Selina Class 6 ICSE Solutions

Selina Concise Physics Class 6 ICSE Solutions – Force

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Selina Concise Physics Class 6 ICSE Solutions – Force

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Physics. You can download the Selina Concise Physics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Physics for Class 6 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 6 Physics Chapter 3 Force

  • Synposis
  • A body not changing position with time with respect to a nearby fixed object is said to be at rest.
  • A body changing position with time with respect to a nearby fixed object is said to be in motion.
  •  Force is a push or pull which can change the state of rest or motion of the body or can change the size and shape of the body (i.e. it can deform a body).
  • A force applied on a body can
    (a) move it, if it is not in motion
    (b) stop it, if it is moving
    (c) increase or decrease its speed
    (d) change its direction of motion
    (e) change its shape or size if it is not free to move.
  • Force is defined as that cause which changes the state of rest or the state of motion of a body and can also deform it.
  • A force has both the magnitude and the direction.
  • When two forces act in opposite directions, the net force is equal to the difference of these forces, in the direction of the bigger force.
  • When two forces act on a body which are equal in magnitude but opposite in direction, the net force on the body is zero.
  • Forces are of two types :
    1. contact forces and
    2. non-contact forces (or forces at a distance).
  • Contact forces are
    1. the muscular force applied as push or pull,
    2. force of friction
    3. the force of reaction normal to the surface and
    4. the force of tension in a string pulled by a load.
  • Non-contact forces are
    1. gravitational force
    2. electrostatic force and
    3. magnetic force
  • The weight of a body is the force with which the earth pulls the body.
  • The unit of weight (or force) is kgf not kg which is unit of mass.
  • Friction is a force that opposes the motion.
  • Friction always acts in a direction opposite to the direction of motion.
  • The cause of friction is the interlocking of the irregular projections on the two surfaces in contact. ‘
  • The force of friction depends on:
    (a) the smoothness (or roughness) of the surfaces in contact, and
    (b) the weight of the sliding (or rolling) body.
  • The force of friction does not depend on the area of the surfaces in contact.
  • The disadvantages of friction are :
    (a) Friction opposes the motion
    (b) Friction produces heat
    (c) Friction causes wear and tear
    (d) Friction reduces efficiency
  • Friction can be reduced by
    (a) making the surfaces smooth
    (b) the use of lubricants
    (c) the use of ball bearings
    (d) streamlining the shape of the moving body.
  • The maximum force exerted by a surface on a body so long as it remains stationary is called the force of static friction.
  • The minimum force required to keep the body moving over a surface such that it moves equal distances in equal intervals of time is called the force of sliding friction.
  • The minimum force required to roll a body on a surface is called the force of rolling friction.
  • Rolling friction is less than the sliding friction and sliding friction is less than the static friction.
  • Friction is advantageous to us in almost all activities of our life.

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) The frictional force acts in the direction of motion of body
Answer. False

(b) The unit of weight is kilogram
Answer. False

(c) A force can change the direction of motion of a moving body.
Answer. True

(d) A force increases the mass of the body when applied on it.
Answer. False

(e) The force of friction is always disadvantageous.
Answer. False

(f) The sliding friction is more than the rolling friction.
Answer. True

(g) Liquids offer more friction than the gases.
Answer. True

(h) A wet oily road offers more friction than a dry rough road.
Answer. False

2. Fill in the blanks

(a) Force is applied as push or pull.
(b) On squeezing a gum tube, its shape changes.
(c) On pulling a string, its length increases.
(d) A moving football when kicked, its direction of motion changes.
(e) On applying brakes on a moving car, its speed slows down.
(f) We use ball bearings to reduce the friction.
(g) Friction opposes the motion.
(h) Lubricants are used to reduce friction.
(i) Friction causes wear and tear of moving parts of machine.

3. Match the following columns

Selina Concise Physics Class 6 ICSE Solutions Chapter 3 Force 1

Answer.

Selina Concise Physics Class 6 ICSE Solutions Chapter 3 Force 2

4. Select the correct alternative 
(a) A body falls downwards because of

  1. electrical force
  2. gravitational force
  3. mechanical force
  4. magnetic force.

(b) A force does not change

  1. mass
  2. length
  3. shape
  4. state of motion.

(c) A force to be expressed correctly requires

  1. only the magnitude
  2. only the direction
  3. both the magnitude and direction
  4. none of the above.

(d) Friction

  1. promotes motion
  2. opposes motion
  3. acts in the direction of motion
  4. is always a nuisance.

(e) Friction is reduced by

  1. making the surfaces wet
  2. making the surfaces dry
  3. making the surfaces rough
  4. sprinkling sand on the surface.

(f) Friction

  1. causes wear and tear
  2. produces heat
  3. stops a moving body
  4. has all the above disadvantages

(g) Friction is increased if

  1. an oil is sprayed
  2. the surfaces are made wet
  3. the surfaces are made dry
  4. the surfaces are polished

B. Short/Long Answer Questions

Question 1.
Name the term used for the push or pull ?
Answer:
Force

Question 2.
Give one example each of a force as

  1. a push
  2. as pull
  3. a stretch and
  4. a squeeze.

Answer:

  1. a push — To1 open a door, we push it.
  2. as pull — To move a grass roller on a lawn, it is pulled by a gardener.
  3. a stretch — Stretching a rubber string.
  4. a squeeze —Change in shape of a sponge on squeezing.

Question 3.
Explain the meaning of the term force.
Answer:
Force: Force is a physical cause that changes or may tend to change the state of rest or the state of motion of an object. The S.I. unit of force is Newton.

Question 4.
What effect can a force have on a stationary body ?
Answer:
When a force is applied on a stationary body, it begins to move.

Question 5.
What effects can a force have on a moving body ?
Answer:
When a force is applied on a moving body, it can be made to stop or it can change the direction of motion.

Question 6.
What effect can a force produce on a body which is not allowed to move ?
Answer:
When a force is applied on a body which is not free to move, it gets deformed i. e., the shape or size of the body changes.

Question 7.
Give one example each to indicate that the application of a force

  1. produces motion
  2. stops motion
  3. slows down motion
  4. changes the direction of motion
  5. deforms a body

Answer:

  1. A car originally at rest when pushed, begins to move.
  2. A moving bicycle is stopped by applying the brakes.
  3. The speed of a moving vehicle is slowed down by applying the brakes.
  4. A player kicks a moving football to change its direction of motion.
  5. On stretching a rubber string, its length increases.

Question 8.
State the effect produced by a force in the following cases :
(a) The sling of a rubber catapult is stretched
(b) A man pushes a heavy cart
(c) A player uses his stick to deflect the ball .
(d) A cyclist applies brakes
(e) A spring is compressed.
Answer:
(a) The shape and size of catapult changes i.e., its length increases.
(b) The heavy cart begins to move.
(c) The direction of the ball changes.
(d) The speed of the moving cycle is slowed down.
(e) There is change in size and shape of spring.

Question 9.
Name the two kinds of forces in nature.
Answer:
Two kinds of forces in nature are :

  1. Contact forces
  2. Non contact forces

Question 10.
Name the type of force which acts in the following cases:
Answer:
(a) A coolie lifts a luggage
Answer. Muscular force
(b) A bicycle comes to rest slowly when the cyclist stops pedalling
Answer. Frictional force
(c) A stone falls from a roof
Answer. Gravitational force.
(d) A comb rubbed with silk attracts the bits of paper
Answer. Electrostatic force
(e) A string hangs with a load
Answer. Force of tension.
(f) A horse moves a cart
Answer. Muscular force
(g) A magnet attracts an iron pin
Answer. Magnetic force
(h) A boy opens the door
Answer. Muscular force
(i) An apple falls from a tree
Answer.Gravitational force
(j) A man rows a boat.
Answer. Muscular force.

Question 11.
What do you mean by the gravitational force ? Give an example to illustrate it.
Answer:
The force of attraction on a body by earth is called gravitational force.
Example : The leaves and fruits fall from a tree downwards towards the ground, water in a river flows down streams, a ball thrown up goes to a height and then returns back on ground are some examples of motion due to gravitational force.

Question 12.
Define the term “weight of a body”
Answer:
Weight: The weight of the body is the force with which the earth attracts it towards the centre. It depends on acceleration due to gravity.

Question 13.
What do you understand by the term friction ?
Answer:
Friction: Friction is that force which opposes the relative motion between the two surfaces that are in contact with each other.

Question 14.
Give an example to illustrate the existence of force of friction.
Answer:
If we stop paddling our bicycle, it gradually slows down and ultimately it stops after travelling a certain distance. This is due to frictional force between bicycle and ground.

Question 15.
What is the cause of friction ?
Answer:
The cause of friction is the interlocking of the irregular projections on the two surfaces in contact.

Question 16.
State two factors which directly affect the force of friction.
Answer:
Two factors which directly affect the force of friction are :

  1. The smoothness of the surface.
  2. The presence of solid, liquid or gas around the moving body.

Question 17.
In which case will there be more friction between the truck and the road : when the truck is empty or when it is loaded ?
Answer:
When the truck is loaded there will be more friction between the truck and the road.

Question 18.
Which offers more friction on a body : a glass surface or a wooden surface ?
Answer:
Wooden surface offers move friction on a body.

Question 19.
Name the three kinds of friction.
Answer:
Friction is of three kinds :

  1. Static friction
  2. Sliding friction,
  3. Rolling friction

Question 20.
List three disadvantages of friction.
Answer:
Disadvantages of friction:

  1. Friction produces heat which damages the moving parts of a machine.
  2. Friction produces wear and tear on the contacting surfaces. This reduces the life of machine parts, tyres and shoe soles.
  3. A lot of energy is wasted due to friction to overcome it before moving.

Question 21.
When you apply the brakes, the bicycle stops and the rim of the wheel becomes hot. Explain the reason.
Answer:
It is due to friction between the brakes and the rim of the wheel that it becomes hot.

Question 22.
The eraser gets smaller and smaller as you use it more and more. Explain the reason.
Answer:
The eraser gets smaller and smaller as we use it more and more due to frictional force causing wear and tear of the eraser.

Question 23.
List three ways of reducing friction.
Answer:
Ways to reduce friction:

  1. Providing ball bearings or wheels between the moving parts of machine or vehicles reduce friction and allow smooth movement as rolling friction is less than sliding friction.
  2. Oiling or lubricating (with graphite or grease) the moving parts of a machine reduces friction. Fine powder like talcum powder also works as a lubricant to reduce friction.
  3. Polishing the rough surface reduces friction offered by it.
  4. Streamlining (giving special shape to experience minimum drag) the bodies of aeroplanes, cars, boats and ships help reduce drag (fluid friction) while travelling through air or water.

Question 24.
It is difficult to open an inkpot with greasy or oily hands. Explain.
Answer:
When the hands are oily, then the oil acts as lubricant and reduces the friction.
As the friction force is less, it is difficult to get grip of the inkpot and it becomes difficult to open it.

Question 25.
It is difficult to walk on a wet road. Explain.
Answer:
When the road becomes wet after rain, friction is reduced and hence, the road becomes slippery.

Question 26.
Give three examples to illustrate that friction is a necessary evil.
Answer:
The examples to illustrate that friction is a necessary evil are:

  1. If friction were absent, we would not be able to walk.
  2. Friction is necessary to burn a matchstick.
  3. It is due to friction that we can write on a board by a chalk.

Question 27.
Define

  1. static friction
  2. sliding friction and
  3. rolling friction

Answer:

  1. The maximum force exerted by a surface on a body so long as it remains stationary is called the force of static friction.
  2. The minimum force required to keep the body moving over a surface such that it moves equal distances in equal intervals of time is called the force of sliding friction..
  3. The minimum force required to roll a body on a surface is called the force of rolling friction.

Question 28.
Arrange the following in descending order :

  1. static friction
  2. sliding friction and
  3. rolling friction?

Answer:
Static friction > Sliding friction > Rolling friction.

Question 29.
A body needs a force F1 just to start motion on a surface, a force F2 to continue its motion and a force F3 to roll on the surface. What is

  1. the static friction
  2. sliding friction and
  3. rolling friction ? State whether F2 is equal, less than or greater than (1) F1 and (2) F3.

Answer:

  1. F1 = Static friction
  2. F2 = Sliding friction
  3. F3 = Rolling friction
    F1 > F2 > F3
    F2 is less than F1 but greater than F3.

Selina Concise Physics Class 6 ICSE Solutions – Light

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Selina Concise Physics Class 6 ICSE Solutions – Light

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Physics. You can download the Selina Concise Physics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Physics for Class 6 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 6 Physics ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 6 Physics Chapter 5 Light

  • Synposis
  • Light itself is not visible, but in the presence of light other objects become visible A.
  • Light is defined as the external physical cause that affects the eye to produce the sensation of vision.
  • Two types of sources of light are:
    (1) Natural sources such as sun, stars andjugnu.
    (2) Artificial sources such as fire, electric lamp, electric tube light, a burning candle, a kerosene lamps heated bodies, etc.
  • The bodies which themselves emit light are called luminous bodies. Examples: torch, electric lamps electric tube light, burning candle, kerosene lamp, sun, stars. Jugnu etc.
  • The bodies which do not emit light by their own, but they become visible because of the light falling on them from a luminous body, are called non-luminous bodies. Examples: moon, earth, table, book, chair etc.
  • A medium which allows the passage of light through it easily, is called a transparent medium. Examples: glass, air, water etc.
  • A medium which allows only a small amount oflight to pass through it, is called a translucent medium. Examples: ground glass, tracing paper etc.
  • A medium which does not allow anydight to pass tough it, is called an opaque medium. Examples: wood, metals etc.
  • Light travels in a straight line path. This is called the rectilinear propagation oflight.
  • The pin hole camera is a simple application of the rectilinear propa – gation oflight.
  • The image (or picture) formed in a pin hole camera is upside down (i.e. inverted). On increasing the distance of screen from the pin hole, the size of image increases.
  • The shadow of an opaque object is the dark patch obtained on the screen when that opaque object is placed in the path of light.
  • Shadow is formed because light travels in a straight line path.
  • The shadow is similar to the shape of the object.
  • The part of the shadow where no light reaches from the source is completely dark and is called the umbra.
  • The part of the shadow where light reaches from only a portion of the source is partially dark and is called the penumbra.
  • There is only umbra in the shadow of an opaque object due to a point source. The umbra is bigger in size than that of the obj ect. The umbra increases in size if the screen is moved away from the object.
  • The shadow of an object due to a light source smaller than the object contains an umbra surrounded by a penumbra. The umbra is bigger in size than that of the object. Both the umbra and penum¬bra increase in size as the screen is moved away from the source.
  • The shadow of an opaque object due to a light source bigger than the object contains an umbra (which is much smaller in size than the object) surrounded by a penumbra. The umbra diminishes while the penumbra increases in size if the screen is moved away from the object.
  • Lunar and solar eclipses are the examples of formation of shadows in nature.
  • A lunar eclipse is caused on a certain full moon night when the earth comes in between the sun and the moon so that the earth casts its shadow on the moon.
  • A solar eclipse is caused on a certain new moon’s day when the moon comes in between the sun and the earth so that the moon casts its shadow

Test yourself

A. Short Answer Questions

1. Write true or false for each statement

(a) The moon is a natural source of light.
Answer. False

(b) The moon is self luminous.
Answer. False

(c) We can see an object through an opaque medium.
Answer. False

(d) Light passes through glass.
Answer. True

(e) Light travels in a straight line path. .
Answer. True

(f) Image formed in a pin hole camera is real.
Answer. True

(g) The image in a pin hole camera gets blurred if the hole is made bigger.
Answer. True

(h) A shadow is formed because light travels in a straight line path.
Answer. True

(i) Solar eclipse occurs when the sun comes in between the earth and the moon.
Answer. False

(j) If the shadow of earth falls on the moon, the eclipse is called the lunar eclipse.
Answer. True

2. Fill in the blanks

(a) Light gives us the sensation of vision.
(b) The sun is a natural source of light.
(c) A medium through which light cannot pass is called the opaque medium.
(d) A medium which allows light to pass through it easily is called the transparent medium.
(e) Moon is a non-luminous body.
(f) Light travels in a straight line path.
(g) In a pin hole camera, the image formed is inverted and real.
(h) The darkest portion of a shadow is called the umbra.
(i) The less dark portion of a shadow is called the penumbra.
(j) Lunar eclipse occurs when the earth comes in between the moon and the sun.

3. Match the following columns
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 1
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 2

4. Select the correct alternative

(i) The natural source of light is

  1. candle flame
  2. electric lamp
  3. sun
  4. kerosene lamp

(ii) The formation of inverted image in a pin hole camera shows that

  1. light enables us to see
  2. light travels in a straight line path
  3. light can pass through the pin hole
  4. light does not pass through the pin hole

(iii) The luminous body is

  1. a lighted bulb
  2. earth
  3. noon
  4. table

(iv) Umbra is a region of

  1. complete darkness
  2. partial darkness
  3. complete brightness
  4. partial brightness

(v) Penumbra is a region of

  1. complete darkness
  2. complete brightness
  3. partial brightness
  4. none of the above

(vi) Solar eclipse occurs on

  1. every new moon’s day
  2. certain new moon’s day
  3. every full moon’s day
  4. certain full moon’s day

(vii) Lunar eclipse occurs on

  1. every full moon’s night
  2. certain full moon’s night
  3. every new moon’s day
  4. certain new moon’s day

B. Short/Long Answer Questions

Question 1.
What is light ? Define it.
Answer:
Light is a form of Energy i. e. The external physical cause that affects our eye to produce the sensation of vision.

Question 2.
How does light make an object visible ?
Answer:
An object becomes visible to us when the light after striking the object reaches our eyes. Light itself is not visible, but light makes objects visible to us.

Question 3.
Name two natural sources of light.
Answer:
Sun, stars,jugnu, firefly.

Question 4.
List two artificial sources of light.
Answer:
Electric bulb, torch, an oil lamp, fluorescent tube, candle.

Question 5.
Differentiate between the luminous and non-luminous bodies. Give two examples of each.
Answer:
Difference Between
Luminous
The bodies which have light of their own e.g. sun stars, bulb, candle, oil lamp, torch, a lantern.
Non-Luminous
The bodies-which do not have their own Iight.e.g. moon, chair, table. When light falls on them, they become visible.

Question 6.
Is the moon a luminous object ?
Answer:
Moon is not a luminous body, it is nbn-luminous body. It has no light of its own.

Question 7.
What do we call a body that shmes on its own ?
Answer:
The bodies that shines onits ownor whichthemselves emittheir own light are called the luminous bodies.

Question 8.
What do we call an electric bulb producing light ?
Answer:
Luminous object.

Question 9.
What is a transparent medium ? Give two examples.
Answer:
Amedium which allows the passage of light through it easily,is called a transparent medium.
Examples: glass, air, water etc.

Question 10.
Explain the difference between a transparent, a translucent and an opaque medium. Give two examples of each.
Answer:

  1. Transparent objects — Those objects through which light can pass easily are called transparent obj ects. e.g. Water, glass, air.
  2. Translucent object— The ohj ect through which light can pass partially are called translucent object, e.g. tracing paper, waxed paper.
  3. Opaque object— The objects which do not allow the light to pass through are called opaque objects, e.g. wood.

Question 11.
What do we call a substance through which we cannot see light ? Give an example of such a substance.
Answer:
A substance through which we cannot see light is called an opaque medium.
Examples:  Wood, metals, butter paper and black paper etc.

Question 12.
What do we call a substance through which light passes ? Give an example of such a substance.
Answer:
A substance through which light passes is called a transparent substance.
Examples: glass, air, water etc.

Question 13.
Can a transparent medium form an image ? Explain your answer.
Answer:
No, a transparent medium cannot form image. All the light that passes through a transparent medium completely pass through the substance. For the formation of image it is must that the light rays get reflected through the surface.

Question 14.
How can you obtain a point source of light ?
Answer:
A point source of light is obtained either by placing a screen having a fine hole, in front of die luminous body or by placing the luminous body inside a box having a fine hole on one of its side.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 3

Question 15.
Define the terms : a ray of light and a beam of light.
Answer:
The light travelling in any one direction in a straight line is called a ray of light.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 4
A group of light rays given out from a source is called a beam of light
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 5

Question 16.
What do you mean by ‘rectilinear propagation of light’ ?
Answer:
Light travels in a straight line path. This is called the rectilinear propagation of light.

Question 17.
Describe an experiment to show that light travels in a straight line path.
Answer:
Take three cardboards A, B and C each about 25 cm square. Take a pin and make a small hole in each cardboard at the same height. Suspend the cardboard pieces by separate threads vertically from a support such that each hole is at the same height, as shown. Pass a string through the holes and pull it taut. This makes the three holes in a straight line. Now take out the string.
Place a lighted candle near one of the cardboards (say A). Look at the candle flame from the other side of the cardboard C. The candle flame is clearly seen.
Now slightly displace one of the cardboards (say B) so that the holes no longer remain in a straight line. Again look at the candle flame from the other side of the cardboard C. You do not see the candle flame. The reason is that light travels in a straight line and now the holes in the cardboards A, B and C are not in a straight line.
Conclusion: Light travels in a straight line path called the rectilinear propagation of light.

Question 18.
In which of the following two arrangements (a) and (b) shown in the diagram, you can see the light of the bulb ? Explain Your answer
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 6
Answer:
We can see the light of the bulb in the arrangement (a).
This is so because in arrangement ‘a’ the rod is straight and light travels in a straight line path.
Whereas in arrangement ‘b’ the rod is bent. So the light cannot pass through it.

Question 19.
Name a simple application of the rectilinear propagation of light
Answer:
The simple applications of rectilinear propagation of light are pin hole camera, formation of shadows and elipses.

Question 20.
What is a pin hole camera ? Draw a neat and labelled diagram to show the formation of image of a lighted candle by it.
Answer:
The pin hole camera is a simple application of the rectilinear propagation of light.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 7

Question 21.
Explain the formation of image of a luminous object in a pin hole camera with the aid of a neat diagram.
Answer:
When a luminous object AB, such as a lighted candle, is placed in front of the pin hole, an inverted picture A’ B’ of the candle is
obtained on the tracing paper. This picture A’ B’ is called the image. The image obtained is upside down (i. e. inverted). The reason is that the light travels in a straight line path. Hence light from the upper point A of the candle passes through the pin hole and strikes
the tracing paper at A’. Similarly, light from the lower point B of the candle passes through the pin hole and strikes the tracing paper
(or screen) at B’. Light from all the other points between A and B, on passing through the pin hole strikes the tracing paper in between
A’ and B’. As a result, an inverted image of the candle is seen on the tracing paper Fig shows the simple ray diagram for die formation of image.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 8

Question 22.
State two factors which affect the size of image formed in a pin hole camera.
Answer:
Factors affecting the size of the image :
The size of image depends on the following two factors:

  1. The distance of screen (i.e. tracing paper) from the pin hole, and
  2. The distance of obj ect (i.e. candle) in front of the pin hole.

Question 23.
Is the image obtained in a pin hole camera erect or inverted ? Give reason for your answer.
Answer:
Image obtained in a pin hole camera is inverted.
The reason is that the light travels in a straight line path. Hence light from the upper point of the candle passes through the pin hole and strikes the tracing paper in the lower point. Similarly light from the lower point of the candle passes through the pin hole and strikes the tracing paper at the upper point.

Question 24.
How is the image affected in a pin hole camera when another fine hole is made near the first pin hole ?
Answer:
If another pin hole is made near the first pin hole, two images are formed on the screen, one due to each of the two pin holes. If the holes are very close, the two images tend to overlap each other. As a result, a blurred image will be seen.

Question 25.
State the effect on the image in a pin hole camera if

  1. The hole is made bigger.
  2. The luminous object is moved towards the pin hole.
  3. The length of the pin hole camera is increased (le. the screen is moved away from the pin hole).

Answer:

  1. If the hole is bigger than a pin hole, again a blurred image in seen. The reason is that a bigger hole is equivalent to a large number of pinholes. Each pin hole produces one image. These images overlap each other resulting in a blurred image.
  2. If the object is moved towards the pin hole the size of the image increases.
    The size of image / The size of object = Distance of screen from pin hole / Distance of object from pin hole
  3. When the length of the pin hole camera is increased. C is the screen is moved away from the pinhole, the size of image also increases.

Question 26.
What is a shadow ? Give a reason for its formation.
Answer:
Shadow : When light falls on an opaque object, light is obstructed and a dark patch on a screen kept behind is called shadow. This is because light propagates in straight line. If distance between obj ect and screen is less, the shad o w will be (umbra) dark and smaller.
If the distance is increased shadow will be dim and larger.

Question 27.
Draw a ray diagram to show the formation of shadow of an opaque object by a point source of light. How is the size of shadow affected if the screen is moved away from the object?
Answer:
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 9
If we move the screen away from the object, the shadow increases in size.

Question 28.
State two differences between an umbra and a penumbra.
Answer:
Umbra

  1. It is the portion of shadow where no light reaches from the source of light due to the opaque object.
  2. It is completely dark.

Penumbra
It is the portion of shadow where a portion of light from the source of light reaches the shadow even in the presence of the opaque object in between them.
It is not completely dark, but is dim (or less bright).

Question 29.
Draw a ray diagram to show the formation of umbra alone.
Answer:
Formation of umbra alone.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 10

Question 30.
Draw a ray diagram to show the formation of umbra and penumbra both. Label the parts umbra and penumbra in your diagram.
Answer:
If your move the screen away from the object, the shadow increases in size. Formation of umbra and penumbra both.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 11

Question 31.
In each of the following diagrams, draw rays to form umbra and penumbra on the screen.
(a)

Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 12
(b)

Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 13
Answer:
(a)A’B’—umbra
Umbra alone is obtained on the screen when the opaque object is illuminated by a point source of light.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 14
(b)
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 15
EF is Penumbra CD is umbra.

Question 32.
State the conditions when only the penumbra of an opaque object is obtained on the screen.
Answer:
If the size of source of light is bigger than the size of the opaque object, the size of umbra is very small. If the screen is moved away from the object, the umbra vanishes and only the penumbra remains.

Question 33.
Why is it that the birds flying in the sky do not cast their shadow on the earth ?
Answer:
We do not see the shadow of a bird flying high up in air because in their shadow, the umbra is absent and the penumbra is too large and too faint that it is not visible as the distance of screen (i.e. earth) is very large from the object (i.e. bird).

Question 34.
Why are shadows at noon shorter than in the morning or in the evening ?
Answer:
At noon the sun is directly overhead. So, the sun rays fall vertically on the body. Hence the shadow is very short. In the morning and evenings, the sun rays fall in an inclined position. So, the shadows are long.

Question 35.
What is an eclipse ? Name the two types of eclipses.
Answer:
Eclipses are the examples of formation of shadows in nature. There are two kinds of eclipses:

  1. Lunar eclipse (the eclipse of the moon), and
  2. Solar eclipse (the eclipse of the sun).
    Lunar eclipse is due to the formation of shadow of earth on moon and solar eclipse is due to the formation of shadow of moon on earth.

Question 36.
When does a lunar eclipse take place ? Does it occur on every full moon’s night ?
Answer:
A lunar eclipse occurs when the earth comes in between the sun and moon and casts its shadow on moon. On a full moon night, the moon rises in the east after sun sets in the west. On such a night, the sun and moon are on the opposite sides of the earth. The shadow of the earth falls on the surface of the moon therefore moon is not visible to us. This is lunar eclipse as shown in the figure.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 16

Question 37.
Draw a diagram to show the formation of lunar eclipse.
Answer:
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 17

Question 38.
When does a solar eclipse take place ? Does it occur on even’ new moon’s day ?
Answer:
Solar eclipse— On a certain moon’s day the moon, happens to come in between the sun and the earth. They come in a straight line. In such a situation, the moon being smaller in size casts its shadow only on a limited region on the earth. In these regions of the earth, the solar eclipse occurs.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 18

Question 39.
Draw a diagram to show the formation of solar eclipse.
Answer:
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 19

Question 40.
What is annular solar eclipse ? Draw a labelled diagram to show its formation.
Answer:
An annular solar eclipse occurs when only the tip of the umbra of the moon falls on the earth. From the point D, the sun will appear to be completely obstructed by the moon, only the outer rim of the sun, called corona, is then visible for a very short time which is known as the diamond ring. The formati on of annular solar eclipse is shown below.
Selina Concise Physics Class 6 ICSE Solutions Chapter 5 Light 20

Selina Concise Chemistry Class 6 ICSE Solutions – Water

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Selina Concise Chemistry Class 6 ICSE Solutions – Water

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Chemistry for Class 6 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 6 Chemistry ICSE SolutionsPhysicsBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 6 Chemistry Chapter  7  Water

POINTS TO REMEMBER

  1. Water cannot be replanished and without water we cannot live.
  2. 4 / 5 th of earth’s surface is covered with water and 1% of this available water is in the form of rivers, lakes, ponds and under ground water. It is utilized for agriculture and living beings consumption.
  3. In the atmosphere water is present in the form of vapour, mist, clouds etc.
    Selina Concise Chemistry Class 6 ICSE Solutions Chapter 7 Water 1
  4. Distillation : The process of removal of the dissolved salts from sea/ocean water.
  5. Purest form of water is rain water as it is formed by evaporation and condensation.
  6. Sea water is unfit for agriculture as plants do not tolerate saline water.
  7. Change of water from water bodies (lakes, rivers and oceans) into vapours in air and condensing of vapours and fall in the form of snow, water is called water cycle and water cycle plays an important role in
    (i) restoring the lost water from earth’s surface
    (ii) in controlling the climatic conditions.
  8. Water vapour is one of the invisible gases which makes up the air we breathe.
  9. Drinking water may contain some minerals and salts but it must be free from suspended impurities, harmful bacteria and germs.
  10. Sometimes water from a source look clear but may contain germs which cause water borne diseases like typhoid, cholera, jaundice, dysentry, gastroenteritis etc.
  11. Water can be made fit for drinking by
    (i) boiling,
    (ii) sterilisation
    (iii) exposing to air and sunlight
    (iv) by chemical treatment like chlorination or ozonisation.
  12. Water is universal solvent i.e. water can dissolve nearly in every substance.
  13. Distilled water is the purest form of water but not good for drinking purposes. It is good for medicinal purposes, laboratories and batteries.
  14. Both stirring and heating help in dissolving the substance in water.
  15. “The amount of solute that dissolves in a given quantity of solvent at a given temperature is called solubility of the solute.”
  16. Air dissolved in water helps aquatic animals and plants to survive.
  17. Eutrophication : “The deficiency of oxygen in the water may cause death of aquatic animals. This loss of dissolved oxygen from water in water bodies is called eutrophication.”

EXERCISE-I

Question 1.
Name:

(a) Three major sources of natural water
(b) Four sources of surface water.
(c) Two underground sources of water.

Answer:

(a) Three major sources of natural water :

  1. Surface water (sea water)
  2. Above surface water (rain)
  3. Underground water (springs)

(b) Four sources of surface water sources :

  1. Ocean water
  2. River water
  3. Lake water
  4. Glaciers

(c) Two sources of underground water :

  1. Well water
  2. Spring water

Question 2.
Answer the following questions in short:

(a) In which form is water present in the atmosphere ?
(b) Which source of water contains the highest concentration of salt in it ?
(c) Why is rain water considered the purest form of natural water?
(d) What possible impurities does rain water contain ?
(e) What is water table ?
(f) Why is spring water pure enough for drinking but unsuitable for laboratory use ?
(g) Why the taste of spring water differs at different places ?

Answer:

(a) Water is present in the form of water vapours in the atmosphere.
(b) Sea water contains highest concentration of salt.
(c) Rain water is the purest form of water as it is distilled water i.e. water from surface of earth has evaporated and then vapours condense in the atmosphere.
(d) The impurities present in rain water are dust and dissolved gases like oxygen, nitrogen and carbon dioxide and these gases are not poisonous, rain water is safe for drinking.
(e) Level of ground water is called water table.
(f) Spring water is free from suspended impurities and germs (harmful bacteria) as water has been filtered through different layers of soil and is fit for drinking. Spring water contains dissolved impurities which are due to the nature of soil surrounding the spring and cannot be used for laboratory purposes, where distilled water free from impurities is needed.
(g) Spring water contains unsolved impurities which have entered the water from surrounding soil and soil impurities (salts) differ from place to place. Hence taste of water differs from place to place.

Question 3.
List three major impurities present in river water.
Answer:
Three major impurities present in river water :

  1. Suspended impurities like clay and sand particles.
  2. Harmful bacteria.
  3. Mineral salts.

Question 4.
Give the percentage of water in the following :

(a) Rice and wheat grains
(b) Eggs
(c) Tomatoes
(d) Bread
(e) Water melon

Answer:
Percentage of water in :

(a) Rice — 3 % to 4% Wheat grain — 3% to 4%
(b) Eggs —75%
(c) Tomatoes — 95%
(d) Bread —25%
(e) Water melon — 97%

Question 5.
What are the three states of water ?
Answer:
Three states of water are :

  1. Ice (solid)
  2. Liquid (water)
  3. Gaseous (steam)

Question 6.
Why are ice, liquid water and steam considered to have the same chemical substance ?
Answer:
Ice, liquid water and steam have same composition i.e. 2 parts hydrogen and one part oxygen by volume and same formula H2O, can be easily transformed from one state to other by heating or cooling.

Question 7.
How is a cloud formed ?
Answer:
Water from the surface of earth, lakes, seas, rivers, formed by plants (transpiration) etc. evaporates and rises high up in the atmosphere and condense into water droplets and collect to form cloud.

Question 8.
What is water cycle ? What is its importance ?
Answer:
Water cycle : The change of water from one form to another in nature which results in continuous circulation of water from earth’s surface to the atmosphere and from the atmosphere back to the earth’s surface is called water cycle.
Importance of water cycle:

  1. It assures a continuous supply of water to us.
  2. In controlling the climatic conditions all over the world.

Question 9.
How are the following formed ?

(a) fog
(b) mist
(c) dew
(d) frost

Answer:

(a) Fog : When water vapours change into tiny droplets of water near the ground, fog is formed.
(b) Mist: When tiny droplets of water remain suspended in air it is called mist.
(c) Dew : When the water vapour condenses on cold objects like grass, leaves and flowers in winter in the form of tiny droplets of water called dew.
(d) Frost: When the dew freezes it is called frost.

EXERCISE-II

Question 1.
Name:

(a) Two chemicals used to destroy germs present in water.
(b) Two diseases which spread through impure water.
(C) A chemical used for loading.
(d) Two substances which add taste to water.
(e) Two household methods to get safe drinking water.

Answer:
(a) Two chemicals used to kill germs in water are:

  1. Chlorine.
  2. Potassium permanganate.
  3. terilizing water with ozone.

(b) Two diseases are cholera, dysentry.
(c) Chemical used for loading is potash alum.
(d) Minerals and carbon dioxide gas adds taste to water.
(e) Methods to get safe drinking water:

  1. Adding chlorine tablets.
  2. By adding potassium permanganate crystals.

Question 2.
Answer in brief:

(a) Why is river water unfit for drinking?
(b) Why is tap water a mixture?
(c) What is mineral water?
(d) What is the purpose of adding bleaching powder to water supplied to the town?
(e) How is chemically pure water obtained in the laboratory?
(f) how is water in swimming pool kept free a germs?

Answer:

(a) River water contains mineral salts, suspended impurities like clay, sand particles, organic matter and bacteria and is not fit for drinking.
(b) Tap water contains, minerals, air, chlorine and other dissolved impurities that varies from place to place, therefore it is a mixture.
(c) Mineral water is pure water fit for drinking. It is collected from natural source and contains air, minerals and salts free from suspended impuiities, harmful bacteria and germs.
(d) Adding bleaching powder to water, kills germs and harmful bacteria and viruses present in water.
(e) Chemically pure water for laboratories is obtained-by distillation. Distilled water does not contain any salt or mineral
(f) Water in swimming pool kept free from infections and germs by chlorination i.e. treating water with chlorine gas.

Question 3.
Define:

(a) Sterilisation
(b) Sedimentation
(c) Loading
(d) Aeration

Answer:

(a) Sterilisation : The process of removal of microorganisms including bacterial spores from water to avoid water borne diseases is known as sterilisation.
(b) Sedimentation : The setting of suspended solid matter at the bottom of a liquid is called sedimentation.
(c) Loading : The process of adding a chemical to an impure liquid in order to increase the speed of sedimentation of suspended particles is called loading.
(d) Aeration : To kill harmful micro-organisms present in filtered water, air underpressure is blown into the filtered water. This process is called aeration.

Question 4.
What is potable water ? List four characteristics of potable water.
Answer:
Potable water : Water fit for drinking purposes is called potable water.
Four characters of drinking water :

  1. It should be transparent.
  2. Should have no colour, no odour.
  3. Should be free from harmful bacteria and germs.
  4. It should contain same salts and minerals needed by the body CO2 to add to taste.

Question 5.
Why is water important for plants and animals ?
Answer:
Importance of water for plants:

  1. Plants need water to prepare their food, for germination and growth to produce fruits, flowers etc.
  2. For conduction of food prepared by plants to other-parts of plant i.e., for translocation.
  3. Large number of plants live in water; water provides nutrients and oxygen for their survival.
    Crops need water for their growth. Water in the form of rain, washes the dust and smoke deposited on leaves by vehicles and helps the stomata in exchanging gases.

Importance of water for animals:

  1. Lot of water is lost by the body of animals by sweating, in the form of urine and evaporation while doing various activities, so to make up for the lost water animals consume a lot of water in the form of drinking.
  2. Water keeps the animals fresh and is natural medicine for their many ailments.
  3. Water is good solvent and helps in the process of digestion, blood circulation, excretion etc. in the body of organisms.
  4. Water is essential for the cleanliness of animals and their surroundings.

Question 6.
What are the three methods of removing germs from natural water. Explain.
Answer:
Three methods to remove germs :

  1. By boiling : Boiling kills germs present in water.
  2. Adding potassium permaganate: By adding potassium permaganate in the well the germs can be killed.
  3. Chlorination : After filtration water is passed through chlorination tank here chlorine kills the germs.

Question 7.
Name the steps involved in the purification of drinking water supplied in cities and towns.
Answer:
The water source for our towns and cities are river, lakes or underground water which contains suspended and dissolved impurities.
To remove these impurities steps involved are :

  1. Loading and sedimentation : to settle the suspended impurities to form a sediment for this purpose potash alum is added.
  2. Filtration : The water still contains lighter suspended impurities which are removed by filtration through sand and gravel. In cities ground water is drawn from tube well or submurcible pumps which have filters fitted in them. This clear water still contains germs.
  3. Chlorination : To kill germs water is passed into chlorination tank where it is treated with chlorine to kill germs.
    Water is now potable i.e. safer for drinking and supplied to homes.

Question 8.
What is the taste of distilled water ? Why is it not potable?
Answer:
Taste of water is tasteless i.e. flat. It is because distilled water does not contain any salt or mineral required for our body. So it is not potable water.

Question 9.
Give reasons :

(a) Ice floats on water.
(b) Marine life is able to survive in colder regions.
(c) Water droplets can be seen outside a chilled water bottle.

Answer:

(a) Water has maximum density at 4°C. Ice is lighter than water therefore it floats on water.
(b) This anomalous property of water enables aquatic plants and animals to survive in colder regions of world because even when the water of ponds, lakes, river freezes it freezes on the top but remains a liquid below ice layer.
(c) Sometimes we see water droplets on the outer surface of the glass containing ice-cold water, this is because the water vapour presents in’air, on coming in contact with the cold glass of water, loses energy and gets converted into liquid state, which we see as water droplets.

EXERCISE-III

Question 1.
Why is water called a universal solvent ?
Answer:
As water can dissolve in most of the substances, solids, liquid and gaseous. Therefore it is called universal solvent.

Question 2.
Define:

(a) Solute
(b) Solvent
(c) Solution
(d) Saturated solutions
(e) Unsaturated solutions

Answer:

(a) Solute : A solute is a substance that dissolves in a medium which can be water or any other substance. Solute is in smaller quantity in a solution.
(b) Solvent : A solvent is a medium in which a solute dissolves. It is in large quantity in a solution. Water is the most common solvent. The other solvents are alcohol, carbon tetrachloride etc.
(c) Solution : A solution is a homogeneous uniform mixture formed by a solute and a solvent.
(d) Saturated solution : When a solution cannot dissolve any more of solute at a given temperature, it is called saturated solution.
(e) Unsaturated solution : A solution that can take up more of the solute at a given temperature, is said to be an unsaturated solution.

Question 3.
State two factors by which solubility of a solute in a solvent can be increased.
Answer:
Two factors are :

  1. Stirring.
  2. Increase in temperature.

Question 4.
Why do aquatic animals die in boiled water ?
Answer:
Boiled water is deprived of oxygen i.e. there is no oxygen. For want of oxygen animals die in boiled water.

Question 5.
State three differences between water and air.
Answer:
Differences between water and air :
Water :

  1. Water is liquid and can exist in three states i.e. solid-ice, liquid-water, gas – vapours.
  2. Is a compound.
  3. H2O is its formula.

Air :

  1. Air is gaseous and can exist in liquid state when cooled under pressure.
  2. Is a mixture.
  3. Has no formula.

EXERCISE -IV

Question 1.
State four ways by which water can be conserved.
Answer:
Four ways to conserve water:

  1. Do not allow water to drip from defective taps.
  2. More dams should be built.
  3. More plantation should be done as plants help in bringing rain.
  4. Wastage of water should be avoided.
  5. Close the tap when you are brushing your teeth. Rather use a mug. Close the tap when you are washing clothes, open the tap only when you need it. Reduce, Reuse and recycle should be our mantra

Question 2.
Explain harvesting of water.
Answer:
When it rains heavily water runs into streets, drains and wasted we can not make use of it. Our purpose is to catch this rain water, store it for future use when we actually need it. For this harvesting of rain water should be done. Rain water is collected from the roofs and verandas of the buildings with the help of pipes and carried to tanks for storage and is used at the time of need when it is not raining.

Question 3.
What are the three main causes of water pollution ?
Answer:
Three main causes of water pollution :

  1. Addition of waste products from homes.
  2. Addition of waste products from agriculture.
  3. Addition of waste from industries.
  4. Addition of sewage in water bodies.

Question 4.
State the main steps to be taken to prevent water pollution.
Answer:
Steps to be taken to prevent water pollution :

  1. Trees and plants be planted along the banks of rivers and canals.
  2. Bathing and cleaning of animals near or in water sources be not allowed.
  3. Use of pesticides, insecticides, fungicides and fertilisers should be reduced.
  4. The polluted water from industries should be treated first and then discharged into water bodies.
  5. Use biodegradable detergents.
  6. Water containing sewage should be passed through sewage treatment plants first and then this water should be used for irrigation.
  7. Wells should be covered properly and washing and cleaning of clothes, utensils and animals should not be done near the well to keep them clean.

Question 5.
What are the causes of floods and drought?
Answer:
Problems cause by floods :

  1. A rise in the level of water in dams, rivers, lakes etc.
  2. Heavy rainfall also causes floods.
  3. Floods cause extensive damage to crops, property, animal and human life.
  4. Crop-fields, villages and”many low-lying areas get submerged under flood water.
  5. Rains/floods also affect smaller animals living in the soil.

Problems cause by drought :

  1. Crops may die, fodder may become scarce.
  2. Living organisms of the soil die.
  3. Animals may die, plants and trees will not survive.
  4. Soil becomes dry, water level in rivers, lakes, dams etc. may fall. The ground water-level falls.
  5. Drought displaces people from a large number of villages and towns.

Question 6.
State some of the ways in which you as an individual can conserve water.
Answer:
The ways in which we can conserve water are as follows :

  1. Use a bucket for taking it.
  2. Make sure, water does not overflow from overhead tanks of your house.
  3. Close the tap when you are brushing your teeth.
  4. Wash fruits, vegetable in a bowl of water, rather than under a running tap. Water used for washing vegetables may be used to water plants in the garden.

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks

(a) Water is a universal; solvent.
(b) Rainwater is the purest form of natural water.
(c) Sand and dust are suspended impurities in water.
(d) Sea water has high concentration of salt (impurity).
(e) Water covers nearly 4 / 5 th of the surface of the earth.
(f) Evaporation of rain water leaves NO residue.
(g) Potash alum is the chemical added to water to remove the tiny suspended particles.
(h) A solution is a uniform mixture of a solute and a solvent.
(i) Ice, water and steam have different physical states but are chemically identical.
(j) Boiling kills most of the germs in water.
(k) The elements present in the molecules of water are hydrogen and oxygen.

2. Write True or False for the following statements

(a) Water is an element.
Answer. False
Correct: Water is aL compound.

(b) Tap water does not contain dissolved impurities.
Answer. False
Correct: Tap water contain dissolved impurities.

(c) Alum is commonly used for removing suspended impurities.
Answer. False
Correct : Alum is commonly used for settling down of suspended impurities.

(d) Distillation is a good method for purifying water for town supply.
Answer. False
Correct : Distillation is a good method-for purifying water for medicinal purpose.

MULTIPLE CHOICE QUESTIONS

Tick (√) the correct alternative from the choice given for the following statements:

1. Water content in human body is

  1. 70%
  2. 75%
  3. 80%
  4. 90%

2. The purest form of natural water is

  1. seawater
  2. river water
  3. rainwater
  4. lake water

3. When the water vapour changes into tiny droplets of water near the ground, it is called

  1. mist
  2. dew
  3. fog
  4. frost

4. Water is a

  1. compound
  2. element
  3. mixture
  4. none of the above

5. Common salt is obtained from sea water by

  1. distillation
  2. crystallisation
  3. evaporation
  4. sublimation

6. Jaundice affects

  1. heart
  2. lungs
  3. liver
  4. kidney

7. Chlorination of water is done

  1. to kill the germs
  2. to remove the suspended impurities
  3. to remove the dissolved impurities
  4. none of the above

Selina Concise Biology Class 6 ICSE Solutions – Health and Hygiene

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Selina Concise Biology Class 6 ICSE Solutions – Health and Hygiene

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Biology. You can download the Selina Concise Biology ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Biology for Class 6 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 6 Biology ICSE SolutionsChemistryPhysicsMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 6 Biology Chapter 6 Health and Hygiene

Review Questions 

Multiple Choice questions:

1. Put a tick mark (✓) against the correct alternative in the following statements:

(a) Malaria is caused by:
(i) Bacteria
(ii) Protozoan
(iii) Fungi
(iv) Virus

(b) Deficiency of calcium causes
(i) Poor growth of teeth and gums
(ii) Goitre
(iii) Anaemia
(iv) Polio

(c) Hay fever and asthma are
(i) Deficiency diseases
(ii) Genetic diseases
(iii) Organic diseases
(iv) Allergy diseases

PQ. Cataract is a disease of:
(i) Ears
(ii) Nose
(iii) Eyes
(iv) Throat

(d) Infectious diseases can be prevented by:
(i) Medicines
(ii) Proper food
(iii) Immunisation
(iv) Exercise

(e) Which one of the following is a genetic disease?
(i) Scurvy
(ii) Leukemia
(lii) Goitre
(iv) Haemophilia

(f) Which one of the following is a degenerative disease?
(i) Thalassemia
(ii) Ben-ben
(iii) Cataract
(iv) Diabetes

(g) Pellagra is one disease caused by the deficiency of:
(i) Vit. B3
(ii) Vit. B1
(iii) Vit. C
(iv) Vit. D

(h) Deficiency of Iodine in one’s food can cause:
(i) Ben-ben
(ii) Goitre
(iii) Scurvy
(iv) Pellagra

(i) Which one of the following mineral deficiency diseases can be cured by eating a diet which include green leafy vegetables, banana, cereals, egg-yolk?
(i) Goitre
(ii) Anaemia
(iii) Brittle bones
(iv) Pain in muscle contraction

(j) Which one of the following vitamin deficiency diseases can be cured by eating a diet which includes carrot, yellow fruits, vegetables, butter, milk, fish ?
(i) Beri-beri
(ii) Dermatitis
(iii) Night blindness
(iv) Scurvy

(k) Which one of the following is a communicable disease?
(i) Measles
(ii) Cancer
(iii) Heart stroke
(iv) Allergy

Short Answer Questions:

Question 1(a).
What is a non-communicable disease ?
Answer:
The diseases which are caused due to improper functioning of the body organs e.g. diabetes, heart attack. They are not caused by germs and not transmitted from one to another.

Question 1(b).
What is a deficiency disease
Answer:
These disease are caused by lack of nutrients, vitamins, minerals as a anaemia, goitre.

Question 1(c).
What are communicable diseases ? How can they be avoided ?
Answer:
Those disease which spread from one person to another by the entry of microorganisms are known as communicable diseases. The disease-causing germs are called Pathogens. These disease can be avoided by proper vaccination, healthy food and hygenic surroundings.

Question 1(d).
Biting nails should be strictly avoided. Give reason.
Answer:
Nail biting may cause many disease as the dirt has many
bacteria causing diseases. Nails should be cut from time to time to save from diseases.

Question 1(e).
Regular exercise and proper rest is a must. Give reason.
Answer:
Regular exercise keeps our body strong and immune to many
diseases, rest refreshes our body.

Question 1(f).
 Children eating more of fast food tend to suffer from obesity (overweight). Comment.
Answer:
Fast food like pizza, burger, patty, oily foods etc. have much carbohydrates and fats. Children eating these become more and more fat and gain weight soon as they do not do much of physical work.

Question 1(g).
How can we control spreading of diseases by mosquitoes and houseflies ?
Answer:
We can control spreading of diseases by mosquitoes and houseflies by using repellants, throwing garbage in covered bins, avoiding stagnation of water and checking breeding of these insects.

2. Name the following:
(a) A bacterial disease caused due to contaminated water
Ans. A bacterial disease caused due to contaminated water Cholera.

(b) A disease caused due to Plasmodium
Ans. A disease caused due to Plasmodium Malaria.

(c) A disease caused due to the bite of female Culex mosquito
Ans. A disease caused due to the bite of female Culex mosquito Elephantiasis.

(d) A viral disease spread by the bite of a dog
Ans. A viral disease spread by the bite of a dog Rabies/ Hydrophobia

(e) Two diseases caused due to deficiency of protein in the diet of a child.
Ans. Kwashiorkor and marasmus.

(f) Any three water-borne disease.
Ans. Amoebiasis, Cholera, Hepatitis A.

(g) A viral disease caused due to unhealthy sexual contact
Ans. A viral disease caused due to unhealthy sexual contact AIDS.

(h) A disease caused due to choking of coronary artery
Ans. A disease caused due to choking of coronary artery Atheraosclerosis.

3. Write short (1-2 sentences) notes on the following:
Disease, immunisation, pathogen, allergy, AIDS.
Answer:
Disease: Disease is a departure from normal health due to structural or functional disorder of the body. Disease may be due to deficiency of nutrients or malfunctioning of organs or genetic disorders, improper metabolic activity, or allergies, or cancer and mental illness as diabetes, haemophilia, leukemia, schizophrenia.

Immunisation: It means, we make the body immune to certain diseases by introducing respective weakened germs into the body. Thus we develop resistance to the concerned disease this process is called immunisation. The germs or the material introduced into the body to make it resistant to the concerned disease is called vaccine. This produces antibodies in the body of the person and the person can be saved by these antibodies. The vaccine can be given by the injection or orally as polio drops, tap vaccine for typhoid, BCG vaccine for tuberculosis.

Pathogens: The germs that cause diseases to human beings and to other animals and plants are called pathogens. They spread the diseases from person to person or through the air or through the articles of the diseased persons. Pathogens may be different kinds of bacteria, viruses, fungi, protozoans or worms.

Allergy: Allergy is an unpredictable reaction to a particular substance. This type of substance is called allergen. Different people are allergic to different substances. A few common allergens are dust, spores, pollen, certain clothes, particular cosmetics, etc. The common areas of the body parts which are affected by allergies are skin, respiratory and digestive tracts. Asthma, eczema, diarrhoea, vomiting, nausea, etc. are some of the common allergic reactions.

AID’S (Acquired Immune Deficiency Syndrome): It is a viral disease caused by the virus called HTV (Human immuno deficiency virus) This virus makes the defence mechanism of the human body very weak. The immune system in the body as W.B.C. becomes weak. Thus the person catches the infectious diseases very easily. This disease spreads through sexual contact as one of the partner may be carrier of the disease. It may spread through the blood transfusion and infected syringes, blades of the barbers, it may infect the developing baby through the blood by the mother. It is very deadly disease.

4. Fill in the blanks by selecting suitable words given below: (clotting, goitre, insuline, rickets, iron, proteins)

  1. Anaemia is caused due to the deficiency of iron
  2. Deficiency of Vit. D causes rickets in children.
  3. Deficiency of iodine in the diet may cause goitre.
  4.  Kwashiorkor is caused due to the deficiency of proteins
  5. Diabetes is caused due to undersecretion of insulin.
  6. Haemophilia is a disease caused due to slow clotting of the blood.

5. Find the odd one out:

(a) Typhoid, cholera, aundice, tuberculosis, tetanus.
Ans. Jaundice is odd one

(b) Cold, malaria, measles, mumps.
Ans. Malaria is odd one.

(c) Scurvy, rickets, polio, pellagra, nightblindness.
Ans. Haemophilia is odd one.

(d) Proteins, carbohydrates, fats, minerals, cancer.
Ans. Cancer is odd one.

6. Fill in the blank in the following table:
Selina Concise Biology Class 6 ICSE Solutions - Health and Hygiene 1
Answer:
Selina Concise Biology Class 6 ICSE Solutions - Health and Hygiene 2

7. Given below is a crossword puzzle. Read the clues across and clues down, and fill up the blank squares. Check up your performance with the correct solution given at the end.
Clues across
1. Category of pathogen that causes diseases, like common cold and mumps.
5. This is the vaccine for preventing tuberculosis.
6. An organ usually affected by tuberculosis.
7. Jumbled spelling of one of the most common insect which visits our exposed foods and contaminates them.
8. Cover this part of your body by a handkerchief while sneezing to prevent droplet infection to others.
9. These may readily grow in your hair, if you do not wash it regularly.
10. A disease that weakens body’s defence system against infections.
Selina Concise Biology Class 6 ICSE Solutions - Health and Hygiene 3
Answer:
Clues down

  1. Germ or germ-substance introduced into the body to prevent occurrence of an infectious disease.
  2. A disease caused by an infected dog, and which affects the central nervous system.
  3. A disease in which the eyes, the skin and the urine turn yellow.
  4. The disease pertussis is popularly known as whooping

Selina Concise Biology Class 6 ICSE Solutions - Health and Hygiene 4

Long Answer Questions
Question 1.
Describe the ways in which communicable diseases are transmitted through various indirect methods.
Answer:
Indirect methods of transmission of communicable diseases:

  1. Using items used by the infected persons:
    The healthy persons may be infected by using the articles like towel, hankey, utensils, bedding used by the patient infected by the communicable diseases. Diseases like tuberculosis, ring worm, common cold, influenza are transmitted by this method.
  2. Contaminated food and water:
    Diseases like dysentery, cholera spread through the contaminated food and water. Flies sitting on the food, if taken by a healthy person may be infected by the germs which may cause vomiting and loose motions. Similarly water and food infected by entamoeba may cause dysentery to persons who may take contaminated food.
  3. Vectors or carriers: Organisms like mosquitoes and house flies, ticks carry germs from the source of infection and pass on the germs to the normal persons and thus they are infected by diseases like malaria, cholera, plague. These organisms which cany the disease are called vectors and are not infected themselves.Mosquitoes suck blood and carry the disease causing protozoans from infected persons to healthy persons. Similarly houseflies carry the germs from garbage and sewage to the food. If this food is taken by the persons they become prey to typhoid and other diseases.
  4. Air: One sneeze from a person infected by cold may give billions of germs which are carried by air and may infect the healthy person. Tuberculosis passes from one person to other by coughing or sneezing of the infected person.These germs remain suspended in the air and persons may be infected by these spores or germs. Common cold, measles, diptheria, chicken pox.

Question 2.
List 3 ways by which you should keep your environmentclean.
Answer:
We should keep our environment clean in following ways:
Cleanliness of Environment:

  1. Maintain a clean environment to prevent the spreading of diseases due to the breeding of mosquitoes, house flies and microorganisms.
  2. Garbage should be kept in covered bins so that flies do not breed on it.
  3. Do not allow water to stagnate outside your house and in your neighbourhood. All the drains should also be covered. This will prevent breeding of mosquitoes.

Question 3.
cine’s are prepared, giving the name of one disease for which each type of vaccine is used.
Answer:
For developing resistance in the body we introduce germs or germ substances in the body to develop resistance in the body against a particular disease. The material introduced into the body is called vaccine, this practice is called prophylaxis. The germ or the germ substance is put into the body orally as polio drops or it is introduced by injection as TAB vaccine. Vaccine or vaccination was attached with small pox, but it is now used in a general sense.
Preparation: 

  1. Killed germs are introduced into the body These act as vaccine for TAB, vaccine for typhoid, Salk’s vaccine for poliomyelitis. Rabies vaccine for dog bite.
  2. Living weakened germs: The living germs are treated in such a way that they become very weak and as such, they cannot cause the disease. They can induce antibody formation such as the vaccine for measles and the freezed dried BCG vaccine for tuberculosis.
  3. Living fully virulent germs: These virulent germs in small doses are introduced into the body as vaccine and these produce antibodies in the body and these do not allow the germs of particular type to cause that disease. In this vaccination the person is inoculated with cowpox virus. It is very similar to small pox virus.
  4. Toxoids: Toxoids are prepared from the extracts of toxins secreted by bacteria. These toxins are poisons and these are made harmless by adding formalin into them. They retain their capacity and as a result when introduced into the body they produce antibodies into the body and do not allow the germs to grow in the body as vaccines for diphtheria and tetanus.

Selina Concise Chemistry Class 6 ICSE Solutions – Pure Substances and Mixtures ; Separation of Mixtures

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Selina Concise Chemistry Class 6 ICSE Solutions – Pure Substances and Mixtures ; Separation of Mixtures

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Chemistry for Class 6 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 6 Chemistry ICSE SolutionsPhysicsBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 6 Chemistry Chapter 5 Pure Substances and Mixtures ; Separation of Mixtures

POINTS TO REMEMBER

  1. Element is made up of atoms of the same kind i.e. atoms of gold are similar in every respect i.e. have same melting point and same boiling pts. same colour, mass and odour.
  2. Molecules of a compound are similar in every respect i.e. same composition taste, colour and odour i.e. compound water is made up of similar molecules.
  3. Pure substances are either elements or compounds have definite set of properties.
  4. Pure substances are required to maintain good health of human beings. Chemists, technologists and scientists need pure substances for manufacture of medicines, chemicals in industry and for scientific purposes.
  5. Milk, air, bronze, sugar and water solution, salt and water solution, salt, iron filings and sand, petrol are all mixtures.
  6. Mixtures, two or more components mixed in any ratio, undergo ho chemical change and retain their individual properties, can be separated by simple physical methods.
  7. Mixture has no formula, no change in volume, no change in mass, energy is neither needed nor produced.
  8. Mixture is impure substance.
  9. Homogeneous mixture : in which constituents are uniformaly distributed throughout its volume, e.g. salt and water solution.
  10. Heterogeneous mixture: The components are not uniformly distributed through its volume and components can be easily seen separately, e.g. water, oil solution.
  11. Milk is emulsion but this mixture cannot be separated. Oh shaking a mixture of mustard oil and water vigorously mixture becomes milky and is called an emulsion after some time water and oil get separated.
  12. Alloys are homogeneous mixtures of metal with metals or metal with non-metals.
  13. Salt and sand can be seperated by solvent extraction method where one of the solid component (salt) is soluble in liquid.
  14. Coagulation or loading : When alum (a chemical) is added to mixture dissolves in water and form clusters with clay and fine dust particles making them heavier, increases the rate of sedimentation.
  15. Centrifugation method is used to separate solids from liquids where mixture is homogeneous. Cream is separated being lighter on churning floats on a liquid (milk).
  16. Loading : The process of adding a chemical substance to help the suspended solid particles in liquid to form a sediment is called loading.

EXERCISE – I

Question 1.
Select homogeneous and heterogeneous mixtures from the following:
Salt solution, petrol and water, sand and charcoal, alcohol and water, air dissolved in water, air, sea water, fruit juices, mist, brass.
Answer:
Homogeneous mixture : Salt solution, alcohol and water, air dissolved in water, sea water, brass.
Heterogeneous mixture: Sand and charcoal, air, fruit juice, mist, petrol and water.

Question 2.
Define the following :

(a) Pure substance
(b) Impure substance
(c) Alloy
(d) Solution
(e) Heterogeneous mixture
(f) Homogeneous mixture

Answer:

(a) Pure substance : “Pure Substance is either element or compound. It contains the same kind of atom or molecules and has a definite set of physical and chemical properties.”
(b) Impure substance : “A substance in which some other substances are also present in smaller or larger amounts is called an impure substance. Mixtures are impure substance.
Example of impure substance is air.
(c) Alloy : “A homogeneous solid mixture of two or more metals or a metal and a non-metal is called an alloy.”
(d) Solution : “The homogeneous mixture of water (or any other solvent) and a substance soluble in it is called a solution.”
(e) Heterogeneous mixture : “A mixture in which the components are not uniformally distributed through its volume and can be easily seen separately is called heterogeneous mixture.”
(f) Homogeneous mixture : “A mixture in which its constituents are uniformly distributed throughout its volume and cannot be seen separately is called a homogenous mixture.”

Question 3.
List four characteristics of a mixture.
Answer:
Four characteristics of a mixture :

  1. Mixture has no fixed composition.
  2. To form a mixture energy is neither produced nor evolved.
  3. Mixture has no fixed melting’point and boiling points.
  4. Mixture retain the properties of its components.
  5. Components of mixtures can be seperated by simple physical methods.

Question 4.
Give reasons :

(a) Why do sugar and water retain their individual properties in a sugar solution ?
(b) Why do petrol and water form a heterogeneous mixture ?
(c) Why sulphur does dissolve when carbon disulphide is added to a mixture of iron and sulphur but not when it is added to iron sulphide ?

Answer:

(a) As sugar solution is a mixture and mixtures has not any specific set of properties. They show the properties of the individual components from which they are formed.
(b) Petrol and water forms a heterogeneous mixture as its constituents can be seen separately and are not uniformly distributed throughout its volume.
(c) Sulphur has the property to dissolve in carbon disulphide whereas iron does not dissolve and retains its individual property. However, On heating Iron and Sulphur, they chemically combined forming Iron Sulphide. In this, Iron and Sulphur particles do not exists separately as such they loose their individual property.

Question 5.
Give two examples for each of the following types of mixture.

(a) solid-solid
(b) solid-liquid
(c) liquid-gas
(d) gas-gas

Answer:
Two examples of :
(a) solid – solid –

  1. sand and sugar
  2. sand and iron filling.

(b) solid – liquid –

  1. salt and water
  2. charcoal and water.

(c) liquid – gas –

  1. coca cola
  2. mist.

(d) gas – gas –

  1. air
  2. helium and hydrogen in air balloon,
  3. perfumes and air.

Question 6.
Name the components present in the following mixtures:

(a) Brass
(b) Duralumin
(c) Tap water
(d) Bronze
(e) Crude petroleum oil .

Answer:

(a) Brass → Copper and Zinc.
(b) Duralumin → Aluminium + Copper with little manganese and magnesium.
(c) Tap water → air, dissolved salts.
(d) Bronze → Copper, Tin and zinc.
(e) Crude petroleum oil → petrol, kerosene, diesel, LPG, mixed with salt, water and earth particles.

Question 7.
State:

(a) Three differences between water and air.
(b) Four differences between compounds and mixtures.

Answer:
(a)
Water :

  1. The components of water are hydrogen and oxygen which are chemically combined in a fixed ratio of 1 : 8 by mass.
  2. The chemical composition of water remains same from whatever source it is obtained.
  3. The properties of water are completely different from the properties of elements from which it is formed i.e. hydrogen and oxygen.
  4. Energy change occurs in the formation of water.
  5. A molecule of water is represented by a definite formula H2O.

Air :

  1. The main components of air are nitrogen, oxygen, carbon -dioxide, water vapour which are not chemically combined.
  2. The composition of air varies from place to place. During rainy season the air becomes humid due to presence of more water vapour. Some impurities like sulphur dioxide, hydrogen sulphide etc. also changes its composition at some places.
  3. The components of air retain their individual properties but not air.
  4. No energy change occurs when components of air are mixed together.
  5. Air cannot be represented by any chemical formula.

(b)
Compound :

  1. A compound is a pure substance.
  2. Compounds are always homogeneous.
  3. A compound has a fixed composition, i.e., it is formed when two or more pure substances chemically combine in a definite ratio by mass.
  4. Formation of a compound involves change in energy.
  5. Compounds have specific set of properties.
  6. Components of compounds can be separated only by complex chemical processes.

Mixture :

  1. A mixture is an impure substance.
  2. Mixtures may be homogeneous or heterogeneous.
  3. A mixture has no fixed composition, i.e., it is formed by mixing two or more substances in any ratio without any chemical reaction.
  4. Formation of a mixture does not involve any change in energy.
  5. Mixtures do not have any specific set of properties.
  6. Components of mixtures can be separated by simple physical methods.

EXERCISE – II

Question 1.
Define:

(a) Filtration
(b) Sublimation
(c) Evaporation
(d) Crystallisation
(e) Miscible liquids
(f) Immiscible liquids

Answer:

(a) Filtration : The process of separating solid particles from liquid by allowing it to pass through a filter paper is called filtration.
(b) Sublimation : The process in which a solid changes directly into its vapours on heating is called sublimation.
(c) Evaporation : Is the process of converting a liquid into its vapours state either by exposing it to air or by heating.
(d) Crystallisation : Evaporation of liquid from a homogeneous liquit-solid mixture and collecting solid in the form of crystals is called
crystallisation.
(e) Miscible liquids: Homogeneous liquid-liquid mixtures are called miscible liquids.
(f) Immiscible liquids : Heterogeneous liquid-liquid mixtures are called immiscible liquids.

Question 2.
Why do we need pure substances?
Answer:
We need pure substances because of the following reasons:

  1. A pure substance has a fixed melting and fixed boiling point.
  2. A pure substance has its characteristic taste, colour and odour.
  3. Pure substances can not be broken further into more simple substances by any physical means.

Question 3.
Give one example for each of the following types of mixtures.

(a) Solid-solid heterogeneous mixture
(b) Solid-liquid heterogenous mixture
(c) Solid-liquid homogeneous mixture

Answer:

(a) Iron and sulphur.
(b) Sand and water, rice and water.
(c) Sugar from its solution in water.

Question 4.
Name the process by which the components of following mixtures can be separated.

(a) Powdered glass and sugar
(b) Chalk powder and iron filings
(c) Chaff and grain
(d) Salt and water
(e) Wheat and sugar
(f) Sand and camphor
(g) Sugar and water

Answer:

(a) FILTRATION : Glass and sugar on dissolving in water and filtering, glass separates out as residue on the filter paper. Filtrate of sugar solution is heated to remove water by evaporation, sugar is collected as crystals.
(b) MAGNETIC SEPERATION : With the help of a magnet, iron filings can be separated leaving behind chalk powder.
(c) WINNOWING : It separates chaff (lighter) from heavier grains in two different heaps.
(d) EVAPORATION : This method is used to separate the components of a homogeneous solid-liquid mixture, like salt from sea water. Sea water is collected in shallow beds and allowed to evaporate in the sun. When all the water is evaporated, salt is left behind.
(e) EVAPORATION : Wheat and sugar are put in water in a beaker. Sugar dissolves and mixture is passed through strainer and separated and dried. Sugar is obtained by evaporating sugar solution.
(f) SUBLIMATION : Camphor sublimes on heating leaving behind sand.
(g) CRYSTALLISATION : Pure sugar is obtained from its solution in water by the process of crystallisation. At first the sugar solution is heated to evaporate Water at a faster speed. When very less of water is left the solution is cooled. On cooling sugar dissolved in it starts separating out in the form of crystals.

Question 5.
Name:

(a) two substances which can sublime
(b) two substances soluble in water
(c) two substances insoluble in water
(d) four substances that can be used as filters.

Answer:

(a) Camphor and Naphthalene
(b) Sugar and salt (NaCl)
(c) Sand and chalk powder.
(d) (i) Filter paper, (ii) A bead of sand, (iii) Charcoal, (iv) A piece of muslin cloth.

Question 6.
Give reasons :

(a) Sand and saw dust cannot be separated by hand picking.
(b) Magnet is used to separate a mixture of iron and sulphur.
(c) Alum is used in purification of river water.

Answer:

(a) Because in hand picking method substances should be large enough in size to be recognized and picked out by hand but sand and saw dust particles are very small in size so they can’t be picked by hand. It can be separated by filtration.
(b) Mixtures of iron and sulphur can be separated by moving a magnet over them. Iron gets attached to the magnet is separated.
(c) Water from a river, pond or lake contains very fine clay particles. To make them settle at a faster rate, a chemical substance called alum in powdered form is added to such mixtures. It dissolves in water and forms clusters with clay and dust particles making them heavier and increasing the rate of sedimentation.

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks

(a) The substances that make a mixture are called its constituents or components.
(b) Evaporation or crystallisation is a process to separate solids dissolved in liquids.
(c) Mist is a heterogeneous (liquid in gas) mixture of droplets of water and air.
(d) Clay is separated from water by the method of loading and decantation.
(e) When cereals are washed before cooking, water is separated from the cereals by decantation.
(f) Crystallisation is a process to obtain a very pure form of a solid dissolved in a liquid.
(g) Ammonium chloride can be separated from common salt by the method of sublimation.
(h) The solid particles which remain on the filter paper are called residue and the liquid which passes through it is called filtrate.
(i) The process of transferring the clear liquid layer above the solid particles which settle at the bottom of the container is known as decantation.
(j) Filtration is a method used for the separation of an insoluble solid from a solid-liquid mixture.

2. Write “true” or “false” for the following statements 

(a) A pure substance consists of only one kind of atom or molecule.
Answer. True

(b) Common salt is separated from its solution in water by decantation.
Answer. False
Correct : Common salt is separated from its solution in water by evaporation.

(c) Winnowing is a process to remove small stones from grains.
Answer. False
Correct : Winnowing is a process to remove husk from grains.

(d) Jewellery gold is a homogeneous mixture of metals.
Answer. False
Correct : Jewellery gold is a heterogeneous mixture of metals.

(e) Air can be separated from water by filtration.
Answer. False
Correct : Air can be separated from water by heating.

(f) Salt and air dissolved in water add taste to water.
Answer. True

(g) Steel is an alloy of iron and aluminium.
Answer. False
Correct: Steel is an alloy of iron and carbon.

MULTIPLE CHOICE QUESTIONS

Tick (√) the correct alternative from the choice given for the following statements:

1. The process of adding a chemical substance to help the suspended solid particles to deposit as sediment fastly is called

  1. loading
  2. sedimentation
  3. decantation
  4. filtration

2. Salt is separated from sea water by

  1. evaporation
  2. sublimation
  3. crystallisation
  4. filtration

3. A mixture of mustard oil and water forms

  1. a compound
  2. a homogeneous mixture
  3. an alloy
  4. a heterogeneous mixture

4. A heterogeneous mixture is

  1. made up of only one kind of atom
  2. made up of only one kind of molecule
  3. made up of different kinds of atoms and molecules.
  4. that looks uniform

5. Example of a homogeneous mixture is

  1. distilled water
  2. tap water
  3. sand and water
  4. sawdust and water

6. A set of mixture is

  1. gold, common salt, water, alloy
  2. alloy, ink, honey, icecream
  3. alloy, mercury, air, sea water
  4. milk, duralumin, brass, silver

7. A gas dissolved in a liquid can be separated by :

  1. filtration
  2. boiling
  3. using magnet
  4. by crystallisation

8. Copper is not a part of the alloy :

  1. brass
  2. bronze
  3. steel
  4. duralumin

9. Which is not a mixture?

  1. sugar solution
  2. tap water
  3. milk
  4. distilled water

10. Give one word name for the following

(a) The solid which is left on the filter paper after filtration residue.
(b) The solid particles which separate out from the solution on slow evaporation crystals.
(c) The solid particles that settles at the bottom of the beaker in a heterogeneous mixture of a solid and a liquid decantation.
(d) The clean liquid which is poured out after sedimentation supernatant liquid.
(e) The technique used to separate the light particles from heavy particles using the flow of wind winnowing.

Selina Concise Chemistry Class 6 ICSE Solutions – Common Laboratory Apparatus and Equipments

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Selina Concise Chemistry Class 6 ICSE Solutions – Common Laboratory Apparatus and Equipments

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Chemistry for Class 6 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 6 Chemistry ICSE SolutionsPhysicsBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 6 Chemistry Chapter 2 Common Laboratory

Points To Remember

  1. Chemistry is an experimental science.
  2. A chemical laboratory is a place to perform experiments and to observe chemical processes.
  3. Knowledge of chemistry is based on experiment and observation.
  4. A chemical laboratory is a place to perform chemical experiments.
  5. Various kinds of apparatus and equipments are required to perform experiments.
  6. Necessary precautions should be taken while performing experiments.
  7. Maintain silence and discipline in the laboratory and concentrate on your experiment.

Exercise

Question 1.
Mention one use of each of the following equipments
(a) Spirit lamp
(b) Test tube
(c) Conical flask
(d) Evaporating dish
(e) Wire gauze
(f) Beaker
(g) Mortar and pestle
(h) Measuring cylinder
(i) Glass tube
(j) Gas jar
(k) Reagent bottle.

Answer:
(a) Spirit lamp is used to heat up substances.
(b) Test tube is used to conduct tests with small quantities of chemicals for heating and boiling purposes.
(c) Conical flask to hold sufficient quantities of substance in the form of solution.
(d) Evaporating dish is used for evaporating liquids.
(e) Wire gauze is used to keep glass apparatus (flask, beaker) on while heating is in progress. It is also used for uniform distribution of heat.
(f) Beaker is used for keeping of solutions.
(g) Morter and pestle is used to grind and crush solid substances into a pov
(h) Measuring cylinder is used to measure the volume of (mainly) liquid substances.
(i) Glass tube is used to transfer fluids or gases from one vessel to another.
(j) Gas jar is used for collecting gases and holding them in captivity vaccum.
(k) Reagent bottle is used for storing chemicals.

Question 2.
From what materials are the following made up of ?
(a) Test tube rack
(b) Test tube holder
(c) Measuring cylinder
(d) Wire gauze
(e) Mortar and pestle

Answer:
(a) Test tube rack is made up of wood or plastic.
(b) Test tube holder is made up of a iron clamp at front and wood or plastic handle at other end.
(c) Measuring cylinder is made of glass.
(d) Wire gauze is made of meshed iron wire and a thin asbestos sheet that is fixed at its centre.
(e) Morter and pestle is made of porcelain.

Question 3.
List any five precautions taken care of while performing an experiment in a chemistry laboratory.
Answer:
Five precautions to be taken in laboratory are :

  1. Do not touch or taste any unknown substance.
  2. Use only small quantity of chemical to carry out experiment.
  3. Do not work alone in the laboratory.
  4. Do not throw hot concentrated acids into the sink directly.
  5. Always wear an apron in the laboratory to protect your clothes.
  6. While heating keep the mouth of test tube away from your eyes and face.
  7. The apparatus to be used in an experiment should be arranged neatly before beginning an experiment.
  8. Do not throw broken glass apparatus or used filter paper in the sink. Throw them in a dustbin.

Question 4.
Answer the following questions in brief:
(a) Why is chemistry known as an experimental science?
(b) Why are most apparatus made of glass ?

Answer:
(a) Chemistry is known as experimental science as an experiment is performed under controlled conditions in an activity and we observe a natural or an artificially created phenomenon.
(b) Most of the laboratory apparatus is made of glass because :

  1. glass is easy to clean.
  2. Glass is transparent material and we can see through it clearly.
  3. It does not react with most of the chemicals used in experiments.
  4. Glass withstands high temperatures.
  5. Pyrex glass or borosil glass is a special type of glass which hardly expands on heating. Such glasses do not break even at high temperatures.

Question 5.
Label the marked equipment s in the diagram given below.
Selina Concise Chemistry Class 6 ICSE Solutions Chapter 2 Common Laboratory Apparatus and Equipments 1
Answer:

  1. Glass tube (delivery tube)
  2. Flask (round bottomed)
  3. Wire gauze
  4. Burner
  5. Tripod (stand)
  6. Gas jar
  7. Water trough
  8. Water

Selina Concise Chemistry Class 6 ICSE Solutions Chapter 2 Common Laboratory Apparatus and Equipments 2

Objective Type Questions

Question 1.
Fill in the blanks :

(a) Experiment and observation are the two important basics of chemistry.
(b) A porcelain dish is used for evaporation.
(c) A test tube holder is used to hold the test tube while-it is heated.
(d) Mortar and pestle is used for grinding and crushing solid substances into a powder.
(e) Glass apparatus is made of Pyrex or borosil glass.

Question 2.
Match the words in Column A with their respective functions in Column B.
Selina Concise Chemistry Class 6 ICSE Solutions Chapter 2 Common Laboratory Apparatus and Equipments 3
Question 3.
Choose the correct alternative from the options given for each of the following statements.

(a) The evaporating dish is made of

  1. porcelain
  2. glass
  3. metal
  4. plastic

(b) The spirit lamp is made of

  1. glass
  2. brass
  3. steel
  4. all of the above

(c) The apparatus to measure an accurate volume of liquid, is

  1. beaker
  2. conical flask
  3. measuring cylinder
  4. test tube

(d) To pass a gas from one vessel to another, you will use

  1. gas jar
  2. delivery tube
  3. glass rod
  4. test tube

(e) To prevent the escape of a gas from a gas jar, you will cover its mouth with

  1. watch glass
  2. crucible
  3. beaker
  4. round bottom flask

Question 4.
Write true or false against the following statements and correct the false ones.

(a) A glass funnel is used to pour off liquids.
(b) A test tube is used to test liquid chemicals.
(c) A mortar and pestle is used for evaporation.
(d) A glass rod is used to stir solutions.
(e) A round-bottom flask is used to store chemicals.

Answer:
(a) True
(b) False
Correct: A test tube is used to test chemicals and for heating and boiling purposes.
(c) False
Correct: Mortar and pestle is used to grind and crush solid substances into powder.
(d) True
(e) False
Correct: A REAGENT bottle is used to store chemicals.

 

Selina Concise Chemistry Class 6 ICSE Solutions – Introduction To Chemistry

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Selina Concise Chemistry Class 6 ICSE Solutions – Introduction To Chemistry

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Chemistry Class 6 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 6 Chemistry Chapter 1 Introduction To Chemistry

Points To Remember

  1. Science : “Science is the systematic effort by human being to study, understand and utilise nature for meaningful purposes. This understanding is slowly developed by careful observations and experiments.”
  2. Chemistry: “The branch of science that deals with the study of the composition, physical and chemical properties of various forms of matter is called chemistry.”
  3. Fertilisers : “Are the chemicals which provide nutrients to crops and increase their yield.” e.g. Urea, calcium nitrate, sodium nitrate, potash, ammonium sulphate etc.
  4. Pesticides : “Are the chemicals used to kill pests which affect the production of crops and fruits.” e.g. Aldrin, malathion, parathion etc.
  5. Insecticides are the chemical used to kill insects, e.g. D.D.T. and B.H.C.
  6. Fungicides are substances which protect the crops from fungi, e.g. Bordeaux mixture and sulphur act as fungicides.
  7. Hormones are group of chemicals secreted in our body to control various activities.
  8. Enzymes are the chemicals secreted by our body and react with food water and oxygen as raw material in mouth and intestines.
  9. Role of Chemistry : Without chemistry our life would have been dull, it has helped us in every field of life. The production of better and faster crops to meet the needs of such a huge population is possible only because of chemicals produced by chemistry, i.e. Pesticides, fungicides, insecticides preservatives help us to preserve food in a better state and of good taste for a long time.
    In Industry : paints, drugs, fibres, soap, toothpaste, dyes plastics even fuels, atomic energy, petrol, diesel, kerosene, wax, paraffin rubber, acids, alkalies metals alloys are the gift of chemistry.
    Medicines: Penicillin, Tetracycline, pain killers and various other antibiotics are used to kill germs and cure diseases. Building and Daily Life: Building materials, items of daily use like, ink, pen, glass, sugar, common salt, paper.
  10. Preservatives are the chemicals which maintain the taste for longer time and do not let the food degrade to produce obnoxious (unpleasant smell), e.g. sugar, common salt, sodium benzoate and sodium meta-bisulphate.

Exercise

Question 1.
Give two examples for each of the following substances :
(a) food preservatives
(b) fuel
(c) fungicides
(d) medicines
(e) building materials
(f) chemical war weapons

Answer:
Two examples of :
(a) Food preservatives :

  1. Sodium Benzoate
  2. Sodium metabisulphate
  3. Sugar
  4. Common salt

(b) Fuel:

  1. L.P.G.
  2. Petrol
  3. Coal

(c) Fungicides :

  1. Sulphur
  2. Bordeaux mixture

(d) Medicines :

  1. Penicillin
  2. Painkillers
  3. Antibiotics

(e) Building materials :

  1. Cement
  2. Steel
  3. Glass

(f) Chemical war weapons :

  1. TNT
  2. RDX

Question 2.
Give short answers :
(a) What is science ?
(b) What is chemistry ?
(c) What is a fuel ? 
(d) How is chemistry helpful in improving the health of human beings ?
(e) What is alchemy?
(f) What kind of experiments did Alchemists do?
(g) What is ‘Philosopher’s stone’ ?
(h) What is the main difference between alchemy and chemistry?
(i) Name the chemicals which help in increasing food production.
(j) Name six such products, which we use daily.
(k) How is the knowledge of chemistry important to mankind ?

Answer:
(a) SCIENCE is the systematic effort by human beings to control nature through experiments and observation for their own use.
OR
SCIENCE is the systematic ongoing effort by human beings to study understand and utilise nature for meaningful purposes. This understanding is slowly developed by careful observations and experiment.
(b) Chemistry : “The branch of science that deals with the study of the composition and the physical and chemical properties of various forms of matter is called Chemistry.”
(c) Fuels : The substances which on burning produce heat energy are called fuels.
(d) Chemistry is very helpful in improving the health of human beings by providing Antibiotics, Pain killers, Pencilin, Tetracycline etc. It has provided us with Vitamins, Enzymes, Minerals and Anesthesia (chloroform, formalene etc.)
(e) The word “Alchemy” has its origin in a Greek word ‘Khemeia’ means “art of transmuting metals”. It was partly based on experimentations and partly on spiritual discipline.
(f) ‘Alchemists’ considered to be early chemists. They used all general techniques of chemistry in healing humans. Their contribution proved valuable to the society and in the advancement of civilization.
They had contributed to an incredible number of future uses of chemicals, metals, ink, paints, cosmetic, medicines, porcelain, etc.
(g) The goal of alchemy was to find a mythical and magical substance called “philosopher’s stone” not a literal stone but wax, liquid or powder with magical power, which on heating with a base, iron and copper metals would turn into gold, the purest form of matter which would bring wealth, health and immortality.
(h) Alchemy was both scientific and spiritual. Alchemists never separated them. It also lacked a common language for its concepts and processes i.e. there was no standardized scientific practice.
Chemistry was completely separated from ancient traditional alchemy. Still modern chemistry in general owes a great deal to alchemy. Alloys are formed by mixing metals with other metals and substances.
(i) Chemicals which help in increasing food production are fertilisers like urea, sodium nitrate, potash, ammonium phosphate, calcium nitrate etc. Pesticides like aldrin, malathion which are used to kill pests. Insecticides like D.D.T., B.H.C. fungicides like sulphur, bordeaux mixture etc.
(j) Six products of daily use are soap, paints, pen, tooth-paste, cooking oil, potable water.
(k) Importance of chemistry to mankind chemistry plays an important role to provide us with things of daily use like toothpaste, soap, detergents, paints, clothes,
medicines, fertilisers, pesticides, plastics, in preparing fuels, consumer products like glass, paper, pencils, pens, in substances used in defence like gunpowder, T.N.T. etc.

Question 3.
What is the contribution of chemistry in the following fields ?
(a) Industry
(b) Clothings
(c) Cosmetics
(d) National Defence
(e) Medicines

Answer:
Contribution of chemistry in the field of :
(a) Industry : To improve efficiency and production of metals, paints, paper, plastics, alloys, textile, pharmaceuticals, electroplating, cosmetics, synthetic fibres etc.
(b) Clothings: Chemistry is widely used in textile industry which manufactures clothing for us. Clothes guard our body from external environment.
Formation of clothing begins with the knowledge of conversion of fibres into fabrics. Fibres can be natural or synthetic. Earlier only natural fibres were known to man such as cotton, jute, silk, wool, etc. which were used to produce dress materials, sarees, bags, sweaters, shawls, etc. With more development, synthetic fibres were also made such as nylon, terylene etc. These fibres are strong, wrinkle resistant and dry quickly. They are used to make towels, bed sheets, bags, curtains, carpets, blankets, dress materials, etc.
(c) Cosmetics: The use of talcum powder, skincare creams, lipsticks, eyes and facial make up, deodorants, lotions, perfumes, bathing oil, body butter, baby products, etc. It is possible to convert various ingredients into usable cosmetics due to knowledge of chemistry.
(d) National Defence : Substances like gunpowder, T.N.T. (trinitrotoluene), phosgene, chemical weapons, laughing gas, etc., are all products of chemistry which contribute to the national defence.
(e) Medicines : Extensive researches by chemists have led to the discovery of number of medicinal drugs. These drugs help in fighting diseases and have thus increased the life span of human beings.
Examples : Aspirin, paracetamol, antibiotics like penicillin, tetracyline, antiseptics and various other medicines used to kill germs and cure diseases and their symptoms.

Question 4.
Who is known as Father of chemistry? Why?
Answer:
Robert William Boyle is known as ‘Father of Modern Chemistry’. He was an Anglo Irish scientist born in Ireland. He was the first to perform experiments under controlled conditions and publish his researches with elaborate details of procedure, apparatus and observations. Robert Boyle put chemistry on a firm scientific footing transforming it from alchemy into one based on measurements. He defined elements, compounds and mixtures.

Question 5.
Name the scientists who discovered the following.
(a) Atoms
(b) Oxygen
(c) Safety lamp
(d) Elements
Answer:
The scientists who discovered
(a) Atoms : John Dalton was a British chemist and physicist. He proved that matter consists of small indivisble called ‘atoms’. For this he proposed the atomic theory which was later on called “Dalton’s atomic theory”.
(b) Oxygen : Joseph Priestly.
(c) Safety lamp : Sir Humphry Davy.
(d) Elements : Antoiene Lavoisier was a French nobleman. He revolutionized chemistry. Lavoisier named the elements carbon, hydrogen and oxygen and discovered the role of oxygen in combustion and respiration for which he is most noted. He established that water is a compound and helped to continue the transformation of chemistry from a qualitative science to a quantitative one.

Objective Type Questions

Question 1.
Fill in the blanks :
(a) Chemistry deals with the study of matter and the changes it undergoes.
(b) Fertilizers help to increase the production of food.
(c) Food items like jams and pickles are protected by using preservatives (salt and sugar).
(d) L.P.G. is used for fuel.
(e) Inert gases were discovered by William Ramsay.

Question 2.
Match the following words in column A with those in column B:
Selina Concise Chemistry Class 6 ICSE Solutions Chapter 1 Introduction To Chemistry 1

Question 3.
Write “True” or “False” against each of the following statements.

(a) Chemistry plays an important role in national economy: True
(b) Antibiotics are used as preservatives : False
(c) D.D.T. is an important fertiliser: False
(d) Gunpowder is an pesticide : False
(e) Enzymes secreted by our body are chemicals : True

Question 4.
Choose the correct alternative from the choices given below for the following statements :
(a) Trinitrotoluene is used as

  1. a preservative
  2. a fertiliser
  3. a fuel
  4. an explosive

(b) Which one of the following is a pesticide ?

  1. benzoic acid
  2. aldrin
  3. sugar
  4. gunpowder

(c) Mortar is used as a

  1. plastic material Urea is an
  2. a building material
  3. an insecticide
  4. as medicine

(d) Used is an important

  1. fuel
  2. preservative
  3. fertiliser
  4. food item

(e) The chemicals prescribed by a doctor in treatment of infectious diseases are called

  1. antigens
  2. lotions
  3. antibiotics
  4. creams

Question 5.
Match the following scientists in column A with their discoveries or contributions in column B.
Selina Concise Chemistry Class 6 ICSE Solutions Chapter 1 Introduction To Chemistry 2

 

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions (Including Evaluation)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions (Including Evaluation)

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Framing Algebraic Expressions Exercise 21 – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Write in the form of an algebraic expression :
(i) Perimeter (P) of a rectangle is two times the sum of its length (l) and its breadth (b).
(ii) Perimeter (P) of a square is four times its side.
(iii) Area of a square is square of its side.
(iv) Surface area of a cube is six times the square of its edge.
Solution:
(i) Let P be the perimeter and / be the length, and b be the breadth.
P = 2 (l + b)
(ii) Let P be the perimeter and a be the side of the square.
P = 4a
(iii) Let A be the area of the square and a be the sides of the square.
A = (a)2
(iv) Let S be the surface area and a be the edges of the cube.
S = 6a2

Question 2.
Express each of the following as an algebraic expression :
(i) The sum of x and y minus m.
(ii) The product of x and y divided by m.
(iii) The subtraction of 5m from 3n and then adding 9p to it.
(iv) The product of 12, x, y and z minus the product of 5, m and n.
(v) Sum of p and 2r – s minus sum of a and 3n + 4x.
Solution:
(i) x + y – m
(ii) \(\frac { xy }{ m }\)
(iii) 3n – 5m + 9p
(iv) 12xyz – 5mn
(v) p + 2r – s – (a + 3n + 4x)

Question 3.
Construct a formula for the following :
Total wages (₹ W) of a man whose basic wage is (₹ B) for t hours week plus (₹ R) per hour, if he Works a total of T hours.
Solution:
Wages for t hours = ₹ B
Wages for overtime = R(T – t)
=> Total wages = Wages for t hours + wages for overtime of (T – t) hours
=> ₹ W = ₹ B + ₹ R (T – t)

Question 4.
If x = 4, evaluate :
(i) 3x + 8
(ii) x2 – 2x
(iii) \(\frac { { x }^{ 2 } }{ 2 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 1

Question 5.
If m – 6, evaluate :
(i) 5m – 6
(ii) 2m2 + 3m
(iii) (2m)2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 2

Question 6.
If x = 4, evaluate :
(i) 12x + 7
(ii) 5x2 + 4x
(iii) \(\frac { { x }^{ 2 } }{ 8 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 3

Question 7.
If m = 2, evaluate :
(i) 16m – 7
(ii) 15m2 – 10m
(iii) \(\frac { 1 }{ 4 } \times { m }^{ 3 }\)
Solution:
16m – 7
= (16 x 2) – 7
= 32 – 7 = 25
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 4

Question 8.
If x = 10, evaluate :
(i) 100x + 225
(ii) 6x2 – 25x
(iii) \(\frac { 1 }{ 50 } \times { x }^{ 3 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 5

Question 9.
If a = – 10, evaluate :
(i) 5a
(ii) a2
(iii) a3
Solution:
(i)5a
= 5 x (-10) = -50
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 6

Question 10.
If x = – 6, evaluate :
(i) 11x
(ii) 4x2
(iii) 2x3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 7

Question 11.
If m = – 7, evaluate :
(i) 12m
(ii) 2m2
(iii) 2m3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 8

Question 12.
Find the average (A) of four quantities p, q, r and s. If A = 6, p = 3, q = 5 and r = 7 ; find the value of s.
Solution:
Given, average of four quantities (A) = 6
and p = 3,q = 5, r = 7 and s = ?
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 9

Question 13.
If a = 5 and b = 6, evaluate :
(i) 3ab
(ii) 6a2b
(iii) 2b2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 10

Question 14.
If x = 8 and y = 2, evaluate :
(i) 9xy
(ii) 5x2y
(iii) (4y)2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 11
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 12

Question 15.
If x = 5 and y = 4, evaluate :
(i) 8xy
(ii) 3x2y
(iii) 3y2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 13

Question 16.
If y = 5 and z = 2, evaluate :
(i) 100yz
(ii) 9y2z
(iii) 5y2
(iv) (5z)3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 14

Question 17.
If x = 2 and y = 10, evaluate :
(i) 30xy
(ii) 50xy2
(iii) (10x)2
(iv) 5y2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 15

Question 18.
If m = 3 and n = 7, evaluate :
(i) 12mn
(ii) 5mn2
(iii) (10m)2
(iv) 4n2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 16

Question 19.
If a = -10, evaluate :
(i) 3a – 2
(ii) a2 + 8a
(iii) \(\frac { 1 }{ 5 }\) x a2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 17

Question 20.
If x = -6, evaluate :
(i) 4x – 9
(ii) 3x2 + 8x
(iii) \(\frac { { x }^{ 2 } }{ 2 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 18
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 19

Question 21.
If m = -8, evaluate :
(i) 2m + 21
(ii) m2 + 9m
(iii) \(\frac { { m }^{ 2 } }{ 4 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 20

Question 22.
If p = -10, evaluate :
(i) 6p + 50
(ii) 3p2 – 20p
(iii) \(\frac { { p }^{ 2 } }{ 50 }\)
Solution:
(i) 6p + 50
= (6 x p) + 50
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 21

Question 23.
If y = -8, evaluate :
(i) 6y + 53
(ii) y2 + 12y
(iii) \(\frac { { y }^{ 3 } }{ 4 }\)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 22
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 23

Question 24.
If x = 2 and 7 = -4, evaluate :
(i) 11xy
(ii) 5x2y
(iii) (5y)2
(iv) 8x2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 24

Question 25.
If m = 9 and n = -2, evaluate
(i) 4mn
(ii) 2m2n
(iii) (2n)3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 25
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 26

Question 26.
If m = -8 and n = -2, evaluate :
(i) 12mn
(ii) 3m2n
(iii) (4n)2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 27

Question 27.
If x = -5 and y = -8, evaluate :
(i) 4xy
(ii) 2xy2
(iii) 4x2
(iv) 3y2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 21 Framing Algebraic Expressions image - 28

Question 28.
Find T, if T = 2a – b, a = 7 and b = 3.
Solution:
T = 2a – b, a = 1 and b = 3
Put the value of a = 1, and b = 3 in above equation
T = (2 x 7) -3
T = 14 – 3 = 11
T = 11

Question 29.
From the formula B = 2a2 – b2, calculate the value of B when a = 3 and b = -1.
Solution:
B = 2a2 – b2
Put the values of a = 3 and b = -1 in above equation
B = 2 x (3)2 – (-1)2
B = 18 – 1
B = 17
Value of B is = 17

Question 30.
The wages ₹ W of a man earning ₹ x per hour for t hours are given by the formula W = xt. Find his wages for working 40 hours at a rate of ₹ 39.45 per hour.
Solution:
T = 40 hours
x = ₹ 39.45
W = xt = 40 x 39.45
W = ₹ 1578

Question 31.
The temperature in Fahrenhiet scale is represented by F and the tempera¬ture in Celsius scale is represented by C. If F = \(\frac { 9 }{ 5 }\) x C + 32, find F when C = 40.
Solution:
F = \(\frac { 9 }{ 5 }\) x C + 32
Given, C = 40
F = \(\frac { 9 }{ 5 }\) x 40 + 32 = 9 x 8 + 32
F = 104°

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution (Including Use of Brackets as Grouping Symbols)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution (Including Use of Brackets as Grouping Symbols)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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IMPORTANT POINTS

  1. Substitution : The value of an expression depends on the value of its variable (s).
  2. Use of Brackets :
    The Symbols —, ( ), { }, [ ] are called brackets.
    If an expression is enclosed within a bracket, it is considered a single quantity, even if it is made up of many terms.
    Keep in Mind :

    • While simplifying an expression containing a bracket, first of all, the terms inside the bracket are operated (combined).
    • ( ) is called a small bracket or Parenthesis.
    • { } is called a middle bracket or Curly bracket.
    • [ ] is called big or square bracket.
    • If one more bracket is needed, then we use the bar bracket.
      i.e. a line ———— is drawn over a group of terms.
      Thus, in \(3x+\bar { 4y-5z }\), the line over 4y – 5z serves as the bar bracket and is called Vinculum.

Substitution Exercise 20A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the following blanks, when :
x = 3,y = 6, z = 18, a = 2, b = 8, c = 32 and d = 0.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 1
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 2
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 44
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 4

Question 2.
Find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 5
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 6
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 7

Question 3.
Find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 8
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 9

Question 4.
If a = 3, b = 0, c = 2 and d = 1, find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 11
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 12

Question 5.
Find the value of 5x2 – 3x + 2, when x = 2.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 13

Question 6.
Find the value of 3x3 – 4x2 + 5x – 6, when x = -1.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 14

Question 7.
Show that the value of x3 – 8x2 + 12x – 5 is zero, when x = 1.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 15

Question 8.
State true and false :
(i) The value of x + 5 = 6, when x = 1
(ii) The value of 2x – 3 = 1, when x = 0
(iii) \(\frac { 2x-4 }{ x+1 }\) = -1,when x = 1
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 16
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 17

Question 9.
If x = 2, y = 5 and z = 4, find the value of each of the following :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 18
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 19

Question 10.
If a = 3, find the values of a2 and 2a.
Solution:
a2 = (3)2 = 3 x 3 = 9
2a = (2)3 = 2 x 2 x 2 = 8

Question 11.
If m = 2, find the difference between the values of 4m3 and 3m4.
Solution:
4m3 = 4 (2)3 = 4 x 2 x 2 x 2 = 32
3m4 = 3 (2)4 = 3 x 2 x 2 x 2 x 2 = 48
Now, a difference 3m4 – 4m3 = 48 – 32 = 16

Substitution Exercise 20B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Evaluate :
(i) (23 – 15) + 4
(ii) 5x + (3x + 7x)
(iii) 6m – (4m – m)
(iv) (9a – 3a) + 4a
(v) 35b – (16b + 9b)
(vi) (3y + 8y) – 5y
Solution:
(i) (23 – 15) + 4 = 8 + 4 = 12
(ii) 5x + (3x + 7x) = 5x + 10x = 15x
(iii) 6m – (4m – m) = 6m – 3m = 3m
(iv) (9a – 3a) + 4a = 6a + 4a = 10a
(v) 35b – (16b + 9b)= 35b – 25b = 10b
(vi) (3y + 8y) – 5y = 11y – 5y = 6y

Question 2.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 20
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 21
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 22
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 23
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 24

Question 3.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 25
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 26
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 27

Substitution Exercise 20C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 28
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 29
(viii) 2t + r – p – q + s = 2t + r – (p + q – s)

Question 2.
Insert the bracket as indicated :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 30
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 31

Substitution Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the value of 3ab + 10bc – 2abc when a = 2, b = 5 and c = 8.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 32

Question 2.
If x = 2, = 3 and z = 4, find the value of 3x2 – 4y2 + 2z2.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 33

Question 3.
If x = 3, y = 2 and z = 1; find the value of:
(i) xy
(ii) yx
(iii) 3x2 – 5y2
(iv) 2x – 3y + 4z + 5
(v) y2 – x2 + 6z2
(vi) xy + y2z – 4zx
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 34

Question 4.
If P = -12x2 – 10xy + 5y2, Q = 7x2 + 6xy + 2y2, and R = 5x2 + 2xy + 4y2 ; find :
(i) P – Q
(ii) Q + P
(iii) P – Q + R
(iv) P + Q + R
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 35
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 36

Question 5.
If x = a2 – bc, y = b2 – ca and z = c2 – ab ; find the value of :
(i) ax + by + cz
(ii) ay – bx + cz
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution image - 37

Question 6.
Multiply and then evaluate :
(i) (4x + y) and (x – 2y); when x = 2 and y = 1.
(ii) (x2 – y) and (xy – y2); when x = 1 and y = 2.
(iii) (x – 2y + z) and (x – 3z); when x = -2, y = -1 and z = 1.
Solution:


Question 7.
Simplify :
(i) 5 (x + 3y) – 2 (3x – 4y)
(ii) 3x – 8 (5x – 10)
(iii) 6 {3x – 8 (5x – 10)}
(iv) 3x – 6 {3x – 8 (5x – 10)}
(v) 2 (3x2 – 4x – 8) – (3 – 5x – 2x2)
(vi) 8x – (3x – \(\bar { 2x-3 }\))
(vii) 12x2 – (7x – \(\bar { 3x^{ 2 }+15 }\))
Solution:

Question 8.
If x = -3, find the value of : 2x3 + 8x2 – 15.
Solution:

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations (Related to Algebraic Expressions)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations (Related to Algebraic Expressions)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

IMPORTANT POINTS

  1. Fundamental Operations : In mathematics, the operations : addition (+), subtraction (-), multiplication (x) and division (÷) are called the four fundamental operations.
  2. Addition and Subtraction :
    • Addition of Like Terms :
      • When all the terms are positive, add their coefficients.
      • When all the terms are negative, add their coefficients without considering their negative signs and then prefix the minus sign to the sum.
    • Addition of Unlike Terms : As discussed above, the sum of two or more like terms is a single like term ; but the two unlike terms cannot be added together to get a single term.
    • Subtraction of Like Terms : The same rules, as those for subtraction of integers, are applied for the subraction of like terms. The result of subtraction of two like terms is also a like term.

Add the positive terms together and negative terms separately together. Then, find the result of two terms obtained.

Fundamental Operations Exercise 19A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
(i) 5 + 4 = ………… and 5x + 4x = ………….
(ii) 12 + 18 = ………… and 12x2y + 18x2y = ………….
(iii) 7 + 16 = ………….. and 7a + 16b = …………
(iv) 1 + 3 = ………… and x2y + 3xy2 = ………..
(v) 7 – 4 = …………… and 7ab – 4ab = …………..
(vi) 12 – 5 = ………… and 12x – 5y = ……………
(vii) 35 – 16 = ………….. and 35ab – 16ba = ………….
(viii) 28 – 13 = …………. and 28ax2 – 13a2x = ………….
Solution:
(i) 5 + 4 = 9 and 5x + 4x = 9x
(ii) 12 + 18 = 30 and 12x2y + 18x2y = 30x2y
(iii) 7 + 16 = 23 and 7a + 16 b = 7a + 16b
(iv) 1 + 3 = 4 and x2y + 3xy2 = x2y + 3xy2
(v) 7 – 4 = 3 and 7ab – 4ab = 3ab
(vi) 12 – 5 = 7 and 12x – 5y = 12x – 5y
(vii) 35 – 16 = 19 and 35ab – 16ba = 19ab
(viii) 28 – 13 = 15 and 28ax2 – 13a2x = 28ax2 – 13a2x

Question 2.
Fill in the blanks :
(i) The sum of – 2 and – 5 = …………. and the sum of – 2x and – 5x = …………….
(ii) The sum of 8 and – 3 = ………….. and the sum of 8ab and – 3ab = ………….
(iii) The sum of – 15 and – 4 = …………….. and the sum of – 15x and -4y = ………………
(iv) 15 + 8 + 3 = ……….. and 15x + 8y + 3x = …………….
(v) 12 – 9 + 15 = …………… and 12ab – 9ab + 15ba = ……………..
(vi) 25 – 7 – 9 = and 25xy – 7xy – 9yx = ……………
(vii) – 4 – 6 – 5 = …………. and – 4ax – 6ax – 5ay = …………….
Solution:
(i) The sum of – 2 and – 5 = – 7 and the sum of – 2x and – 5x = -7x
(ii) The sum of 8 and -3 = 5 and the sum of 8ab and – 3ab = 5ab
(iii) The sum of – 15 and – 4 = – 19 and the sum of – 15x and – 4y = – 15x – 4y
(iv) 15 + 8 + 3 = 26 and 15x + 8y + 3x = 18x + 8y
(v) 12 – 9 + 15 = 18 and 12ab – 9ab + 15ba = 18ab
(vi) 25 – 7 – 9 = 9 and 25xy – 7xy – 9yx = 9xy
(vii) – 4 – 6 – 5 = – 15 and – 4ax – 6ax – 5ay = – 10ax – 5ay

Question 3.
Add:
(i) 8xy and 3xy
(ii) 2xyz, xyz and 6xyz
(iii) 2a, 3a and 4b
(iv) 3x and 2y
(v) 5m, 3n and 4p
(vi) 6a, 3a and 9ab
(vii) 3p, 4q and 9q
(viii) 5ab, 4ba and 6b
(ix) 50pq, 30pq and 10pr
(x) – 2y, – y and – 3y
(xi) – 3b and – b
(xii) 5b, – 4b and – 10b
(xiii) – 2c, – c and – 5c
Solution:
(i) 8xy + 3xy = 11xy
(ii) 2xyz + xyz + 6xyz = (2 + 1 + 6) xyz = 9xyz
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 1
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 2

Question 4.
Evaluate :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 3
Solution:
(i) 6a – a – 5a – 2a = 6a – (1 + 5 + 2).a
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 4

Question 5.
Evaluate :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 5
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 6

Question 6.
Subtract the first term from the second :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 7
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 8

Question 7.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 9
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 11
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 12
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 13
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 14

Fundamental Operations Exercise 19B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the sum of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 15
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 16
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 17
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 18

Question 2.
Add the following expressions :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 19
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 20
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 21

Question 3.
Evaluate :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 22
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 23
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 24

Question 4.
Subtract :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 25
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 26
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 27

Question 5.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 28
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 29

Question 6.
From the sum of x + y – 2z and 2x – y + z subtract x + y + z.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 30

Question 7.
From the sum of 3a – 2b + 4c and 3b – 2c subtract a – b – c.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 31

Question 8.
Subtract x – 2y – z from the sum of 3x – y + z and x + y – 3z.
Solution:
(3x – y + z) + (x + y – 3 z) – (x – 2y – z)
= 3x – y + z + x + y – 3z – x + 2y + z
= 3x + x – x – y + y + 2y + z + z – 3z
= 3x + 2y – z

Question 9.
Subtract the sum of x + y and x – z from the sum of x – 2z and x + y + z
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 32

Question 10.
By how much should x + 2y – 3z be increased to get 3x ?
Solution:
3x – (x + 2y – 3z)
= 3x – x – 2y + 3z
= 2x – 2y + 3z

Question 11.
The sum of two expressions is 5x2 – 3y2. If one of them is 3x2 + 4xy – y2, find the other.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 33

Question 12.
The sum of two expressions is 3a2 + 2ab – b2. If one of them is 2a2 + 3b2, find the other.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 34

Fundamental Operations Exercise 19C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 35
Solution:
(i) 6 x 3 = 18 and 6x x 3x = 6 x 3x x x x = 18x2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 36

Question 2.
Fill in the blanks :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 37
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 38
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 39

Question 3.
Find the value of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 40
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 41

Question 4.
Multiply :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 42
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 43
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 44

Question 5.
Multiply :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 45
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 46
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 47

Question 6.
Multiply :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 48
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 49

Question 7.
Multiply :
(i) x + 2 and x + 10
(ii) x + 5 and x – 3
(iii) x – 5 and x + 3
(iv) x – 5 and x – 3
(v) 2x+ y and x+ 3y
(vi) (3x – 5y) and (x + 6y)
(vii) (x + 9y) and (x – 5y)
(viii) (2x + 5y) and (2x + 5y)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 50
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 51

Question 8.
Multiply :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 52
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 53
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 54

Question 9.
Find the product of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 55
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 56
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 57
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 58

Fundamental Operations Exercise 19D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Divide :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 59
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 60

Question 2.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 61
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 62
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 63
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 64

Question 3.
Divide :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 65
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 66
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 67
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 68

Question 4.
Simplify :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 69
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 70

Question 5.
Divide :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 71
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 72
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 73
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 19 Fundamental Operations image - 74

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts

by

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 6 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

IMPORTANT POINTS

  1. Algebra : Algebra is a generalized form of Arithmetic. In Arithmetic, we use numbers, such as : 3, – 8, 0.63, etc., each of which has one definite value ; whereas in Algebra, we use letters along with numbers.
    For Example : 5x, 3x – 4, 7a + b, 3y – 5x, x + 3y – 9z, etc.
    The letters used in Algebra are called variables or literal numbers or simply literals.
  2. Signs and Symbols : In Algebra, the signs +, -, x and ÷ are used with the same meaning as in Arithmetic.
    Following sign and symbols are frequently used in algebra and have the same meanings as they have in any other branch of Mathematics.
    = means, “is equal to”
    ≠ means, “is not equal to”
    < means, “is less than” > means, “is greater than”
    \(\nless\) means, “is not less than”
    \(\ngtr\) means, “is not greater than”
    ∴ means, “therefore”
    ∵ means, “because” or “since”
    ~ means, “difference between”
    ⇒ means, “implies that”.
  3. To Write a Given Statement in Algebraic Form
    Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 1

Fundamental Concepts Exercise 18A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express each of the following statements in algebraic form :
(i) The sum of 8 and x is equal to y.
(ii) x decreased by 5 is equal to y.
(iii) The sum of 2 and x is greater than y.
(iv) The sum of x and y is less than 24.
(v) 15 multiplied by m gives 3n.
(vi) Product of 8 and y is equal to 3x.
(vii) 30 divided by b is equal to p.
(viii) z decreased by 3x is equal to y.
(ix) 12 times of x is equal to 5z.
(x) 12 times of x is greater than 5z.
(xi) 12 times of x is less than 5z.
(xii) 3z subtracted from 45 is equal to y.
(xiii) 8x divided by y is equal to 2z.
(xiv) 7y subtracted from 5x gives 8z.
(xv) 7y decreased by 5x gives 8z.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 2

Question 2.
For each of the following algebraic expressions, write a suitable statement in words:
(i) 3x + 8=15
(ii) 7 – y > x
(iii) 2y – x < 12
(iv) 5 ÷ z = 5
(v) a + 2b > 18
(vi) 2x – 3y= 16
(vii) 3a – 4b > 14
(viii) b + 7a < 21
(ix) (16 + 2a) – x > 25
(x) (3x + 12) – y < 3a
Solution:
(i) 3x plus 8 is equal to 15
(ii) 1 decreased by y is greater than x
(iii) 2y decreased by x is less than 12
(iv) 5 divided by z is equal to 5
(v) a increased by 2b is greater than 18
(vi) 2x decreased by 3y is equal to 16
(vii) 3a decreased by 4b is greater than 14
(viii) b increased la is less than 21
(ix) The sum of 16 and 2a decreased by x is greater than 25
(x) The sum of 3x and 12 decreased by y is less than 3a.

Fundamental Concepts Exercise 18B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Separate the constants and the variables from each of the following:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 3
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 4

Question 2.
Group the like terms together :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 5
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 6
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 7

Question 3.
State whether true or false :
(i) 16 is a constant and y is a variable but 16y is variable.
(ii) 5x has two terms 5 and x.
(iii) The expression 5 + x has two terms 5 and x
(iv) The expression 2x2 + x is a trinomial.
(v) ax2 + bx + c is a trinomial.
(vi) 8 x ab is a binomial.
(vii) 8 + ab is a binomial.
(viii) x3 – 5xy + 6x + 7 is a polynomial.
(ix) x3 – 5xy + 6x + 7 is a multinomial.
(x) The coefficient of x in 5x is 5x.
(xi) The coefficient of ab in – ab is – 1.
(xii) The coefficient of y in – 3xy is – 3
Solution:
(i) True
(ii) False
(iii) True
(iv) False
(v) True
(vi) False
(vii) True
(viii) True
(ix) True
(x) False
(xi) True
(xii) False

Question 4.
State the number of terms in each of the following expressions :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 8
Solution:
(i) 2 terms
(ii) 2 terms
(iii) 2 terms
(iv) 2 terms
(v) 3 terms
(vi) 2 term
(vii) 2 terms
(viii) 3 terms
(ix) 3 terms

Question 5.
State whether true or false:
(i) xy and – yx are like terms.
(ii) x2y and – y2x are like terms.
(iii) a and – a are like terms.
(iv) – ba and 2ab are unlike terms.
(v) 5 and 5x are like terms.
(vi) 3xy and 4xyz are unlike terms.
Solution:
(i) True
(ii) False
(iii) True
(iv) False
(v) False
(vi) True

Question 6.
For each expression, given below, state whether it is a monomial, or a binomial or a trinomial.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 9
Solution:
(i) Monomial
(ii) Binomial
(iii) Monomial
(iv) Monomial
(v) Trinomial
(vi) Binomial
(vii) Trinomial
(viii) Binomial
(ix) Trinomial

Question 7.
Write down the coefficient of x in the following monomial :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 10
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 11

Question 8.
Write the coefficient of :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 12
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 13

Question 9.
State the numeral coefficient of the following monomials :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 14
(vii) – 7x ÷ y
(viii) – 3x ÷ (2y)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 15

Question 10.
Write the degree of each of the following polynomials :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 16
Solution:
(i) 2
(ii) 2
(iii) 10
(iv) 20
(v) 3
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 17

Fundamental Concepts Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express each of the following statements in algebraic form :
(i) The sum of 3x and 4y is 8.
(ii) 5x decreased by 7 gives y.
(iii) 31 added to 4x gives 6x.
(iv) 3x subtracted from 89 gives 44.
Solution:
(i) 3x + 4y = 8
(ii) 5x – 7 = y
(iii) 4x + 37 = 6x
(iv) 89 – 3x = 44

Question 2.
Group the like terms :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 18
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 19

Question 3.
Write the number of terms in each of the following polynomials :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 20
Solution:
(i) 2 terms
(ii) 3 terms
(iii) 3 terms
(iv) 4 terms
(v) 3 terms

Question 4.
For each expression, given below, state whether it is a monomial, or a binomial or a trinomial:
(i) x + y
(ii) 5x – 4y
(iii) 7x2 + 5x + 8
(iv) 64 + 3 ÷ 6
(v) 9 ÷ a x b
(vi) 8a ÷ b
Solution:
(i) binomial
(ii) binomial
(iii) trinomial
(iv) 6a + 3 ÷ b = 6a + \(\frac { 3 }{ b }\)
It has two terms
It is binomial
(v) 9 ÷ a x b = \(\frac { 9b }{ a }\)
It has one term
It is monomial.
(vi) monomial

Question 5.
Write the coefficient of x2y in :
(i) -7x2yz
(ii) 8abx2y
(iii) – x2y
Solution:
(i) – 7z
(ii) 8ab
(iii) -1

Question 6.
Write the coefficient of :
(i) x2 in – 8x2y
(ii) y in -4y
(iii) x in – xy2
Solution:
(i) – 8y
(ii) – 4
(iii) – y2

Question 7.
Write the numeral coefficient in :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 21
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 22

Question 8.
Write the degree of each of the following polynomials :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 23
Solution:
(i) 8
(ii) 4
(iii) 2
(iv) 1
(v) 3
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 24
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 25

Question 9.
Write each statement, given below in algebraic form :
(i) 28 more than twice of x is equal to 45.
(ii) 3y reduced by 5z is greater than 8x.
(iii) 6x divided by 13y is less than 17.
(iv) 9 multiplied by 5x is equal to 2y.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 18 Fundamental Concepts image - 26

Question 10.
State whether true or false :
(i) If 23 is a constant and x is a variable, 23 + x is constant.
(ii) If 23 is a constant and x is a variable, 23x is a variable.
(iii) If y is a variable and 57 is a constant, y – 57 is a variable.
(iv) If 3x and 2y are variable; each of 3x + 2y, 3x – 2y, 3x ÷ 2y and 3x x 2y is a variable.
Solution:
(i) False
Sum of a constant and a variable is also variable.
(ii) True
Product of a constant and a variable is variable.
(iii) True
Constant subtracted from a variable is also variable.
(iv) True
Sum, difference product or quotient of two variables is also variable.

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage)

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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IMPORTANT POINTS

  1. Percent: Out of one hundred, is called percent and it is denoted as (%) e.g. 10%, 15% ….
    • Percent can be expressed in fraction and decimal such as given below :
      Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 1
    • A fraction or decimal can be expressed in percent.
    • We use percentage in profit and loss and also in finding interest etc.
  2. Percentage : When a quantity is expressed in the percent form, it is called percentage.
  3. To convert a given Fraction or Decimal into Percentage (Percent Form) : Multiply the given fraction or the given decimal by 100 and at the same time write the sign of percentage.
  4. To convert a given Percentage into a Fraction or Decimal: Remove the sign of percentage and at the same time divide by 100. Then reduce the fraction obtained to its lowest terms or decimal as required.
  5. To Express one Quantity (number) as a percentage of the other : Divide first quantity by the second and at the same time multiply the result by 100%.
    Keep in Mind :

    • Percent or percentage has no unit.
    • In order to express one quantity as a percentage of another quantity ; both the quantities must have same units.
  6. To find the Increase or Decrease Percent :
    Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 2

Percent Exercise 16A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express each of the following statements in the percentage form :
(i) 13 out of 20
(ii) 21 eggs out of 30 are good
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 3

Question 2.
Express the following fractions as percent :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 4
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 37
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 6

Question 3.
Express as percent:
(i) 0.10
(ii) 0.02
(iii) 0.7
(iv) 0.15
(v) 0.032
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 7

Question 4.
Convert into fractions in their lowest terms:
(i) 8%
(ii) 20%
(iii) 85%
(iv) 250%
(v) 12\(\frac { 1 }{ 2 }\) %
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 8

Question 5.
Express as decimal fractions :
(i) 25%
(ii) 108%
(iii) 95%
(iv) 4.5%
(v) 29.2%
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 9

Question 6.
Express each of the following natural numbers as percent :
(i) 7
(ii) 2
(iii) 19.5
(iv) 5.37
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 10

Percent Exercise 16B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Express :
(i) Rs 5 as a percentage of Rs 25.
(ii) 80 paise as a percent of Rs 4.
(iii) 700 gm as a percentage of 2.8 kg.
(iv) 90 cm as a percent of 4.5 m.
Solution:
(i) \(\frac { 5 }{ 25 }\) x 100 = 20%
(ii) 80 paise as a percent of 400 paise (as/rupee = 100 paise)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 11

Question 2.
Express the first quantity as a percent of the second :
(i)) 40 P, ₹ 2
(ii) 500 gm, 6 kg
(iii) 42 seconds, 6 minutes
Solution:
40 p, ₹ 2 = 40 p to 200 p
(1 Rupee = 100 paise)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 12

Question 3.
Find the value of each of the following:
(i) 20% of ₹ 150
(ii) 90% of 130
(iii) 15% of 2 minutes
(iv) 7.5 % of 500 kg.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 13

Question 4.
If a man spends 70% of his income, what percent does he save ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 14

Question 5.
A girl gets 65 marks out of 80. What percent marks did she get ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 15

Question 6.
A class contains 25 children, of which 6 are girls. What percentage of the class are the boys.
Solution:
Total number of students = 25
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 16

Question 7.
A tin contains 20 litres of petrol. Due to leakage, 3 litres of petrol is lost. What percent is still present in the tin ?
Solution:
Total petrol in tin = 20 litres
last due to leakage = 3 litres
Balance petrol in tin = (20 – 3) = 17 litres
Percentage of petrol in tin = \(\frac { 17 }{ 20 }\) x 100 = 85%

Question 8.
An alloy of copper and zinc contains 45% copper and the rest is zinc. Find the weight of zinc in 20 kg of the alloy.
Solution:
Total weight of alloy = 20 kg
Weight copper = 20 x 45% = 20 x \(\frac { 45 }{ 100 }\) = 9 kg
Weight of zinc = (total weight of alloy – weight of copper) = 20 – 9 = 11 kg

Question 9.
A boy got 60 out of 80 in Hindi, 75 out of 100 in English and 65 out of 70 in Arithmetic. In which subject his percentage of marks the best ? Also, find his overall percentage.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 17
Now total marks he gets = 60 + 75 + 65 = 200
Total marks = 80 + 100 + 70 = 250
Percent marks obtained = \(\frac { 200 }{ 250 }\) x 100 = 80%

Question 10.
In a camp, there were 500 soldiers. 60 more soldiers joined them. What percent of the earlier (original) number have joined the camp.
Solution:
Number of soldiers = 500
More joined them = 60
Percentage to join the earlier = \(\frac { 60 }{ 500 }\) x 100 = 12%

Question 11.
In a plot of ground of area 6000 sq. m, only 4500 sq. m is allowed for construction. What percent is to be left without construction ?
Solution:
Total ground area = 6000 sq. m.
Allowed for construction = 4500 sq.m.
Area left without construction = 6,000 sq. m – 4500 sq. m = 1500 sq. m
Percentage of construction left = \(\frac { 1500 }{ 6000 }\) x 100 = 25%

Question 12.
Mr. Sharma has a monthly salary of ₹ 8,000. If he spends ₹ 6,400 every month; find :
(i) his monthly expenditure as percent.
(ii) his monthly savings as percent.
Solution:
Monthly salary of Mr. Sharma = ₹ 8000
He spends every month = ₹ 6400
His savings = ₹ 8000 – 6400 = ₹ 1600
(i) Percent expenditure = \(\frac { 6400}{ 8000 }\) x 100% = 80%
(ii) Percent savings = \(\frac { 1600}{ 8000 }\) x 100% = 20%

Question 13.
The monthly salary of Rohit is ₹ 24,000. If his salary increases by 12%, find his new monthly salary
Solution:
Salary = ₹ 24000
New salary = ₹ 24000 + 12% of 24000
= ₹ 24000 + \(\frac { 12 }{ 100 }\) x 24000
= ₹ 24000 + 2880 = ₹ 26880
New salary = ₹ 26880

Question 14.
In a sale, the price of an article is reduced by 30%. If the original price of the article is ₹ 1,800, find :
(i) the reduction in the price of the article
(ii) reduced price of the article.
Solution:
(i) Original price of article = ₹ 1800
Reduction = 30%
Reduction in price = 30% of 1800
= \(\frac { 30 }{ 100 }\) x 1800 = ₹ 540
(ii) Reduced price of the article = Original price – Reduction = ₹ 1800 – ₹ 540 = ₹ 1260

Question 15.
Evaluate :
(i) 30% of 200 + 20% of 450 – 25% of 600
(ii) 10% of ₹ 450 – 12% of ₹ 500 + 8% of ₹ 500.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 18
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 19

Percent Exercise 16C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The price of rice rises from Rs. 30 per kg to Rs. 36 per kg. Find the percentage rise in the price of rice.
Solution:
First price of rice = Rs. 30 per kg
Rised price = Rs. 36 per kg
Rise per kg = 36 – 30 = Rs. 6
Percent rise = \(\frac { 6 }{ 30 }\) x 100 = 20%

Question 2.
The population of a small locality was 4000 in 1979 and 4500 in 1981, By what percent had the population increase ?
Solution:
Year 1979 population = 4,000
Year 1981 population = 4,500
Increase in population = (4,500 – 4,000) = 500
percentage of increase in population = \(\frac { 500 }{ 4000 }\) x 100 = 12.5%

Question 3.
The price of a scooter was ₹ 8000 in 1975. It came down to ₹ 6000 in 1980. By what percent had the price of the scooter came down ?
Solution:
Original cost of scooter = ₹ 8,000
Reduced cost of scooter = ₹ 6000
Reduction in price of scooter = ₹ 8,000 – ₹ 6,000 = ₹ 2,000
Percentage of reduction = \(\frac { 2000 }{ 8000 }\) x 100 = 25%

Question 4.
Find the resulting quantity when :
(i) ₹ 400 is decreased by 8%.
(ii) 25 km is increased by 5%.
(iii) a speed of 600 km/h is increased by 12\(\frac { 1 }{ 2 }\) %
(iv) there is 2.5% increase in a salary of ₹ 62, 500.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 20
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 21
= ₹ 1562.50
Resulting quantity (salary) = ₹ 62500 + ₹ 1562.50 = ₹ 64062.50

Question 5.
The population of a village decreased by 12%. If the original population was 25,000, find the population after decrease ?
Solution:
Original population = 25,000
Decrease in population = 12% Population after decrease
= 25,000 – 12% of 25,000
= 25,000 – \(\frac { 12 }{ 100 }\) x 25,000
= 25,000 – 3,000 = 22,000

Question 6.
Out of a salary of Rs. 13,500,1 keep 1/3 as savings. Of the remaining money, I spend 50% on food and 20% on house rent. How much do I spend on food and house rent ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 22

Question 7.
A tank can hold 50 litres of water. At present, it is only 30% full. How many litres of water shall I put into the tank so that it becomes 50% full ?
Solution:
Capacity of tank = 50 litres
30% of capacity = 30% of 50 litres
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 23

Question 8.
In an election, there are a total of 80,000 voters and two candidates, A and B. 80% of the voters go to the polls out of which 60% vote for A. How many votes does B get.
Solution:
Member of voters = 80,000
Total vote polled = 80% of 80,000 = \(\frac { 80 }{ 100 }\) x 80,000 = 64,000
Vote polled to A = 60% of 64,000 = \(\frac { 60 }{ 100 }\) x 64000 = 38,400
Vote polled to B = Total vote polled – vote polled to A = 64,000 – 38,400 = 25,600

Question 9.
70% of our body weight is made up of water. Find the weight of water in the body of a person whose body weight is 56 kg.
Solution:
Water in human body = 70%
Weight of a man = 56 kg
Quantity of water in him = 70% of 56
= \(\frac { 70 }{ 100 }\) = x 56 = 39.2 kg

Question 10.
Only one-fifth of water is available in liquid form. This limited amount of water is replenished and used by man recurrently. Express this information as percent, showing :
(i) water available in liquid form.
(ii) water available in frozen form.
Solution:
Let total quantity of water = 1
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 24

Question 11.
By weight, 90% of tomato and 78% of potato is water. Find :
(i) the weight of water in 25 kg of tomato.
(ii) the total quantity, by weight, of water in 90 kg of potato and 30 kg of tomato
(iii) the weight of potato which contains 39 kg of water.
Solution:
Water in tomato = 90% and water in potato = 78%
(i) Weight of water in 25 kg of tomato
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 25

Percent Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Rohit’s age is 12 years and Geeta’s age is 15 years. Express :
(i) Rohit’s age as a percent of Geeta’s age.
(ii) Geeta’s age as a percent of Rohit’s age.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 26

Question 2.
A class has 30 boys and 20 girls. Find:
(i) the percentage of girls in the class
(ii) the percentage of boys in the class
(iii) percentage of number of boys as compared with number of girls.
Solution:
Total students in class = Number of boys + Number of girls = 30 + 20 = 50
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 27

Question 3.
Mrs. Sharma went to the market with ₹ 800 in her purse. When she returned to her home, ₹ 240 were still left in her purse. What percent of her money did she spend in the market ?
Solution:
Money in her purse = ₹ 800
Balance in her purse = ₹ 240
Money spent = ₹ 800 – ₹ 240 = ₹ 560
Percentage of money spent = \(\frac { 560 }{ 800 }\) x 100 = 70%

Question 4.
In a mixture of two liquids A and B, 35% is liquid B. If the total quantity of the mixture is 20 kg, find the quantity of A, by weight.
Solution:
Total quantity of A and B = 20 kg
Quantity of B = 35% of 20 = \(\frac { 35 }{ 100 }\) x 20 = 7kg
Quantity of A = Total quantity – Quantity of B = 20 kg – 7 kg = 13 kg
Hence, quantity of A = 13 kg.

Question 5.
A girl got 375 marks out of 500 in the first term examination, 560 marks out of 800 in the second term examination and 840 marks out of 1200 in the third term examination. Find:
(i) her percentage score in the first term examination.
(ii) her percentage score in the second term examination.
(iii) her percentage score in the third term examination.
(iv) the total marks secured in all the three examinations.
(v) the total marks scored in all the three examinations.
(vi) her percentage score on the whole in all the three examinations.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 28

Question 6.
Out of his monthly income of ₹ 2,500; a man spends ₹ 1,750. What percent of his income does he save every month ?
Solution:
Monthly income = ₹ 2,500
Spending = ₹ 1,750
Saving = monthly income – spending = 2,500- 1,750 = ₹ 750
Percentage of income he saves = \(\frac { 750 }{ 2500 }\) x 100 = 30%

Question 7.
Mr. Singh’s monthly salary is ₹ 15,000. This month he was promoted with an increment of ₹ 3,000 in his salary. Express his increment as a percent of his original salary.
Solution:
Monthly salary = ₹ 15,000
Increment on promotion = ₹ 3,000
Percentage of increment to monthly salary = \(\frac { 3000 }{ 15000 }\) x 100 = 20%

Question 8.
(i) The price of an article increased from ₹ 16 to ₹ 20 ; find the percentage increase.
(ii) The price of an article decreased from Rs 20 to Rs 16 ; find the percentage decrease.
Solution:
(i) Original price = ₹ 16
Increased price = ₹ 20
Amount of increase = 20 – 16 = ₹ 4
Percentage of increase = \(\frac { 4 }{ 16 }\) x 100 = 25%
(ii) Original price = ₹ 20
Decrease price = ₹ 16
Amount of decrease = 20 – 16 = ₹ 4
Percentage of decrease = \(\frac { 4 }{ 20 }\) x 100 = 20%

Question 9.
(i)) The salary of a man is ₹ 7,200 per month, which is now increased by 8%. Find his new salary per month.
(ii) The salary of Mr. Sahni is ₹ 8,400 per month, which is now decreased by 8%. Find his new salary per month.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 29

Question 10.
Find the percentage change from the first quantity to the second :
(i) ₹ 80, ₹ 120
(ii) 75 kg, 60 kg
(iii) 50 cm, 45 cm
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 30

Question 11.
The original price of an article is ₹ 640. Find its new price when its price is :
(i) increased by 30%
(ii) decreased by 20%
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 31

Question 12.
Find the number that is :
(i) 50% more than 48
(ii) 30% less than 70
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 32

Question 13.
Evaluate :
(i) 8% of 900 – 12% of 750 + 20% of 165.
(ii) 70% of 70 + 90% of 90 – 120% of 120.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 33

Question 14.
Approximately 97.3% water on the earth is not fit for drinking. Find :
(i) the percentage of water on the earth that is fit for drinking.
(ii) The total volume of water available in certain part of the earth where there is 21,600 m3 of drinking water.
Solution:
Approximately water on earth which is not fit for drinking = 97.3%
(i) Water fit for drinking = 100 – 97.3 = 2.7%
(ii) At a certain place, the water which is fit for drinking = 21600 m3
Volume of total water on that place
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 34

Question 15.
Air is an important inexhaustible natural resource. It is essential for the survival of human beings, microbes, plants and animals. The following table shows the percentage of various gases in air.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 35
(i) In 800 m3 of air, calculate the approximate quantities of nitrogen, oxygen and other gases.
(ii) If a certain quantity (by volume) of air contains 4,200 litres of oxygen, find the total quantity of air taken and the amount of nitrogen in it.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 16 Percent (Percentage) imag - 36

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