in y-axis is point P’ (- x, y).
We can say, when a point (x,y) is reflected in y-axis, the sign of its first component (abscissa) change i.e. the sign of x changes and so the image of (x, y) in y- axis is (-x,y).
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 11
6. Reflection in Origin : When a point P (x, y) is reflected in origin, the sign of both of its components change i.e. the image of P (x, y) is P’ (- x, -y) as shown along side in the figure.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 12

Symmetry Exercise 17A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
For each figure, given below, draw the line (s) of symmetry, if possible :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 13
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 14

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 15

Question 2.
Write capital letters A to Z of English alphabet ; and in each case, if possible, draw the largest number of lines of symmetry.
Solution:
Line or lines of symmetry’ is possible in the following alphabets. For others alphabets it is not possible
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 16
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 17

Question 3.
By drawing a free hand sketch of each of the following, draw in each case, the line (s) of symmetry,
if any:
(i) a scalene triangle
(ii) an isosceles right angled triangle
(iii) a rhombus
(iv) a kite shaped figure triangle.
(v) a rectangle
(vi) a square
(vii) an isosceles

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 18.

Question 4.
Draw a triangle with :
(i) no line of symmetry,
(ii) only one line of symmetry,
(iii) exactly two lines of symmetry,
(iv) exactly three lines of symmetry,
(v) more than three lines of symmetry.
In each case, if possible, represent the line/ lines of symmetry by dotted lines. Also, write the special name of the triangle drawn.

Solution:
(i) Scalene triangle : It has no line of symmetry.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 19
(ii) Isosceles Triangle : It has one line of symmetry as shown.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 20
(iii) It is not possible.
(iv) Equilateral Triangle : It has three lines of symmetry as shown
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 21
(v) It is not possible.

Question 5.
Draw a quadrilateral with :
(i) no line of symmetry.
(ii) only one line of symmetry.
(iii) exactly two lines of symmetry.
(iv) exactly three lines of symmetry.
(v) exactly four lines of symmetry.
(vi) more than four lines of symmetry.
In each case, if possible, represent the line/ lines of symmetry by dotted lines. Also, write the special name of the quadrilateral drawn.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 22

Question 6.
Construct an equilateral triangle with each side 6 cm. In the triangle drawn, draw all the possible lines of symmetry.

Solution:
Steps of Construction :
(i) Draw a line segment BC = 6 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 23
(ii) With centres B and C and radius 6 cm, draw two arcs intersecting each other at A.
(iii) Join AB and AC
∆ ABC is the required equilateral triangle,
(iv) Draw the angle bisectors of ∠A, ∠B and ∠C.
These are the lines of symmetry which are three in numbers as the triangle is equilateral.

Question 7.
Construct a triangle ABC in which AB = AC = 5cin and BC = 5.6 cm. If possible, draw its lines of symmetry.
Solution:
Steps of Construction :
(i) Draw a line segment BC = 5.6 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 24
(ii) With centres B and C and radius 5 cm, draw two arcs intersecting each other at A.
(iii) Join AB and AC.
∆ ABC is an isosceles triangle.
(iv) Draw the bisector of ∠A. This is the only one line of symmetry as the triangle is an isosceles.

Question 8.
Construct a triangle PQR such that PQ = QR = 5 .5 cm and angle PQR = 90°. If possible, draw its lines of symmetry.
Solution:
∴ ∠PQR = 90°, and ∠P = ∠R
(opposite sides are equal)
∴∠P + ∠R = 90°
Hence ZP = ZR = \(\frac { 99\circ }{ 2 }\)  = 45°
Steps of Construction:
(i) Draw a line segment QR = 5.5 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 25
(ii) At Q, draw a ray making an angle of 90° and cut off QP = 5.5 cm.
(iii) Join PR.
∆PQR is an isosceles triangle.
(iv) Draw the angle bisector of ∠PQR. It is the line of symmetry. Since the triangle is an isosceles.
∴It has only one line of symmetry.

Question 9.
If possible, draw a rough sketch of a quadrilateral which has exactly two lines of symmetry.

Solution:
Since the quadrilateral has exactly two lilies of symmetry
∴It must be a rectangle or a rhombus
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 26

Question 10.
A quadrilateral ABCD is symmetric about its diagonal AC. Name tire sides of this quadrilateral which are equal.

Solution:
The quadrilateral ABCD is symmetric about its diagonal AC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 27
∴It must be a kite shaped.
Hence side AB = AD and BC = DC.

Symmetry Exercise 17B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
In each figure, given below, find the image of the point P in the line AB :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 28
Solution:
Steps of Construction : Fig. (i) and (ii)
(i) From P, draw a perpendicular to the given line AB meeting it at O.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 29
(ii) Produce PO to P’ such that OP’ = PO.
P’ is the required image of P in AB.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 30

Question 2.
In each figure, given below, find the image of the line segment AB in the line PQ :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 31
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 32
Steps of Construction :
(i) From A and B, draw perpendiculars on PQ intersecting PQ at L and M.
(ii) Produce AL to A’ such that AL = LA’ and produce BM to B’ such that BM = MB’ A’B’ is the image of the line segment AB in PQ.

Question 3.
Complete the following table :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 33
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 34

Question 4.
A point P (7,3)|is reflected in x-axis to point P’. The point P’ is further reflected in v-axis to point P” Find :
(i) the co-ordinates of P’
(ii) the co-ordinates of P”
(iii) the image of P (7, 3) in origin.
Solution:
(i) Image of point P (7,3) when reflected in x-axis is P’ whose co-ordinates will be (7,-3)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 35
(ii) Image of point P’ (7,-3) when reflected in y-axis, is P” whose co-ordinates will be (- 7,-3)
(iii) The image of P (7, 3) in origin is P” whose co-ordinates are (- 7, – 3).

Question 5.
A point A (- 5, 4) is reflected in y-axis to point B. The point B is further reflected in origin to point C. find :
(i) the co-ordinates of B
(ii) the co-ordinates of C
(iii) the image of A (- 5, 4) in x-axis.
Solution:
(i) Image of point A (- 5,4) when reflected in y-axis is B whose co-ordinates will be (5,4)
(ii) Image of B (5, 4) when reflected in origin is C whose co-ordinates will be (- 5, – 4)
(iii) Image of A (- 5,4) in x-axis is C whose co-ordinates are (- 5, – 4)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 36

Question 6.
The point P (3, – 8) is reflected in origin to point Q. The Point Q is further reflected in x-axis to point R. Find :
(i) the co-ordinates of Q
(ii) the co-ordinates of R
(iii) the image of P (3, – 8) in y-axis.
Solution:
(i) The image of the given point P (3, – 8) when reflected in origin is Q whose co-ordinates will be (- 3, 8).
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 37
(ii) The image of Q (- 3, 8) when reflected in x-axis is R whose co-ordinates will be (-3,-8)
(iii) Tlie image of P (3, 8) in y-axis is R whose co-ordinates are (- 3, – 8).

Question 7.
Each of the points A (3, 0), B (7, 0), C (- 8, 0), D (- 7, 0) and E (0, 0) is reflected in x-axis to points A’, B’, C’, D’ and E’ respectively. Write the co-ordinates of each of the image points A’, B’, C’, D’ and E’.
Solution:
The points are given :
A (3, 0), B (1, 0), C (-8, 0), D (- 7, 0) and E (0, 0)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 38
This images will be when reflected in x-axis. A’ (3, 0), B’ (7, 0), C’ (- 8, 0) D’ (- 7, 0) and E’ (0, 0) as the given points lie on x-axis.

Question 8.
Each of the points A (0, 4), B (0, 10), C (0, – 4), D (0, – 6) and E (0, 0) is reflected in y-axis to points A’, B’, C’, D’ and E’ respectively. Write the co-ordinates of each of the image points A’, B’, C’, D’ and E’.
Solution:
The given points
A (0, 4), B (0, 10), C (0, – 4), D (0, – 6) and E (0, 0) are reflected in y-axis. The co – ordinates of their images will be A’ (0, 4), B’ (0, 10), C’ (0, – 4) D’ (0, – 6) and E’ (0, 0) as they all lie on y-axis.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 39

Question 9.
Each of the points A (0, 7), B (8. 0), C (0, -5), D (- 7, 0) and E (0, 0) are reflected in origin to points A’, B’, C’, D’ and E’ respectively. Write the co-ordinates of each of the image points A’, B’, C’, D’ and E’.
Solution:
The points A (0, 7), B (8, 0), C (0, – 5). D (- 7.0) and E (0,0) are reflected in origin.
So, the co-ordinates of their images will be A’ (0,-7), B’ (- 8, 0), C’ (0,5), D’ (7, 0) and E’ (0, 0)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 40

Question 10.
Mark points A (4, 5) and B (- 5, 4) on a graph paper. Find A’, the image of A in x-axis and B’, the image of B in x-axis.
Mark A’ and B’ also on the same graph paper.
(ii) Join AB and A’ B’ and
find if AB = A’ B’ ?
Solution:
The given points :
A (4, 5) and B (- 5, 4) have been marked on the graph.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 41
The image of A in x-axis A (4, – 5) and image of B in x-axis is B’ (-5, -4) which have been also plotted on the same graph.
AB and A’ B’ are joined. We see that AB = A’ B’.

Question 11.
Mark points A (6, 4) and B (4, – 6) on a graph paper.
Find A’, the image of A in y-axis and B’, the image of B in y-axis. Mark A’ and B’ also on the same graph paper.
Solution:
The given points are
A (6, 4) and B (4, – 6)
The images of A and B is y-axis are A’ (- 6, 4) and B’ (- 4, – 6) respectively as shown in the same graph.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 42

Question 12.
Mark points A (- 6, 5) and B (- 4, – 6) on a graph paper. Find A’, the image of A in origin and B’, the image of B in origin. Mark A’ and B’ also on the same graph paper. Join AB and A’ B’. Is AB = A’ B’ ?
Solution:
The given points are A (- 6, 5) and B (- 4, – 6). The images of A and B in the origin are A’ and B’ where co-ordinates are A’ (6, – 5) and B’ (4, 6) which have been plotted on the same graph.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 43
AB and A’ B’ are joined we see that AB = A’ B’.

Symmetry Exercise 17C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
How many lines of symmetry does a rhombus have?
Solution:
It has two lines of symmetry.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 44

Question 2.
What is the order of rotational symmetry of a rhombus?
Solution:
The order of rotational symmetry can be defined as the number of times that a shape appears exactly the same during a full 360° rotation. Order of rotational symmetry of a rhombus is 2.

Question 3.
Show that each of the following figures has two lines of symmetry and a rotational.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 45
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 46

Question 4.
Name a figure that has a line of symmetry but does not have any roational symmetry.
Solution:
Isosceles triangle has only line symmetry and no rotational symmetry.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 47

Question 5.
In each of the following figures, draw all possible lines of symmetry and also write the order of rotational symmetry:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 48
Solution:
(i) It has no line of symmetry and order of rotational symmetry is 3
(ii) It has 2 line of symmetry and order of rotational symmetry is 4.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 49
(iii) It has 2 line of symmetry and order of rotational symmetry is 2.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 50
(iv) It has no line of symmetry and order of rotational symmetry is 0.
(v) It has 1 line of symmetry and order of rotational symmetry is 0.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 51

 

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 13 Set Concepts

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 13 Set Concepts (Some Simple Divisions by Vedic Method)

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 13 Set Concepts (Some Simple Divisions by Vedic Method)

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POINTS TO REMEMBER
1. Definition of a Set : In our day to day life, different collective nouns are used to describe collection of objects ; such as : a group of students playing cricket, a pack of cards, a bunch of flowers, etc
In mathematics, such collections of objects are named as sets.
A set is a collection of well-defined objects, things or symbols, etc.
The phrase ‘well-defined’ means ; it must be possible to know, without any doubt, whether a given object (thing or symbol) belongs to the set under consideration or not.
For example :
“The set of tall boys of Class 10” is not well-defined ; since it is not possible to know that which boys are to be included and exactly what is the limit.
But when we say, “The set of boys of Class 10, which are taller than Peter”, now we can compare the heights of different boys with the height of Peter and can know exactly, that which boys are to be included in the required set. Thus, the objects are well-defined.
2. Elements of a set: The objects (things, symbols, etc.) used to form a set are called elements or members of the set.
In general, a set is denoted by a capital letter of English alphabet with its elements written inside curly braces and separated by commas.
e.g., Set A = {5, 10, 12, 15}
3. Use of Symbol ‘∈’ or Symbol ‘∉ ‘ : The symbol ‘∈’ stands for ‘belongs to’ or ‘ is an element of’ or ‘is a member of’; whereas the symbol ‘∉’ stands for ‘does not belong to’ or ‘is not an element of or ‘is not a member of’.
e.g., For set P = {3, 6, 8, 13, 18} ; 3 ∈ P, 5∉ P and so on.
(i) The elements in a set can be written in any order.
Thus, {a, b, c, d] is the same set as {b, d, a, c} or {c, b, d, a}, etc.
(ii) The elements in a set should not be repeated, i.e. if any element occurs many times, it should be written only once.
Thus, set of letters of the word ‘crook’ = {c, r, o, k}.
There are two os in the given word “crook” ; but in the set, it is written only once.
4. Representation of A Set: A set, in general, is represented in :
(i) Description method (form)
(ii) Tabular or Roster method (form)
(iii) Set-builder or Rule method
For example :
N is the set of natural numbers [Description method]
N = {1, 2, 3, 4, 5, ……. } [Roster or Tabular method]
N = {x : x is a natural number}, or {x : x ∈ N} [Set-builder or Rule method]
[The symbol ‘ : ’ stands for such that and the set {x : x ∈ N} is read as, “the set of x such that x is a natural number”].
It is clear from the example given above that:
(i) in description method a well-defined description about the set is given.
(ii) in roster or tabular method the elements of the set are written inside a pair of curly braces and are separated by commas.
(iii) in set-builder or ruler method the actual elements of the set are not written, but a rule or a statement or a formula is written in the briefest possible way.
5. Cardinal Number : The cardinal number of a set is the number of elements in it.
Thus, if a set A has 5 elements ; its cardinal number is 5 and we represent it by writing n (A) = 5 .
Similarly, if set B = Set of even natural numbers less than 10 then, B = {2, 4, 6, 8} and n (B) = 4.
If B = {0}, then n (B) = 1. Since, 0 is an element of set B.
6. Types of Sets :
(i) Finite Set: A set is said to be a finite set, if it has a limited (countable) number of elements in it.
For example :
(a) S = Set of natural numbers between 10 and 15 = {11, 12, 13, 14}
(b) P = {0, 1,2, , 20} = {x : x ∈ W and x ≤ 20} and so on.
(ii) Infinite set: A set is said to be an infinite set, if it has an unlimited (uncountable) number of elements in it.
For example :
(a) P = Set of prime numbers = {2, 3, 5,…. }
(b) B = (x : x ∈ N and x ≥ 21} = (21, 22, 23, } ans so on.
(iii) Empty set or Null set: The set, with no element in it, is called the empty set or the null set.
The empty set is represented by a pair of braces with no element in it or by the Danish letter Φ, which is pronounced as ‘oe’
Thus, the empty set = { } = Φ
Note : For empty set, it is wrong to call ‘an empty set’ or ‘a null set’ as there is one and only one empty set though it may have many descriptions.
Therefore, it is always called “the empty set or the null set”.
Some examples of the empty set :
(a) Let A = {a man of age more than 400 years}.
Since there can not be any man with the age more than 400 years; the set A will have no element in it i.e. It is the empty set. And we write : A = { } or Φ
(b) If B = (Triangles with 4 sides} ; it is clear that B = Φ
Note :
1. Φ ≠{0}, since {0} is a set with 0 as its element whereas Φ has no element.
2. {Φ} ≠ {0}, since both the sets have different elements.
3. The cardinal number of the empty set is 0 i.e. n (Φ) = 0.
(iv) Disjoint sets : Sets having no element in common are called disjoint sets.
For example :
Sets P = (5, 7, 9} and O = (4, 6, 10, 12} are disjoint; as they do not have any element in common.
(v) Joint (overlapping) sets : Sets having atleast one element in common are called joint or overlapping sets.
For example:
Set B = (4,6, 8,10,12} and set C = (3,6,9,12,15} are joint sets; as they have elements 6 and 12 common.
(vi) Equal sets : Two sets are said to be equal, if the elements of both the sets are the same.
For example :
If set A = {x, y, z} and set B = {last three letters of English alphabet}.
Clearly, sets A and B have the same elements and so set A = set B.
(vii) Equivalent sets : Two sets are said to be equivalent, if they have equal number of elements in them, i.e. the cardinal numbers of both the sets are equal.
For example :
Let A = (3, 6, 9} and B = {a, b, c}.
Since, set A has 3 elements and set B also has 3 elements i.e., n (A) = n (B); therefore, sets A and B are equivalent and for this, we write : A ↔ B.
Note:
1. Equal sets are always equivalent; but the converse is not always true (i.e. it is not necessary that equivalent sets are equal also).
2. In equivalent sets, the number of elements (cardinal number) are equal, whereas in equal sets the element are the same.
3. Two infinite sets are always equivalent;

Set Concepts Exercise 13A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find, whether or not, each of the following collections represent a set:
(i) The collection of good students in your school.
(ii) The collection of the numbers between 30 and 45.
(iii) The collection of fat-people in your colony.
(iv) The collection of interesting books in your school library.
(v) The collection of books in the library and are of your interest.
Solution:
(i) It is not a set as it is not well defined.
(ii) It is a set
(iii) It is not a set as it is not well defined.
(iv) It is not a set as it is not well defined.
(v) It is a set.

Question 2.
State whether true or false :
(i) Set {4, 5, 8} is same as the set {5, 4, 8} and the set {8, 4, 5}
(ii) Sets {a, b, m, n} and {a, a, m, b, n, n) are same.
(iii) Set of letters in the word ‘suchismita’ is {s, u, c, h, i, m, t, a}
(iv) Set of letters in the word ‘MAHMOOD’ is {M, A, H, O, D}.
Solution:
(i) True
(ii) True
(iii) True as it has the same elements
(iv) True as it has the same elements.

Question 3.
Let set A = {6, 8, 10, 12} and set B = {3, 9, 15, 18}.
Insert the symbol ‘ ∈ ’ or ‘ ∉ ’ to make each of the following true :
(i) 6 …. A
(ii) 10 …. B
(iii) 18 …. B
(iv) (6 + 3) …. B
(v) (15 – 9) …. B
(vi) 12 …. A
(vii) (6 + 8) …. A
(viii) 6 and 8 …. A
Solution:
(i) 6 ∈ A
(ii) 10 ∉ B
(iii) 18 ∈B
(iv) (6 + 3) or 9 ∈ B
(v) 15 – 9 or 6 g B
(vi) 12 ∈ A
(vii) 6 + 8 or 14 ∉ A
(viii) 6 and 8 ∈ A

Question 4.
Express each of the following sets in
roster form :
(i) Set of odd whole numbers between 15 and 27.
(ii) A = Set of letters in the word “CHITAMBARAM”
(iii) B = {All even numbers from 15 to 26}
(iv) P = {x : x is a vowel used in the word ‘ARITHMETIC’}
(v) S = {Squares of first eight whole numbers}
(vi) Set of all integers between 7 and 94; which are divisible by 6.
(vii) C = {All composite numbers between 2 and 20}
(viii) D = Set of Prime numbers from 2 to 23.
(ix) E = Set of natural numbers below 30 which are divisible by 2 or 5.
(x) F = Set of factors of 24.
(xi) G = Set of names of three closed figures in Geometry.
(xii) H = {x : x eW and x < 10}
(xiii) J = {x: x e N and 2x – 3 ≤17}
(xiv) K = {x : x is an integer and – 3 < x < 5}
Solution:
(i) {17, 19, 21, 23, 25}
(ii) A = (C, H, I, T, A, M, B, R}
(iii) B = {16, 18, 20, 22, 24, 26}
(iv) P = {a, e, i}
(v) S = {0, 1, 4, 9, 16, 25, 36, 49}
(vi) {12, 18, 24; 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90}
(vii) C = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18}
(viii) D = {2, 3, 5, 7, 11, 13, 17, 19,23}
(ix) E = {2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26, 28}
(x) F={l,2, 3, 4, 6, 8, 12, 24}
(xi) G = {Triangle, quadrilateral, circle}
(xii) H = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(xiii) 2x – 3 ≤ 17
⇒ 2x ≤ 17 + 3 2 x ≤ 20
⇒ x ≤ \(\frac { 20 }{ 2 }\)
x ≤ 10
∴ J = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(xiv) ∵ – 3 < x < 5
∴x lies between – 3 and 5
∴K = {- 2, – 1, 0, 1, 2, 3, 4}

Question 5.
Express each of the following sets in set- builder notation (form) :
(i) {3, 6, 9, 12, 15}
(ii) {2, 3, 5, 7, 11, 13 …. }
(iii) {1, 4, 9, 16, 25, 36}
(iv) {0, 2, 4, 6, 8, 10, 12, …. }
(v) {Monday, Tuesday, Wednesday}
(vi) {23, 25, 27, 29, … }
(vii) {\(\frac { 1 }{ 3 }\),\(\frac { 1 }{ 4 }\),\(\frac { 1 }{ 5 }\),\(\frac { 1 }{ 6 }\),\(\frac { 1 }{ 7 }\),\(\frac { 1 }{ 8 }\)}
(viii) {42, 49, 56, 63, 70, 77}
Solution:
(i) {3, 6, 9, 12, 15}
= {x: x is a natural number divisible by 3 ;x< 18}
(ii) {2, 3, 5, 7, 11, 13, }
= {x : x is a prime number}
(iii) {1,4,9, 16,25,36}
= {x : x is a perfect square ; x < 36}
(iv) {0, 2, 4, 6, 8, 10, 12, }
= {x : x is a whole number divisible by 2}
(v) {Monday, Tuesday, Wednesday}
= {x : x is one of the first three days of 3 week}
(vi) {23, 25, 27, 29, }
= {x : x is an odd natural number; x ≥ 23}
(vii) {\(\frac { 1 }{ 3 }\),\(\frac { 1 }{ 4 }\),\(\frac { 1 }{ 5 }\),\(\frac { 1 }{ 6 }\),\(\frac { 1 }{ 7 }\),\(\frac { 1 }{ 8 }\)}
= {x: x = \(\frac { 1 }{ n }\) when n is a natural number: 3 ≤ n ≤ 8}
(viii) {42, 49, 56, 63, 70, 77}
= (x: x is a natural number divisible by 7 ; 42 ≤x ≤ 77}

Question 6.
Given : A = {x : x is a multiple of 2 and is less than 25}
B = {x : x is a square of a natural number and is less than 25}
C = {x : x is a multiple of 3 and is less than 25}
D = {x: x is a prime number less than 25}
Write the sets A, B, C and D in roster form.
Solution:
A = {x: x is a multiple of 2 and is less than 25}
= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
B = {x : x is a square of natural number and is less than 25}
= {1,4,9,16}
C = {x : x is a multiple of 3 and is less than 25}
= {3, 6, 9, 12, 15, 18,21,24}
D = {x : x is a prime number less than 25}
= {2, 3, 5, 7, 11, 13, 17, 19, 23}

Set Concepts Exercise 13B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Write the cardinal number of each of the following sets:
(i) A = Set of days in a leap year.
(ii) B = Set of numbers on a clopk-face.
(iii) C = {x : x ∈ N and x ≤ 7}
(iv) D = Set of letters in the word “PANIPAT”.
(v) E = Set of prime numbers between 5 and 15.
(vi) F = {x : x ∈ Z and – 2 < x ≤ 5}
(vii) G = {x : x is a perfect square number, x ∈N and x ≤ 30}.
Solution:
(i) n A = 366
(ii) n B = 12
(iii) n C =7
(iv) n D = 5
(v) n E = 3
(vi) n F =7
(vii) n G = 5

Question 2.
For each set, given below, state whether it is finite set, infinite set or the null set :
(i) {natural numbers more than 100}
(ii) A = {x : x is an integer between 1 and 2}
(iii) B = {x : x ∈ W ; x is less than 100}.
(iv) Set of mountains in the world.
(v) {multiples of 8}.
(vi) {even numbers not divisible by 2}.
(vii) {squares of natural numbers}.
(viii) {coins used in India}
(ix) C = {x | x is a prime number between 7 and 10}.
(x) Planets of the Solar system.
Solution:
(i) {Natural numbers more than 100}
= It is an infinite set
(ii) A = {x : x is an integer between 1 and 2}
It is a null set
(iii) B = {x : x ∈ W, x is less than 100}
It is finite set as it has 100 elements i.e. from 0 to 99.
(iv) Set of mountains in the world.
∴ It is an infinite set
(v) {Multiples of 8}
It is an infinite set
(vi) {Even numbers not divisible by 2}
It is a null set
(vii) {Squares of natural numbers}
∴ It is an infinite set
(viii) {Coins used in India}
∴It is a finite set as these are countable
(ix) {x | x is a prime number between 7 and 10}
As there is not such prime number between 7 and 10.
Hence it is null set
(x) Planets of two Solar system.
It is finite set as there are countable.

Question 3.
State, which of the following pairs of sets are disjoint :
(i) {0, 1, 2, 6, 8} and {odd numbers less than 10.
(ii) {birds} and {tress}
(iii) {x : x is a fan of cricket} and {x : x is a fan of football}.
(iv) A = {natural numbers less than 10} and B = {x : x is a multiple of 5}.
(v) {people living in Calcutta} and {people living in West Bengal}.
Solution:
(i) {0, 1, 2, 6, 8} and {odd numbers less than 10}
⇒ {0, 1,2, 6, 8} and {1,3, 5, 7, 9}
∴There sets are not disjoint sets as there is one element (1) is common.
(ii) {Birds} and {trees}
These are disjoint sets as there is no common element in term
(iii) {x : x is a fan of cricket} and {x : x is a fan of football}
These are not disjoint sets as there can be a person who is fan of both the games.
(iv) A = {Natural numbers less than 10} and B = {x : x is a multiple of 5}
⇒ A = {1, 2, 3, 4, 5, 6, 7, 8, 9} and B = {5, 10, 15 }
These are hot disjoint sets as there is one element 5, which is common.
(v) {People living in Calcutta} and {People living in West Bengal}.
These are not disjoint sets as people of Calcutta are the people of West Bengal as Calcutta is a city of West Bengal.
So, only (ii) is a pair of disjoint sets.

Question 4.
State whether the given pairs of sets are equal or equivalent.
(i) A = {first four natural numbers} and B = {first four whole numbers}.
(ii) A = Set of letters of the word “FOLLOW” and B = Set of letters of the word “WOLF”.
(iii) E = {even natural numbers less than 10} and O = {odd natural numbers less than 9}
(iv) A = {days of the week starting with letter S} and B = {days of the week starting with letter T}.
(v) M = {multiples of 2 and 3 between 10 and 20} and N = {multiples of 2 and 5 between 10 and 20}.
(vi) P = {prime numbers which divide 70 exactly} and Q = {prime numbers which divide 105 exactly}
(vii) A = {0², 1², 2², 3², 4²} and = {16, 9,4, 1, 0}.
(viii) E = {8,JO, 12, 14, 16} and F = {even natural numbers between 6 and 18}.
(ix) A = {letters of the word SUPERSTITION} and B = {letters of the word JURISDICTION}.
Solution:
(i) A = {first four natural numbers}
= {1,2, 3, 4}
B = {first first whole number)
= {0, 1,2,3}
These are equivalent sets as both have equal number of elements but not same.
(ii) A = Set of letters of the word ‘FOLLOW’
= {F, O, L, W}
B = Set of letters of the word ‘WOLF’
= {W, O, L, F}
These are equal sets as these have same and equal elements.
(iii) E = {even natural numbers less than 10}
= {2, 4, 6, 8}
O = {odd natural numbers less than 9}
= {L3, 5, 7}
These are equivalent sets as both have equal number of elements but not the same.
(iv) A = {Days of the week starting with letter S}
= {Sunday, Saturday}
B = {Days of the week starting with letter T}
= {Tuesday, Thursday}
These are equivalent sets as both have equal number of elements.
(v) M = {Multiples of 2 and 3 between 10 and 20}
= {12, 14, 15, 16, 18}
N = {Multiples of 2 and 5 between 10 and 20}
= {12, 14, 15, 16, 18}
These are equal sets as these have same and equal number of elements.
(vi) P = {Prime numbers which divide 70 exactly}
= {2, 5, 7}
Q = {Prime numbers which divide 105 exactly}
= {3, 5, 7}
These are equivalent sets as these have equal number of elements.
(vii) A = {02, l2, 22, 32, 42} = {0, 1, 4, 9, 16} B = {16, 9, 4, 1,0}
These are equal sets as these have same and equal number of elements.
(viii) E = {8, 10, 12, 14, 16}
F = {even natural numbers between 6 and 18}
= {8, 10, 12, 14, 16}
These sets are equal as these have same and equal number of elements
(ix) A = {Letters of the word SUPERSTITION}
= {S, U, P, E, R, T, I, O, N}
B = Letters of the word JURISDICTION.
= (J, U, R, I, S, D, C, T, O, N}
These are neither equal nor equivalent sets as these have different and unequal elements.

Question 5.
Examine which of the following sets are the empty sets :
(i) The set of triangles having three equal sides.
(ii) The set of lions in your class.
(iii) {x 😡 + 3 = 2 and xeN}
(iv) P = {x : 3x = 0}
Solution:
(i) The set of triangle having three equal sides. This is not an empty set
(ii) The set of lions in your class This is an empty set
(iii) {x : x +3 = 2 and x ∈N}
x ≠ 3 = 2 ⇒ x = 2-3= -l
which is not a natural number.
∴ It is an empty set.
(iv) P = {x : 3x = 0} = {0} which is not an empty set.
Hence (ii) and (iii) are empty sets.

Question 6.
State true or false :
(i) All examples of the empty set are equal.
(ii) All examples of the empty set are equivalent.
(iii) If two sets have the same cardinal number, they are equal sets.
(iv) If n (A) = n (B) then A and B are equivalent sets.
(v) If B = {x : x + 4 = 4}, then B is the empty set.
(vi) The set of all points in a line is a finite set.
(vii) The set of letters in your Mathematics book is an infinite set.
(viii) If M = {1, 2, 4, 6} and N = {x : x is a factor of 12} ; then M = N.
(ix) The set of whole numbers greater than 50 is an infinite set.
(x) If A and B are two different infinite sets, then n (A) = n (B).
Solution:
(i) True
(ii) True
(iii) False
(iv) True
(v) False
(vi) False
(vii) False
(viii) False
( ix )True
(x) False

Question 7.
Which of the following represent the null set ?
φ, {0}, 0, { }, {φ}

Solution:
φ and { } are the null sets other are not as there have same element.

Set Concepts Exercise 13C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Fill in the blanks :
(i) If each element of set P is also an element of set Q, then P is said to be …… of Q and Q is said to be of P.
(ii) Every set is a ….. of itself.
(iii) The empty set is a …… of every set.
(iv) If A is proper subset of B, then n (A) …. n (B).
Solution:
(i) If each element of set P is also an element of set Q then P is said to be subject of Q ; and Q is said to be super set of P.
(ii) Every set is a subset of itself.
(iii) The empty set is subset of every set.
(iv) If A is proper subset of B, then n (A) is less than n (B)

Question 2.
If A = {5, 7, 8, 9} ; then which of the following are subsets of A ?
(i) B = {5, 8}
(ii) C = {0}
(iii) D = {7, 9, 10}
(iv) E = { }
(V) F = {8, 7, 9, 5}
Solution:
(i) B = {5, 8,}
∴B ⊂ A
(ii) C = {0}
∴ C φ A
(iii) D = {7, 9, 10}
∴D ⊄ A
(iv) E = { }
∴E ⊂A (An empty set is subset of every set)
(v) F = (8, 7, 9, 5}
∴F ⊂A
∵ Every set is subset of it self.
Hence (i), (iv)
and (v) are subsets of A.

Question 3.
If P = {2, 3, 4, 5} ; then which of the following are proper subsets of P ?
(i) A = {3, 4}
(ii) B = { }
(iii) C = {23, 45}
(iv) D = {6, 5, 4}
(v) E = {0}
Solution:
P = {2, 3, 4, 5}
(i) A = {3,4},
(ii) B = { }, C = {23, 45},
D = {6, 5, 4} and E = {0}.
We see that only A and B are the proper subset of P.

Question 4.
If A = {even numbers less than 12},
B = {2, 4},
C = {1, 2, 3},
D = {2, 6} and E = {4}
State which of the following statements are true :
(i) B⊂A
(ii) C⊆A
(iii) D⊂C
(iv) D ⊄ A
(v)E⊇B
(vi) A⊇B⊇E
Solution:
A = {Even number less than 12} = {2, 4, 6, 8, 10}
B = {2, 4}, C = {1, 2, 3},
D = {2, 6} and E = {4}
(i) B ⊂ A: It is true
(ii) C ⊆ A: It is false
(iii) D ⊂ C : It is false
(iv) D ⊄ A
(v) E ⊇ B : It is false
(vi) A ⊇ B ⊇ E : It is true

Question 5.
Given A = {a, c}, B = {p, q, r} and C = Set of digits used to form number 1351.
Write all the subsets of sets A, B and C.
Solution:
(i) A = {a, c}
∴ Subsets are : { } or φ, {a}, {c} and {a, c}
(ii) B = {p,q, r)
∴ subsets are : { } or φ, {p}, {q}, {r}, {p, q}, ip, r}, {q, r} and {p, q, r}
(iii) C = Set of digits used in 135, = {1,3,5}
∴ Subsets are = { }
or φ, {1}, {3}, {5}, {1,3}, {1,5}, {2,5} and {1, 3, 5}

Question 6.
(i) If A = {p, q, r}, then number of subsets of A = ……
(ii) If B = {5, 4, 6, 8}, then number of proper subsets of B = ……
(iii) If C = {0}, then number of subsets of C = …..
(iv) If M = {x : x ∈ N and x < 3}, then M has …… proper subsets.
Solution:
(i) If A = {p, q, r},
then number of subsets of A = 2³ = 2×2×2 = 8
(ii) If B = {5, 4, 6, 8},
then number of proper subsets of B = 24 – 1 = 2 × 2 × 2× 2 – 1 = 16 – 1 = 15
(iii) If C = {0},
then number of subsets of C = 21 = 2
(iv) If M = {x: x ∈ N and x < 3}, = {1, 2}
Then M has proper subsets = 22—1 = 4 — 1 = 3

Question 7.
For the universal set {4, 5, 6, 7, 8, 9, 10, 11,12,13} ; find its subsets A, B, C and D such that
(i) A = {even numbers}
(ii) B = {odd numbers greater than 8}
(iii) C = {prime numbers}
(iv) D = {even numbers less than 10}.
Also, find compliments of each set i.e., find A’, B’, C’ and D’.
Solution:
(i) A = {even numbers}
= {4, 6, 8, 10, 12}
(ii) B = {odd numbers greater then 8}
= {9, 11, 13}
(iii) C = {Prime numbers}
= {5, 7, 11, 13}
(iv) D = {even numbers less than 10}
= {4, 6, 8}
A’= {5, 7, 9, 11, 13},
B’ = {4, 5, 6, 7, 8, 10, 12}
C’ = { 4, 6, 8, 9, 10, 12}
and D’ = {5, 7, 9, 10, 11, 12, 13}

Set Concepts Exercise 13D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
If A = {4, 5, 6, 7, 8} and B = {6, 8, 10, 12}, find :
(i) A∪B
(ii) A∩B
(iii) A-B
(iv) B-A
Solution:
(i) A∪B
= [All the elements from set A and all the elements from set B]
= {4, 5, 6, 7, 8, 10, 12}
(ii) A∩B
= [elements common to both the sets A and B]
= {6, 8}
(iii) A-B
= [elements of set A which are not in set B]
= {4, 5, 7}
(iv) B-A
= [elements of set B which are not in set A]
= {10, 12}

Question 2.
If A = {3, 5, 7, 9, 11} and B = {4, 7, 10}, find:
(i) n(A)
(ii) n(B)
(iii) A∪B and n(A∪B)
(iv) A∩B and n(A∩B)
Solution:
(i) n(A) = (3, 5, 7, 9, 11) = 5
(ii) n(B) = (4, 7, 10) = 3
(iii) A ∪ B = {3, 4, 5, 7, 9, 10, 11} n(A ∪ B) = 7
(iv) A∩B = {6} n(A∩B)=l

Question 3.
If A = {2, 4, 6, 8} and B = {3, 6, 9, 12}, find:
(i) (A ∩ B) and n(A ∩ B)
(ii) (A – B) and n(A – B)
(iii) n(B)
Solution:
(i) (A ∩ B) = {2, 4, 8} n( A ∩ B) = 3
(ii) (A – B) and n(A – B)
⇒ (A – B) = (2, 4, 8)
⇒ n(A-B) = 3
(iii) n(B) = {3, 6, 9, 12} = 4

Question 4.
If P = {x : x is a factor of 12} and Q = {x: x is a factor of 16}, find :
(i) n(P)
(ii) n(Q)
(iii) Q – P and n(Q – P)
Solution:
(i) n(P) = Factors of 12 are
= 1, 2, 3, 4, 6, 12
∴ n(P) = 6
(ii) n(Q) = Factors of 16 are = 1. 2, 4, 8, 16
∴n(Q) = 5
(iii) Q – P and n(Q – P)
Elements of set P = {1, 2, 3, 4, 6, 12}
Elements of set Q = {1, 2, 4, 8, 16}
∴ Q – P = 8, 16
n(Q-P) = 2

Question 5.
M = {x : x is a natural number between 0 and 8) and N = {x : x is a natural number from 5 to 10}. Find :
(i) M – N and n(M – N)
(ii) N – M and n(N – M)
Solution:
Natural numbers between 0 and 8 M = {0, 1, 2, 3, 4, 5, 6, 7} and Natural numbers between 5 to 10 N = {6, 7, 8, 9, 10}
(i) M – N = {1, 2, 3, 4} and n(M – N) = 4
(ii) N – M = {8, 9, 10} and n (N – M) = 3

Question 6.
If A = {x: x is natural number divisible by 2 and x< 16} and B = {x:x is a whole number divisible by 3 and x < 18}, find :
(i) n(A)
(ii) n(B)
(iii) A∩B and n(A∩B)
(iv) n(A – B)
Solution:
(i) A = {x : x is natural number divisible-by 2 and x < 16}
A = {2, 4, 6, 8, 10, 12, 14}
n(A) = 7
(ii) B = {x: x is a whole number divisible by 3 and x < 18}
B = {3, 6, 9, 12, 15, 18}
n(B) = 6
(iii) A n B = {2, 4, 6, 8, 10, 12, 14} n {3, 6, 9, 12, 15, 18}
A∩B = {6,12} n(A ∩ B) = 2
(iv) A – B = {2, 4, 6, 8, 10, 12, 14} – {3, 6, 9, 12, 15, 18}
A-B = {2,4, 8, 10, 14} n(A – B) = 5

Question 7.
Let A and B be two sets such that n(A) = 75, M(B) = 65 and n(A ∩ B) = 45, find :
(i) n(A∪ B)
(ii) n(A – B)
(iii) n(B – A)
Solution:
n(A ∩ B)
n(A) = 75, n(B) = 65 and n(A ∩ B) = 45
(i) We know that,
n( A ∪B) = n(A) + n(B) – n( A ∩ B)
n(A ∪B) =75 + 65 – 45
n(A∪B) = 140-45 = 95
(ii) We know that,
n(A – B) = n(A) – n(A ∩ B)
n(A – B) = 75 – 45 = 30
(iii) We know that,
n(B – A) = n(B) – n(A ∩ B)
n(B – A) = 65 – 45 = 20

Question 8.
Let A and B be two sets such that n(A) = 45, n(B) = 38 and n(A ∪B) = 70, find :
(i) n(A∩B)
(ii) n(A-B)
(iii) n(B – A)
Solution:
n(A) = 45, n(B) = 38 and n(A∪ B) = 70
(i) We know that,
n(A ∩ B) = n(A) + M(B) – n(A ∪B)
n(A ∩ B) = 45 + 38 – 70 = 83 – 70 = 13
(ii) We know that,
n(A-B) = n(A ∪B)-n(B)
n(A – B) = 70 – 38 = 32
(iii) We know that,
n(B – A) = n(A ∪ B) – n(A)
n(B – A) = 70 – 45 = 25

Question 9.
Let n(A) 30, n(B) = 27 and n(A∪B) = 45, find :
(i) n(A∩B)
(ii) n(A-B)
Solution:
n(A) = 30, n(B) = 27 and n(A ∪ B) = 45
(i) We know that,
n(A ∩ B) = n( A) + n(B) – n( A∪ B)
n(A ∩ B) = 30 + 27 – 45
n(A ∩ B) = 57 – 45 = 12
(ii) We know that,
n(A-B) = n(A ∪B) – n(B)
n(A – B) = 45 – 27 = 18

Question 10.
Let n(A) = 31, n(B) = 20 and n(A ∩ B) = 6, find:
(i) n(A-B)
(ii) n(B – A)
(iii) n(A ∪B)
Solution:
n(A) = 31, n(B) = 20 and n(A ∩ B) = 6
(i) We know that,
n(A – B) = n(A) – n(A ∩ B)
n(A -B) = 31 -6 = 25
(ii) We know that,
n(B – A) = n(B) – n(A n B)
n(B – A) = 20 – 6 = 14
(iii) We know that,
n(A ∪B) = n(A) + n(B) – n(A ∪ B) n(A∪B) = 31 +20-6 = 45

 

Selina Concise Chemistry Class 7 ICSE Solutions – Atoms, Molecules and Radicals

Selina Concise Chemistry Class 7 ICSE Solutions – Atoms, Molecules and Radicals

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Chemistry for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 4 Atoms, Molecules and Radicals

Points to Remember :

  1. Atoms of the most elements are composed of still smaller particles known as fundamental particles or subatomic particles. They are protons, electrons and neutrons.
  2. Protons are the positively charged particles present in an atom.
  3. Electrons are the negatively charged particles present in an atom and its mass is 1/1837 the mass of one hydrogen atom.
  4. Neutrons are particles with no electrical charge. Its mass is equal to one atomic mass unit.
  5. The central part of an atom is called nucleus, which contains both protons and neutrons.
  6. Shells or orbits around the nucleus.
  7. Atomic number refers to the number of protons present in an atom. It is denoted by the alphabet Z.
  8. Mass number— Mass number is the sum of number of protons and neutrons present in the nucleus of an atom. Mass number (A) = Number of protons + Number of neutrons.
  9. Atomic mass— The mass of an atom is known as its atomic mass.
  10. Relative atomic mass ip the mass of an atom of an element as a multiple of the standard atomic mass unit.
  11. Atoms of different elements combine to form molecules so as to attain chemical stability.
  12. A positively charged ion is called a cation and a negatively charged ion is called anion.
  13. The valency of an element is the number of electrons donated or accepted by its ‘atom’ during chemical combination.
  14. There are some elements with more than one valency e.g., iron, copper, tin, lead.
  15. Two or more different non-metals that collectively accept or donate one or more electrons and become negatively or positively charged in the process are called radicals.

EXERCISE

Question 1.
Define the following terms :
Answer:

  1. Atom : An atom is the smallest indivisible unit of an
    OR
    Atom is the smallest unit of matter.
  2. Molecule : Molecule is the smallest unit of a compound (or an element) which always has an independent existance.
  3. Radicals : A radical is an atom of an element or a group of atoms of different elements that behaves as a single unit with a positive or negative charge on it.
  4. Valency : It is the number of electrons donated or accepted by the valence shell of an atom during chemical combination.
  5. Periodic table represents the tabular arrangment of elements in horizontal rows called periods and vertical columns called groups in order to classify the elements and their systematic study.

Question 2.
Write the names of the elements present in the following compounds.
Answer:

  1. Common salt : Sodium, chlorine.
  2. Ammonia : Nitrogen, hydrogen.
  3. Sulphuric acid : Hydrogen, sulphur, oxygen.
  4. Glucose : Carbon, hydrogen, oxygen.
  5. Sodium hydroxide : Sodium, oxygen, hydrogen.
  6. Acetic acid : Carbon, hydrogen, oxygen.

Question 3.
What does each of the following represent ?
Answer:

  1. 2C02 = 2 molecules of carbon dioxide.
  2. 2H2S = 2 molecules of hydrogen sulphide.
  3. 5H2S04 = 5 molecules of sulphuric acid.
  4. 6NaNO3 = 6 molecules of sodium nitrate.

Question 4.
Write the symbols and valencies of the following radicals:
Answer:
Selina Concise Chemistry Class 7 ICSE Solutions - Atoms, Molecules and Radicals- 4
Selina Concise Chemistry Class 7 ICSE Solutions - Atoms, Molecules and Radicals- 4.
Question 5.
Name the following radicals :
Answer:

  1. SO42- = Sulphate
  2. HC03 = Bicarbonate
  3. OH- = Hydroxide
  4. Cr2072-  = Dichromate

Question 6.

  1. Name one ion for  each of the valencies +1, +2  and +3.
  2. Name one ion for  each of the valencies-1, -2  and -3.

Answer:

  1. +1 = Sodium Na+
    +2 = Calcium Ca+2
    +3 = Aluminium Al+3
  2. -1 = Chlorine Cl-
    -2 = Oxygen 0-2
    -3 = Nitrogen N3_

Question 7.
The valency of calcium is 2. Write the valencies of other radical in the following :

  1. CaO
  2. Ca(OH)2
  3. CaC03
  4. CaCl2

Answer:

  1. O= 2
  2. OH = 1
  3. CO3 = 2
  4. Cl = l

Question 8.
Write the names of the following compounds :
Answer:
Selina Concise Chemistry Class 7 ICSE Solutions - Atoms, Molecules and Radicals- 8

Question 9.
Write the molecular formulae of:
Answer:
Selina Concise Chemistry Class 7 ICSE Solutions - Atoms, Molecules and Radicals- 9

Question 10.
The valency of sodium is one, write the molecular formula for the following compounds of sodium.

  1. sodium oxide : Na20
  2. sodium sulphate : Na2S04
  3. sodium carbonate : Na2CO3
  4. sodium hydroxide : NaOH
  5. sodium nitrate : NaN03

 

Question 11.
What is variable valency ? Give two examples of elements showing variable valency.
Answer:
There are some elements with more than one valency. They are said to have variable valency, e.g. Iron, copper.

Question 12.
Give the group number of following elements present in periodic table

  1. Magnesium : IIA
  2. Carbon : IVA
  3. Sulphur : VIA
  4. Neon : Zero

 

Question 13.
An element belongs to group VA. What would be its valency? Name two such elements.
Answer:
Elements of group VA has valency 3.
Two elements : Nitrogen and phosphorus.

Question 14.
An element belongs to group II. What would be its valency? Write the formula of molcules of compounds it will form with elements in VA, VIA and VIIA groups.
Answer:
Selina Concise Chemistry Class 7 ICSE Solutions - Atoms, Molecules and Radicals- 14
Selina Concise Chemistry Class 7 ICSE Solutions - Atoms, Molecules and Radicals- 14 a

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks:
Answer:

  1. Atoms are neutral.
  2. An ion with positive charge is called cation.
  3. An ion with negative charge is called anion.
  4. 2H2 means two atoms of hydrogen.
  5. Ozone is a triatomic molecule.
  6. Metals have variable valency.
  7. Chemical name of caustic soda is sodium hydroxide NaOH.

2. Tick (√) the correct answer.

(a) The valency of iron in Fe203 is

  1. 1
  2. 2
  3. 3
  4. 6

(b) Which of the following has valency 4 ?

  1. aluminium
  2. oxygen
  3. carbon
  4. phosphorus

(c) The sulphate radical is written as S042-. What is the formula of calcium sulphate ?

  1. Ca(S04)2
  2. Ca2(S04)
  3. Ca(S04)3
  4. CaS04

(d) Which of the following exhibit variable valency ?

  1. calcium
  2. copper
  3. carbon
  4. chlorine

3. State the term for the following:

  1. The number of atoms present in a molecule of an element atomicity.
  2. The symbolic representation of a molecule molecular formula.
  3.  A group of atoms that react as a single unit molecule.
  4. The combining capacity of an element valency.
  5. The tabular arrangement of elements in horizontal rows and vertical columns periodic table.

Selina Concise Mathematics class 7 ICSE Solutions – Fundamental Concepts (Including Fundamental Operations)

Selina Concise Mathematics class 7 ICSE Solutions – Fundamental Concepts (Including Fundamental Operations)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 7 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

POINTS TO REMEMBER

  1. Constants and Variables : The numbers which has fixed value is called constant and same at English alphabet which can be assigned any value according to the requirement is called variables.
  2. Term : A term is a number, (constant), a variable or a combination of numbers and variables.
  3. Algebraic Expression : An algebraic expression is a collection of one or more terms, which are separated from each other by addition (+) or subtraction (-) signs.
  4. Types of algebraic expressions :
    (i) Monomial : It has only one term
    (ii) Binomial : It has two terms
    (iii) Trinomial : It has three terms
    (iv) Multinomial : It has more than three terms
    (v) Polynomial : It has two or more than two terms.
    Note : An expression of the type \(\frac { 2 }{ 5 }\) does not form a monomial unless JC is not equal to zero.
  5. Product: When two or more quantities are multiplied together, the result is called their product.
  6. Factors : Each of the quantities (numbers or variables) multiplied together to form a term is called a factor of the given term.
  7. Co-efficient: In a monomial, any factor or group of factors of a term is called the co-efficient of the remaining part of the monomial.
  8. Degree of a monomial: The degree of a monomial is the exponent of its variable or the sum of the exponents of its variables.
  9. Degree of a polynomial: The degree of a polynomial is the degree of its highest degree term.
  10. Like and unlike terms : Terms having the same literal co-efficients or alphabetic letters are called like terms ; whereas the terms with different literal co-efficients are called unlike terms.
  11. Addition and subtraction : Addition and subtraction of only like terms is possible by adding or subtracting the numerical co-efficients.
  12. Multiplication and division :
    (A) Multiplication :
    (i) Multiplications of monomials.
    (a) Multiply the numerical co-efficient together
    (ii) Multiply the literal co-efficients separately together.
    (iii) Combine the like terms.
    (B) Division :
    (i) Dividing a polynomial by a monomial Divide each term of the polynomial by monomial and simplify each fractions.
    (ii) While dividing one polynomial by another polynomial ; arrange the terms of both the dividend and the divisior both in descending or in ascending order of their powers and then divide.

SOME IMPORTANT POINTS

TYPES OF BRACKETS:
The name of different types of brackets and the order in which they are removed is shown below:
(a) ____ ; Bar (Vinculum) bracket
(b) ( ); Circular bracket .
(c) { } ; Curly bracket and then
(d) [ ]; square bracket

EXERCISE 11 (A)

Question 1.
Separate constant terms and variable terms from tile following :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 1

Solution:

Constant is only 8 others are variables

Question 2.
Constant is only 8 others are variables
(i) 2x ÷ 15
(ii) ax+ 9
(iii) 3x2 × 5x
(iv) 5 + 2a-3b
(v) 2y – \(\frac { 7 }{ 3 }\) z÷x
(vi) 3p x q ÷ z
(vii) 12z ÷ 5x + 4
(viii) 12 – 5z – 4
(ix) a3 – 3ab2 x c

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 3

Question 3.
Write the coefficient of:
(i) xy in – 3axy
(ii) z2 in p2yz2
(iii) mn in -mn
(iv) 15 in – 15p2

Solution:
(i) Co-efficient of xy in – 3 axy = – 3a
(ii) Co-efficient of z2 in p2yz2 = p2y
(iii) Co-efficient of mn in – mn = – 1
(iv) Co-efficient of 15 in – 15p2 is -p2

Question 4.
For each of the following monomials, write its degree :
(i) 7y
(ii) – x2y
(iii) xy2z
(iv) – 9y2z3
(v) 3 m3n4
(vi) – 2p2q3r4

Solution:
(i) Degree of 7y = 1
(ii) Degree of – x2y = 2+1=3
(iii) Degree of xy2z = 1 + 2 + 1 = 4
(iv) Degree of – 9y2z3 = 2 + 3 = 5
(v) Degree of 3m3n4 = 3 + 4 = 7
(vi) Degree of – 2p2q3r4 = 2 + 3 + 4 = 9

Question 5.
Write the degree of each of the following polynomials :
(i) 3y3-x2y2 + 4x
(ii) p3q2 – 6p2q5 + p4q4
(iii) – 8mn6+ 5m3n
(iv) 7 – 3x2y + y2
(v) 3x – 15
(vi) 2y2z + 9yz3

Solution:
(i) The degree of 3y3 – x2y2+ 4x is 4 as x2
y2 is the term which has highest degree.
(ii) The degree of p3q2 – 6p2q5-p4q4 is 8 as p4 q4 is the term which has highest degree.
(iii) The degree of- 8mn6 + 5m3n is 7 as – 8mx6 is the term which has the highest degree.
(iv) The degree of 7 – 3x2 y + y2 is 3 as – 3x2y is the term which has the highest degree.
(v) The degree of 3x – 15 is 1 as 3x is the term which is highest degree.
(vi) The degree of 2y2 z + 9y z3 is 4 as 9yz3 has the highest degree.

Question 6.
Group the like term together :
(i) 9x2, xy, – 3x2, x2 and – 2xy
(ii) ab, – a2b, – 3ab, 5a2b and – 8a2b
(iii) 7p, 8pq, – 5pq – 2p and 3p

Solution:
(i) 9x2, – 3x2 and x2 are like terms
xy and – 2xy are like terms
(ii) ab, – 3ab, are like terms,
– a2b, 5a2b, – 8a2b are like terms
(iii) 7p, – 2p and 3p are like terms,
8pq, – 5pq are like terms.

Question 7.
Write numerical co-efficient of each of the followings :
(i) y
(ii) -y
(iii) 2x2y
(iv) – 8xy3
(v) 3py2
(vi) – 9a2b3

Solution:
(i) Co-efficient of y = 1
(ii) Co-efficient of-y = – 1
(iii) Co-efficient of 2x2y is = 2
(iv) Co-efficient of – 8xy3 is = – 8
(v) Co-efficient of Ipy2 is = 3
(vi) Co-efficient of – 9a2b3 is = – 9

Question 8.
In -5x3y2z4; write the coefficient of:
(i) z2
(ii) y2
(iii) yz2
(iv) x3y
(v) -xy2
(vi) -5xy2z
Also, write the degree of the given algebraic expression.

Solution:
-5x3y2z4
(i) Co-efficient of z2 is -5x3y2z2
(ii) Co-efficient of y2 is -5x3z4
(iii) Co-efficient of yz2 is -5x3yz2
(iv) Co-efficient of x3y is -5yz4
(v) Co-efficient of -xy2 is 5x2z4
(vi) Co-efficient of -5xy2z is x2z3
Degree of the given expression is 3 + 2 + 4 = 9

EXERCISE 11 (B)

Question 1.
Fill in the blanks :
(i) 8x + 5x = ………
(ii) 8x – 5x =……..
(iii) 6xy2 + 9xy2 =……..
(iv) 6xy2 – 9xy2 = ………
(v) The sum of 8a, 6a and 5b = ……..
(vi) The addition of 5, 7xy, 6 and 3xy = …………
(vii) 4a + 3b – 7a + 4b = ……….
(viii) – 15x + 13x + 8 = ………
(ix) 6x2y + 13xy2 – 4x2y + 2xy2 = ……..
(x) 16x2 – 9x2 = and 25xy2 – 17xy2=………

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 4

Question 2.
Add :
(i)- 9x, 3x and 4x
(ii) 23y2, 8y2 and – 12y2
(iii) 18pq – 15pq and 3pq

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 5

Question 3.
Simplify :
(i) 3m + 12m – 5m
(ii) 7n2 – 9n2 + 3n2
(iii) 25zy—8zy—6zy
(iv) -5ax2 + 7ax2 – 12ax2
(v) – 16am + 4mx + 4am – 15mx + 5am

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 6

Question 4.
Add : 
(i) a + i and 2a + 3b
(ii) 2x + y and 3x – 4y
(iii)- 3a + 2b and 3a + b
(iv) 4 + x, 5 – 2x and 6x

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 7

Question 5.
Find the sum of:
(i) 3x + 8y + 7z, 6y + 4z- 2x and 3y – 4x + 6z
(ii) 3a + 5b + 2c, 2a + 3b-c and a + b + c.
(iii) 4x2+ 8xy – 2y2 and 8xy – 5y2 + x2
(iv) 9x2 – 6x + 7, 5 – 4x and 6 – 3x2
(v) 5x2 – 2xy + 3y2 and – 2x2 + 5xy + 9y2
and 3x2 -xy- 4y2
(vi) a2 + b2 + 2ab, 2b2 + c2 + 2bc
and 4c2-a2 + 2ac
(vii) 9ax – 6bx + 8, 4ax + 8bx – 7
and – 6ax – 46x – 3
(viii) abc + 2 ba + 3 ac, 4ca – 4ab + 2 bca
and 2ab – 3abc – 6ac
(ix) 4a2 + 5b2 – 6ab, 3ab, 6a2 – 2b2
and 4b2 – 5 ab
(x) x2 + x – 2, 2x – 3x2 + 5 and 2x2 – 5x + 7
(xi) 4x3 + 2x2 – x + 1, 2x3 – 5x2– 3x + 6, x2 + 8 and 5x3 – 7x

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 8
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 9
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 10
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 11

Question 6.
Find the sum of:
(i) x and 3y
(ii) -2a and +5
(iii) – 4xand +7x
(iv) +4a and -7b
(v) x3+3x2y and 2y2
(vi) 11 and -by

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 12

Question 7.
The sides of a triangle are 2x + 3y, x + 5y and 7x – 2y, find its perimeter.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 13

Question 8.
The two adjacent sides of a rectangle are 6a + 96 and 8a – 46. Find its, perimeter.

Solution
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 14

Question 9.
Subtract the second expression from the first:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 15

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 16
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 17

Question 10.
Subtract:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 18

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 19
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 20

Question 11.
Subtract – 5a2 – 3a + 1 from the sum of 4a2 + 3 – 8a and 9a – 7.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 21
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 22

Question 12.
By how much does 8x3 – 6x2 + 9x – 10 exceed 4x3 + 2x2 + 7x -3 ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 23

Question 13.
What must be added to 2a3 + 5a – a2 – 6 to get a2 – a – a3 + 1 ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 24

Question 14.
What must be subtracted from a2 + b2 + lab to get – 4ab + 2b2 ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 25

Question 15.
Find the excess of 4m2 + 4n2 + 4pover m2+ 3n2 – 5p2

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 26

Question 16.
By how much is 3x3 – 2x2y + xy2 -y3 less than 4x3 – 3x2y – 7xy2 +2y3

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 27

Question 17.
Subtract the sum of 3a2 – 2a + 5 and a2 – 5a – 7 from the sum of 5a2 -9a + 3 and 2a – a2 – 1

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 28

Question 18.
The perimeter of a rectangle is 28x3+ 16x2 + 8x + 4. One of its sides is 8x2 + 4x. Find the other side

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 29
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 30

Question 19.
The perimeter of a triangle is 14a2 + 20a + 13. Two of its sides are 3a2 + 5a + 1 and a2 + 10a – 6. Find its third side.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 31.

Question 20.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 32

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 33

Question 21.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 34

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 35

Question 22.
Simplify:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 36

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 37
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 38
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 39

EXERCISE 11 (C)

Question 1.
Multiply:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 40

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 41
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 42
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 43

Question 2.
Copy and complete the following multi-plications :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 44

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 45
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 46

Question 3.
Evaluate :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 47
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 49

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 50
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 51
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 52
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 53
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 54

Question 4.
Evaluate:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 55

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 56

Question 5.
Evaluate :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 57

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 58
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 59

Question 6.
Multiply:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 60
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 61
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 62

Question 7.
Multiply:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 63

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 64
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 65

EXERCISE 11 (D)

Question 1.
Divide:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 66
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 67

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 68
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 69
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 70

Question 2.
Divide :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 71

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 72
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 73
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 74
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 75
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 76
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 77
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 78
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 79

Question 3.
The area of a rectangle is 6x2– 4xy – 10y2 square unit and its length is 2x + 2y unit. Find its breadth

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 80
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 81

Question 4.
The area of a rectangular field is 25x2 + 20xy + 3y2 square unit. If its length is 5x + 3y unit, find its breadth, Hence find its perimeter.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 83

Question 5.
Divide:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 84

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 85
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 86

EXERCISE 11 (E)

Simplify
Question 1.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 87

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 88

Question 2.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 89

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 90

Question 3.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 91

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 92

Question 4.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 93

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 94

Question 5.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 95

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 96

Question 6.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 97

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 98

Question 7.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 99

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 100

Question 8.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 101

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 102

Question 9.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 104

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 105

Question 10.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 106

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 107

Question 11.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 108

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 109

Question 12.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 110

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 111
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 112

Question 13.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 113

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 114

Question 14.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 115

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 116

Question 15.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 118

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 119

Question 16.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 120

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 121

Question 17.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 122

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 123

Question 18.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 124

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 125

Question 19.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 126

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 127

Question 20.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 128

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 129

Question 21.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 130

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 131

Question 22.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 132

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 133

Question 23.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 134

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 135

Question 24.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 136

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 137

Question 25.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 138

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 139

Question 26.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 140

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 141
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 142

EXERCISE 11 (F)

Enclose the given terms in brackets as required :

Question 1.
 x – y – z = x-{…….)

Solution:
x – y – z = x – (y + z)

Question 2.
x2 – xy2 – 2xy – y2 = x2 – (…….. )

Solution:
x– xy– 2xy – y2
= x2 – (xy2 + 2xy + y2)

Question 3.
4a – 9 + 2b – 6 = 4a – (…….. )

Solution:
4a – 9 + 2b – 6
= 4a – (9 – 2b + 6)

Question 4.
x2 -y2 + z2 + 3x – 2y = x2 – (…….. )

Solution:
x2 – y2 + z2 + 3x – 2y
= x2 – (y2 – z2 – 3x + 2y)

Question 5.
– 2a2 + 4ab – 6a2b2 + 8ab2 = – 2a (……… )

Solution:
 – 2a2 + 4ab – 6a2b2 + 8ab2
= – 2a (a – 2b + 3ab2 – 4b2)

Simplify :

Question 6.
2x – (x + 2y- z)

Solution:
2x-(x + 2y-z) = 2x – x – 2y + z
= x – 2y + z

Question 7.
p + q – (p – q) + (2p – 3q)

Solution:
p + q – (p – q) + (2p- 3q)
= p + q – p + q + 2p – 3q = 2p – q

Question 8.
9x – (-4x + 5)

Solution:
9x – (-4x + 5) = 9x + 4x – 5
= 13x- 5

Question 9.
6a – (- 5a – 8b) + (3a + b)

Solution:
6a – (- 5a – 8b) + (3a + b)
= 6a + 5a + 8b + 3a + b
= 6a + 5a + 3a + 8b + b
= 14a + 9b

 Question 10.
(p – 2q) – (3q – r)

Solution:
(p-2q) – (3q – r) =p – 2q – 3q + r =p – 5q + r

Question 11.
9a (2b – 3a + 7c)

Solution:
9a (2b – 3a + 7c)
= 18ab – 27a2 + 63ca

Question 12.
-5m (-2m + 3n – 7p)

Solution:
-5m (-2m + 3n- 7p)
= – 5m x (-2m) + (-5m) (3n) – (-5m) (7p)
= 10m2 – 15mn + 35 mp.

Question 13.
-2x (x + y) + x2

Solution:
– 2x (x + y) + x2
= -2x x x + (-2x)y + x2
= – 2x2 – 2xy + x2
= – 2x2 + x2 – 2xy = – x2 – 2xy

Question 14.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 143

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 144

Question 15.
8 (2a + 3b – c) – 10 (a + 2b + 3c)

Solution:
8 (2a + 3b -c)- 10 (a + 2b + 3c)
= 16a + 24b – 8c – 10a – 20b- 30c
= 16a – 10a + 24b – 20b – 8c – 30c
= 6a + 4b – 38c

Question 16.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 145

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 146

Question 17.
5 x (2x + 3y) – 2x (x – 9y)

Solution:
5x (2x + 3y) – 2x (x – 9y)
= 10x2 + 15xy – 2x2 + 18xy
= 10x– 2x2+ 15xy+ 18xy
= 8x2 + 33 xy

Question 18.
a + (b + c – d)

Solution:
a + (b + c – d) = a + (b + c – d)
= a + b + c – d

Question 19.
5 – 8x – 6 – x

Solution:
5 – 8x – 6 – x
= 5 – 6 –  8x – x
= -1 -7x

Question 20.
2a + (6- \(\overline { a-b }\) )

Solution:
2a + (6 – \(\overline { a-b }\) )
= 2a + (b – a + b)
= 2a + b – a + b
= a + 2b

Question 21.
3x + [4x – (6x – 3)]

Solution:
3x + [4x – (6x – 3)]
= 3x + [4x – 6x + 3]
= 3x + 4x – 6x + 3
= 3x + 4x – 6x + 3
= 7x – 6x + 3= x + 3

Question 22.
5b – {6a + (8 – b – a)}

Solution:
5b- {6a + 8- 6-a}
= 5b – 6a – 8 + b + a
= -6a + a + 5b +b – 8
= -5a + 6b-8

Question 23.
2x-[5y- (3x -y) + x]

Solution:
2x – [5y- (3x – y) + x]
= 2x – {5y – 3x +y + x}
= 2x – 5y + 3x -y – x
= 2x + 3x – x – 5y – y
= 4x – 6y

Question 24.
6a – 3 (a + b – 2)

Solution:
6a – 3 (a + b – 2)
=
6a – 3a – 3b + 6
= 3a -3b + 6

Question 25.
8 [m + 2n-p – 7 (2m -n + 3p)]

Solution:
8 [m + 2n-p -1 (2m – n + 3p)]
8 [m + 2n-p- 14m + 7n-21p]
= 8m+ 16n -8p- 112m + 56n – 168p
= 8m – 112m + 16n + 56n -8p – 168p
= -104m + 72n – 176p

Question 26.
{9 – (4p – 6q)} – {3q – (5p – 10)}

Solution:
{9 – {4p – 6q)} – {3q – (5p – 10)}
{9 – 4p + 6q} – {3q -5p+ 10}
= 9 – 4p + 6q – 3q + 5p – 10
= 9 – 4p +
5p + 6q – 3q – 10
= p + 3q – 1

Question 27.
2 [a – 3 {a + 5 {a – 2) + 7}]

Solution:
2 [a – 3 {a + 5 {a – 2) + 7}]
= 2 [a- 3 {a + 5a- 10 + 7}]
= 2 [a -3a- 15a + 30 -21]
= 2a-6a- 30a + 60-42
= 2a- 36a + 60-42
= -34a + 18

Question 28.
5a – [6a – {9a – (10a – \(\overline { 4a-3a }\)  )}]

Solution:
5a – [6a – {9a – (10a – 4a + 3a)}]
= 5a – [6a – {9a – (10a – 4a + 3a)}]
= 5a – [6a – {9a – 10a + 4a – 3a}]
= 5a- [6a – 9a + 10a – 4a + 3a]
= 5a – 6a + 9a – 10a + 4a – 3a
= 5a + 9a + 4a – 6a – 10a – 3a
= 18a – 19a = – a

Question 29.
9x + 5 – [4x – {3x – 2 (4x – 3)}]

Solution:
9x + 5 – [4x – {3x – 2 (4x – 3)}]
= 9x + 5 – [4x – {3x – 8x + 6}]
= 9x + 5 – [4x – 3x + 8x – 6]
= 9x + 5-4x + 3x-8x + 6
= 9x + 3x-4x-8x + 5 + 6
= 12x- 12x+ 11 = 11

Question 30.
(x + y – z)x + (z + x – y)y – (x + y – z)z

Solution:
(x + y – z)x + (z + x -y )y – (x + y -z)z
= x+ xy – zx + yz + xy -y– zx – yz + z2
= x2 -y2 + z2 + 2xy – 2zx

Question 31.
-1 [a-3 {b -4 (a-b-8) + 4a} + 10]

Solution:
– 1 [a – 3 {b – 4(a – b – 8) + 4a} + 10]
= -1 [a-3 {b-4{a-b-8) + 4a} + 10]
= -1[a-3 {b-4a + Ab +32 + 4a} + 10]
= -1 [a-3b+ 12a- 126-96- 12a + 10]
= -a + 3b – 12a + 12b + 96 + 12a – 10
= -a-12a + 12a+ 3b+ 12b-96-10
= – a + 15b – 106

Question 32.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 148

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 149

Question 33.
10 – {4a – (7 – \(\overline { a-5 }\)) – (5a – \(\overline { 1+a }\))}

Solution:

10 – {4a – (7 – \(\overline { a-5 }\)) – (5a – \(\overline { 1+a }\))}
= 10 – {4a – (7 – a + 5) – (5a – 1 – a)}
= 10- {4a -(12 -a) -(4a- 1)}
= 10 – {4a – 12 + a- 4a + 1}
= 10 – 4a + 12 – a + 4a- 1
= 10 + 12 – 1 – 4a – a + 4a
= 21 -a

Question 34.
7a- [8a- (11a-(12a- \(\overline { 6a-5a }\))}]

Solution:
7a – [8a – {1 la – (12a \(\overline { 6a-5a }\))}]
= 7a-[8a-{11a-(12a-6a + 5a)}]
= 7a -[8a -{11a -(17a -6a)}]
= 7a- [8a- {11a-(11a)}]
= 7a- [8a- {11a- 11a}]
= 7a – 8a = -a

Question 35.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 150

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 151

Question 36.
x-(3y- \(\overline { 4z-3x }\) +2z- \(\overline { 5y-7x }\))

Solution:
x-(3y- \(\overline { 4z-3x }\) +2z- \(\overline { 5y-7x }\))
= x – (3y – 4z + 3x  + 2z -5y + 7x)
= x-(-2y-2z+10x)
= x + 2y + 2z- 10x
= -9x + 2y + 2z