Selina Class 8 ICSE Solutions – Page 3

Selina Concise Physics Class 8 ICSE Solutions – Heat Transfer

August 11, 2022January 27, 2019 by Kalyan

Selina Concise Physics Class 8 ICSE Solutions – Heat Transfer

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 6 Heat Transfer. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 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 8 Physics Chapter 6 Heat Transfer

Heat is a form of energy. When two bodies are in contact heat flows from body at higher temperature to body at lower temperature till the lower temperature of both is same.
When a body is heated, its molecules move faster about their means position and kinetic energy increases and with fall in temperature their K.E. decreases.
When a substance is heated
(i) It expands i.e. a change in size takes place
(ii) Change in temperature takes place.
(iii) Change in state takes place.
CHANGE OF STATE : “The process of change from one state to another at a constant temperature is called change of state.”
Solid on heating changes into LIQUID. LIQUID on absorbing heat changes to VAPOURS some SOLIDS on heating DIRECTLY change in vapours called SUBLIMATION. Substance is called SUBLIMATE.
SOLIDIFICATION on cooling when a vapours change into SOLID. GAS OR VAPOURS on cooling \(\xrightarrow { Condensation }\) changes to LIQUID also called LIQUIFACTION.
MELTING: Change of solid into liquid at constant temperature. FUSION ⇒ FREEZING is change of LIQUID into SOLID at constant temperature and change of solid into liquid at a constant _ temperature is called FUSION.
EVAPORATION: “Change liquid to gas at ALLTEMPERATURES” It is surface phenomenon. “
VAPOURIZATION : “Change of liquid into vapours at fixed temperature”.
METING POINT: “Is the temperature at which a solid starts melting and remains constant till the whole of solid melts.”
M.P. is same as freezing point.
M.P. of ice is 0°C or freezing point, of water is 0°C.
BOILING POINT: “Is the temperature of a liquid at which it start, boiling i.e. change into vapours or gaseous state.”
B .P. of pure water is 100°C.
ABSOLUTE ZERO: “The temperature at which molecular motion completely ceases.”
FACTORS EFFECTING THE RATE OF EVAPORATION :
(i) Temperature: Increases with increase in temperature
(ii) S.A.: Increases with increase in S.A.
(iii) BLOWING AIR—Renewal of air increases evaporation.
(iv) NATURE—Some liquids like spirit, alcohol, petrol evaporate easily.
EVAPOURATION → produces coolness, BOILING produces Hotness.
LINEAR EXPANSION: When a solid rod (metal) is heated change in length takes place, which depends upon
(i) original length (L₀)
(ii) Increase in temperature
(iii) Material of rod.
Let L0 be the original length at 0°C, when heated to T°C final length becomes L
Increase in length (L_t – L₀) a L₀ (T – 0)
Or
Coefficient of linear expension a which depends upon material of rod.
L_t– L₀ = L₀ α T
α = L_t – L₀/ L₀T = increase in length / original length × Rise in temperature
When a metal plate is heated, change in area takes place and the expansion is called SUPERFICIAL expansion.
When a solid of volume v₀is heated change in volume called cubical expansion takes place.
α : β : γ = 1 : 2 : 3

Test your self

A. Objective Questions

1. Write true or false for each statement

(a) Evaporation is rapid on a wet day.
Answer. False.

(b) Evaporation takes place only from the surface of liquid.
Answer. True.

(c) All molecules of a liquid take part in the process of evaporation.
Answer. False.

(d) Temperature of a liquid rises during boiling or vaporization
Answer. False.

(e) All molecules of a liquid take part in boiling.
Answer. True.

(f) Boiling is a rapid phenomenon.
Answer. True.

(g) All solids expand by the same amount when heated to the same rise in temperature.
Answer. False.

(h) Telephone wires are kept tight between the two poles in winter.
Answer. True.

(i) Equal volumes of different liquids expand by the different amount when they are heated to the same rise in temperature.
Answer. True.

(j) Solids expand the least and gases expand the most on being heated.
Answer. True.

(k) A mercury thermometer makes use of the property of expansion of liquids on heating.
Answer. True.

(l) Kerosene contracts on heating.
Answer. False.

2. Fill in the blanks

(a) Boiling occurs at a fixed temperature.
(b) Evaporation takes place at all temperature.
(c) The molecules of liquid absorb heat from surroundings in evaporation.
(d) Heat is absorbed during boiling.
(e) Cooling is produced in evaporation.
(f) A longer rod expands more than a shorter rod on being heated to the same temperature.
(g) Liquids expand more than the solids.
(h) Gases expand more than the liquids.
(i) Alcohol expands more than water.
(j) Iron expands less than copper.

3. Match the following

Selina Concise Physics Class 8 ICSE Solutions Chapter 6 Heat Transfer 2

4. Select the correct alternative

(a) In evaporation

all molecules of liquid begin to escape out
only the molecules at the surface escape out
the temperature of liquid rises by absorbing heat from surroundings.
the molecules get attracted within the liquid.

(b) The rate of evaporation of a liquid increases when :

temperature of liquid falls
liquid is poured in a vessel of less surface area
air is blown above the surface of liquid
humidity increases.

(c) During boiling or vaporization

all molecules take part
temperature rises
no heat is absorbed
the average kinetic energy of molecules increases.

(d) The boiling point of a liquid is increased by

increasing the volume of liquid
increasing the pressure, on liquid
adding ice to the liquid
decreasing pressure on liquid.

(e) Two rods A and B of the same metal, but of length 1 m and 2 m respectively, are heated from 0°C to 100°C. Then

both the rods A and B elongate the same
the rod A elongates more than the rod B
the rod B elongates more than the rod A
the rod A elongates, but the rod B contracts.

(f) Two rods A and B of the same metal, same length, but one solid and the other hollow, are heated to the same rise in temperature.
Then

the solid rod A expands more than the hollow rod B
the hollow rod B expands more than the solid rod A
the hollow rod B contracts, but the solid rod A expands
both the rods A and B expand the same.

(g) A given volume of alcohol and the same volume of water are heated from the room temperature to the same temperature then.

alcohol contracts, but water expands
water contracts, but alcohol expands
water expands more than alcohol
alcohol expands more than water.

(h) The increase in length of a metal rod depends on

the initial length of the rod only
the rise in temperature only
the material of rod only
all the above three factors.

(i) The correct statement is

Iron rims are cooled before they are placed on the cart wheels.
A glass stopper gets tighten on warming the neck of the bottle.
Telephone wires sag in winter, but become tight in summer.
A little space is left between two rails on a railway track.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers)

August 11, 2022January 27, 2019 by Veerendra

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 2 Exponents (Powers). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Exponents Exercise 2A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 1
Solution:

Question 2.
If 1125 = 3^m x 5ⁿ; find m and n.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 8

Question 3.
Find x, if 9 × 3^x = (27)^2x-3
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 9

Exponents Exercise 2B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Compute:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 10

Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 12

Question 2.
Simplify:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 48
Solution:

Question 3.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 22
Solution:

Question 4.
Simplify:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 24
Solution:

Question 5.
Simplify:
(x^a+b)^a-b.(x^b+c)^b-c.(x^c+a)^c-a Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 28

Question 6.
Simplify:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 49
Solution:

Question 7.
(i) (a^-2)^-2. (ab)^-3
(ii) (xⁿy^-m)⁴× (x³y^-2)^-n

Solution:

Question 8.
Show that:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 34
Solution:

Question 9.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 36
Solution:

Question 10.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 38
Solution:

Question 11.
(m+n)^-1(m^-1+ n^-1) = (mn)^-1
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 40

Question 12.
Prove that:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 41
Solution:

Question 13.
Find the values of n, when:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 52
Solution:

Question 14.
Simplify:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 46
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 9 Simple and Compound Interest. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Simple and Compound Interest Exercise 9A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the interest and the amount on:
(i) ₹ 750 in 3 years 4 months at 10% per annum.
(ii) ₹ 5,000 at 8% per year from 23rd December 2011 to 29th July 2012.
(iii) ₹ 2,600 in 2 years 3 months at 1% per month.
(iv) ₹ 4,000 in 1\(\frac { 1 }{ 3 }\) years at 2 paise per rupee per month.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -1

Question 2.
Rohit borrowed Rs. 24,000 at 7.5 percent per year. How much money will he pay at the end of 4th years to clear his debt ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -3

Question 3.
The interest on a certain sum of money is Rs. 1,480 in 2 years and at 10 per cent per year. Find the sum of money.
Solution:
Let P = Rs. x
Time (T) = 2 years
Rate (R) = 0%
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -4

Question 4.
On what principal will the simple interest be Rs. 7,008 in 6 years 3 months at 5% per year ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -5

Question 5.
Find the principal which will amount to Rs. 4,000 in 4 years at 6.25% Per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -6

Question 6.
(i) At what rate per cent per annum will Rs. 630 produce an interest of Rs. 126 in 4 years ?
(ii) At what rate per cent per year will a sum double itself in 6\(\frac { 1 }{ 4 }\) years ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -8

Question 7.
(i) In how many years will Rs.950 produce Rs.399 as simple interest at 7% ?
(ii) Find the time in which Rs.1200 will amount to Rs.1536 at 3.5% per year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -9

Question 8.
The simple interest on a certain sum of money is \(\frac { 3 }{ 8 }\) of the sum in 6\(\frac { 1 }{ 4 }\) years. Find the rate percent charged.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -11

Question 9.
What sum of money borrowed on 24th May will amount to Rs. 10210.20 on 17th October of the same year at 5 percent per annum simple interest.
Solution:
A = Rs. 10210.20
R = 5% P.A.
T=May + June + July + August + Sept.+ Oct.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -12

Question 10.
In what time will the interest on a certain sum of money at 6% be \(\frac { 5 }{ 8 }\) of itself ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -13

Question 11.
Ashok lent out Rs.7000 at 6% and Rs.9500 at 5%. Find his total income from the interest in 3 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -15

Question 12.
Raj borrows Rs.8,000; out of which Rs. 4500 at 5% and remainder at 6%. Find the total interest paid by him in 4 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -16

Question 13.
Mohan lends Rs.4800 to John for 4\(\frac { 1 }{ 2 }\) years and Rs.2500 to Shy am for 6 years and receives a total sum of Rs.2196 as interest. Find the rate percent per annum, it being the same in both the cases.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -18

Question 14.
John lent Rs. 2550 to Mohan at 7.5 per cent per annum. If Mohan discharges the debt after 8 months by giving an old black and white television and Rs. 1422.50; find the price of the television.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -20

Simple and Compound Interest Exercise 9B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
The interest on a certain sum of money is 0.24 times of itself in 3 years. Find the rate of interest.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -21

Question 2.
If ₹ 3,750 amount to ₹ 4,620 in 3 years at simple interest. Find:
(i) the rate of interest
(ii) the amount of Rs. 7,500 in 5\(\frac { 1 }{ 2 }\) years at the same rate of interest
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -22

Question 3.
A sum of money, lent out at simple interst, doubles itself in 8 years. Find :
(i) the rate of interest
(ii) in how many years will the sum become triple (three times) of itself at the same rate per cent ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -24

Question 4.
Rupees 4000 amount to Rs.5000 in 8 years ; in what time will Rs.2100 amount to Rs.2800 at the same rate ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -25

Question 5.
What sum of money lent at 6.5% per annum will produce the same interest in 4 years as Rs.7500 produce in 6 years at 5% per annum ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -27

Question 6.
A certain sum amounts to Rs.3825 in 4 years and to Rs.4050 in 6 years. Find the rate percent and the sum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -29

Question 7.
At what rafepercent of simple interest will the interest on Rs.3750 be one-fifth of itself in 4 years ? To what will it amount in 15 years ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -30

Question 8.
On what date will ₹ 1950 lent on 5th January, 2011 amount to ₹ 2125.50 at 5 percent per annum simple interest?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -32

Question 9.
If the interest on Rs.2400 be more than the interest on Rs.2000 by Rs.60 in 3 years at the same rate percent ; find the rate.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -34

Question 10.
Divide Rs. 15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at 4\(\frac { 1 }{ 2 }\) per cent for 6 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -36

Simple and Compound Interest Exercise 9C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A sum of Rs. 8,000 is invested for 2 years at 10% per annum compound interest. Calculate:
(i) interest for the first year.
(ii) principal for the second year.
(iii) interest for the second year.
(iv) final amount at the end of second year
(v) compound interest earned in 2 years.
Solution:
(i) Here Principal (P) = Rs. 8,000
Rate of interest = 10%
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -37

Question 2.
A man borrowed Rs. 20,000 for 2 years at 8% per year compound interest. Calculate :
(i) the interest of the first year.
(ii) the interest of the second year.
(iii) the final amount at the end of second year.
(iv) the compound interest of two years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -38

Question 3.
Calculate the amount and the compound interest on Rs. 12,000 in 2 years and at 10% per year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -39

Question 4.
Calculate the amount and the compound interest on Rs. 10,000 in 3 years at 8% per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -40

Question 5.
Calculate the compound interest on Rs. 5,000 in 2 years ; if the rates of interest for successive years be 10% and 12% respectively.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -42

Question 6.
Calculate the compound interest on Rs. 15,000 in 3 years ; if the rates of interest for successive years be 6%, 8% and 10% respectively.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -43

Question 7.
Mohan borrowed Rs. 16,000 for 3 years at 5% per annum compound interest. Calculate the amount that Mohan will pay at the end of 3 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -45

Question 8.
Rekha borrowed Rs. 40,000 for 3 years at 10% per annum compound interest. Calculate the interest paid by her for the second year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -47

Question 9.
Calculate the compound interest for the second year on Rs. 15000 invested for 5 years at 6% per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -48
= 159 x 6 = Rs. 954

Question 10.
A man invests Rs. 9600 at 10% per annum compound interest for 3 years. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of the first year.
(iii) the interest for the second year.
(iv) the interest for the third year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -49

Question 11.
A person invests Rs. 5,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. 5,600. Calculate :
(i) the rate of interest per year.
(ii) the amount at the end of the second year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -50

Question 12.
Calculate the difference between the compound interest and the simple interest on ₹ 7,500 in two years and at 8% per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -52

Question 13.
Calculate the difference between the compound interest and the simple interest on ₹ 8,000 in three years and at 10% per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -53

Question 14.
Rohit borrowed ₹ 40,000 for 2 years at 10% per annum C.I. and Manish borrowed the same sum for the same time at 10.5% per annum simple interest. Which of these two gets less interest and by how much?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -54

Question 15.
Mr. Sharma borrowed ₹ 24,000 at 13% p.a. simple interest and an equal sum at 12% p.a. compound interest. Find the total interest earned by Mr. Sharma in 2 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -56

Question 16.
Peter borrows ₹ 12,000 for 2 years at 10% p.a. compound interest. He repays ₹ 8,000 at the end of first year. Find:
(i) the amount at the end of first year, before making the repayment.
(ii) the amount at the end of first year, after making the repayment.
(iii) the principal for the second year.
(iv) the amount to be paid at the end of second year, to clear the account.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -58

Question 17.
Gautam takes a loan of ₹ 16,000 for 2 years at 15% p.a. compound interest. He repays ₹ 9,000 at the end of first year. How mucH must he pay at the end of second year to clear the debt?
Solution:
Loan taken (P) = ₹ 16000
Rate (R) = 15% p.a.
Time (T) = 2 years
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -59

Question 18.
A certain sum of money, invested for 5 years at 8% p.a. simple interest, earns an interest of ₹ 12,000. Find:
(i) the sum of money.
(ii) the compound interest earned by this money in two years and at 10% p.a. compound interest.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -60

Question 19.
Find the amount and the C.I. on ₹ 12,000 at 10% per annum compounded half-yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -62

Question 20.
Find the amount and the C.I. on ₹ 8,000 in 1\(\frac { 1 }{ 2 }\) years at 20% per year compounded half- yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -63

Question 21.
Find the amount and the compound interest on ₹ 24,000 for 2 years at 10% per annum compounded yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -65

Question 22.
Find the amount and the compound interest on ₹ 16,000 for 3 years at 5% per annum compounded annually.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -66

Question 23.
Find the amount and the compound interest on ₹ 20,000 for 1\(\frac { 1 }{ 2 }\) years at 10% per annum compounded half-yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -68

Question 24.
Find the amount and the compound interest on ₹ 32,000 for 1 year at 20% per annum compounded half-yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -69

Question 25.
Find the amount and the compound interest on ₹ 4,000 in 2 years, if the rate of interest for first year is 10% and for the second year is 15%.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -71

Question 26.
Find the amount and the compound interest on ₹ 10,000 in 3 years, if the rates of interest for the successive years are 10%, 15% and 20% respectively.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -72

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots (Including use of tables for natural numbers)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots (Including use of tables for natural numbers)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 4 Cubes and Cube-Roots (Including use of tables for natural numbers). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Cubes and Cube-Roots Exercise 4A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the cube of :
(i) 7
(ii) 11
(iii) 16
(iv) 23
(v) 31
(vi) 42
(vii) 54
Solution:
(i) (7)³ = 7 x 7 x 7 = 343
(ii) (11)³ =11 x 11 x 11 = 1331
(iii) (16)³ = 16 x 16 x 16 = 4096
(iv) (23)³ = 23 x 23 x 23 = 12167
(v) (31)³ = 31 x 31 x 31 = 29791
(vi) (42)³ = 42 x 42 x 42 = 74088
(vii) (54)³ = 54 x 54 x 54 = 157464

Question 2.
Find which of the following are perfect cubes :
(i) 243
(ii) 588
(iii) 1331
(iv) 24000
(v) 1728
(vi) 1938
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -1

1938 = 2 x 3 x 17 x 19
1938 is not a perfect cube.

Question 3.
Find the cubes of :
(i) 2.1
(ii) 0.4
(iii) 1.6
(iv) 2.5
(v) 0.12
(vi) 0.02
(vii) 0.8
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -4

Question 4.
Find the cubes of :
(i) \(\frac { 3 }{ 7 }\)
(ii) \(\frac { 8 }{ 9 }\)
(iii) \(\frac { 10 }{ 13 }\)
(iv) \(1\frac { 2 }{ 7 }\)
(v) \(2\frac { 1 }{ 2 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -7

Question 5.
Find the cubes of :
(i) -3
(ii) -7
(iii) -12
(iv) -18
(v) -25
(vi) -30
(vii) -50
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -8

Question 6.
Which of the following are cubes of:
(i) an even number
(ii) an odd number
216, 729, 3375, 8000, 125, 343, 4096 and 9261.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -10

(i) Cubes of an even number are 216, 8000, 4096.
(ii) Cubes of an odd number are 729, 3375, 125, 343, 9261.

Question 7.
Find the least number by which 1323 must be multiplied so that the product is a perfect cube.
Solution:
The prime factor of 1323 are =3 x 3 x 3 x 7 x 7
= (3 x 3 x 3) x 7 x 7
Clearly, 1323 must be multiplied by 7.

Question 8.
Find the smallest number by which 8768 must be divided so that the quotient is a perfect cube.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -13

Question 9.
Find the smallest number by which 27783 be multiplied to get a perfect square number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -14

Question 10.
With what least number must 8640 be divided so that the quotient is a perfect cube?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -15

Question 11.
Which is the smallest number that must be multiplied to 77175 to make it a perfect cube?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -17

Cubes and Cube-Roots Exercise 4B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the cube-roots of :
(i) 64
(ii) 343
(iii) 729
(iv) 1728
(v) 9261
(vi) 4096
(vii) 8000
(viii) 3375
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -18

Question 2.
Find the cube-roots of :
(i) \(\frac { 27 }{ 64 }\)
(ii) \(\frac { 125 }{ 216 }\)
(iii) \(\frac { 343 }{ 512 }\)
(iv) 64 x 729
(v) 64 x 27
(vi) 729 x 8000
(vii) 3375 x 512
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -23

Question 3.
Find the cube-roots of :
(i) -216
(ii) -512
(iii) -1331
(iv) \(\frac { -27 }{ 125 }\)
(v) \(\frac { -64 }{ 343 }\)
(vi) \(\frac { -512 }{ 343 }\)
(vii) -2197
(viii) -5832
(ix) -2744000
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -24

Question 4.
Find the cube-roots of :
(i) 2.744
(ii) 9.261
(iii) 0.000027
(iv) -0.512
(v) -15.625
(vi) -125 x 1000
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -28

Question 5.
Find the smallest number by which 26244 may be divided so that the quotient is a perfect cube.
Solution:
The prime factors of 26244 are
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -31
Clearly, 26244 must be divided by
3 x 3 x 2 x 2 = 36

Question 6.
What is the least number by which 30375 should be multiplied to get a perfect cube?
Solution:
The prime factors of 30375 are
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -32

Question 7.
Find the cube-roots of :
(i) 700 x 2 x 49 x 5
(ii) -216 x 1728
(iii) -64 x -125
(iv) \(\frac { -27 }{ 343 }\)
(v) \(\frac { 729 }{ -1331 }\)
(vi) 250.047
(vii) -175616
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -33

Selina Concise Physics Class 8 ICSE Solutions – Energy

August 11, 2022January 27, 2019 by Kalyan

Selina Concise Physics Class 8 ICSE Solutions – Energy

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 4 Energy. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 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 8 Physics Chapter – 4 – Energy

When we are pushing a wall we are not doing any work as the position of wall is not change i.e. wall has not moved in the direction of force.
WORK: “is said to be done if on applying force on a body, the body moves (or changes it position) from it place in the direction of force. W = F × d
Or
“Work is said to be done by a force applied on a body, if it changes its size or shape.”
FACTORS AFFECTING THE AMOUNT OF WORK DONE : W = F × d
(i) Magnitude of force applied.
(ii) Distance moved by the body in the direction of force.
UNIT OF WORK : W = F × d
s.i unit W= 1N × 1m = Nm = joule (J)
1 kgf = 9.8 N is force on 1 kg ∴ F = mg
Work done – 1 kgf × m = 9.8 N m = 9.8 J = 10 J nearly
A cooli standing with a box on his head, does no work as distance moved is zero. ~
A cooli with a box on his head and walking is doing no work as force is acting vertically downward and direction of motion is at right angle.
ENERGY: “is capacity of doing work.”
Or
“The work done on a body in changing its state is called energy.”
s.1. unit of energy = S.I. unit of work = (J)
JOULE: “A body is said to possess a energy of one joule. If a force of 1 Newton moves the body by a distance of 1 metre in the direction of force.”
MECHANICAL ENERGY: “The energy possessed by a body due to its state of rest or state of motion is called mechanical energy.
Potential energy and kinetic energy are mechanical energies.
POTENTIAL Energy (P.E.) : “Is energy possessed by body due to its state of rest or position.” P.E. = mgh
KINETIC Energy (K.E.) : “Is energy possessed by body due to its motion.”
K.E. = 1/2 M V²
GRAVITATIONAL POTENTIAL ENERGY: “When a stone or water is raised (lifted) from ground to a height, work is done against the force of gravity. This work is stored in the stone or water in the form of GRAVITATIONAL POTENTIAL ENERGY.”
A stretched bow, due to change in position possesses potential energy. When stretched bow is relreased the arrow comes in motion and due to motion possesses the kinetic energy and hits the body on which it strikes.
When a body at a hight, it possess P.E. = mgh. When it falls, height decreases and speed increases
∴ its P.E. decreases and K.E. increases.
Powder : “Rate of doing work”. P = W/t

Test your self

A.Objective Questions

1. Write true or false for each statement

(a) A coolie does no work against the force of gravity while carrying a luggage on a road.
Answer. True.

(b) The energy stored in water of a dam is the kinetic energy.
Answer. False.
The energy stored in water of a dam is the potential energy.

(c) The energy of a flying kite is kinetic energy.
Answer. True.

(d) Work done by a boy depends on the time in which he does work.
Answer. False.

(e) Power spent by a body depends on the time for which it does work.
Answer. True.

2. Fill in the blanks

(a) Work is said to be done by a forte only when the body moves.
(b) Work done = Force × distance moved in direction of force.
(c) The energy of a body is its capacity to do work.
(d) The S.I. unit of energy is joule.
(e) The potential energy of a body is due to its state of rest or position and kinetic energy of body is due to its state of motion.
(f) Gravitational potential energy U = mass × force of gravity on unit mass × height.
(g) Kinetic energy = 1/2 × mass × (speed)²
(h) Power P=work done/time taken.
(i) The S . i. unit of power is watt
(j) I H.P. = 746 W

3. Match the following
Selina Concise Physics Class 8 ICSE Solutions Chapter 4 Energy 2

4. Select the correct alternative

(a) The S.I. unit of work is

second
metre
joule
newton

Answer:
joule

(b) No work is done by a force if the body

moves in direction of force
does not move
moves in opposite direction
none of the these

Answer:
does not move

(c) Two coolies A and B do some work in time 1 minute and 2 minute respectively. The power spent is

same by both coolies
is more by coolie A than by B
is less by coolie A than by B
nothing can be said.

Answer:
is more by coolie A than by B

(d) The expression of power P is

P = mgh
P = P = 1/2 Mv²
P = F × d
P = F × d/t

Answer:
P = F × d/t

(e) I H.P. ¡s equal to

1 W
1 J
746 J
746 W

Answer:
746 W

(f) When a boy doubles his speed, his kinetic energy becomes

half
double
four times
no change

Answer:
four times

(g) A boy lifts a luggage from height 2 m to 4 m. The potential energy will become

half
double
one-third
one-fourth

Answer:
double

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes (Including Polygons)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes (Including Polygons)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 16 Understanding Shapes (Including Polygons). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths Solutions Physics Chemistry Biology Geography History & Civics

Understanding Shapes Exercise 16A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
State which of the following are polygons :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 1
If the given figure is a polygon, name it as convex or concave.
Solution:
Only Fig. (ii), (iii) and (v) are polygons.
Fig. (ii) and (iii) are concave polygons while
Fig. (v) is convex.

Question 2.
Calculate the sum of angles of a polygon with :
(i) 10 sides
(ii) 12 sides
(iii) 20 sides
(iv) 25 sides
Solution:
(i) No. of sides n = 10
sum of angles of polygon = (n – 2) x 180°
= (10 – 2) x 180° = 1440°
(ii) no. of sides n = 12
sum of angles = (n – 2) x 180°
= (12 – 2) x 180° = 10 x 180° = 1800°
(iii) n = 20
Sum of angles of Polygon = (n – 2) x 180°
= (20 – 2) x 180° = 3240°
(iv) n = 25
Sum of angles of polygon = (n – 2) x 180°
= (25 – 2) x 180° = 4140°

Question 3.
Find the number of sides in a polygon if the sum of its interior angles is :
(i) 900°
(ii) 1620°
(iii) 16 right-angles
(iv) 32 right-angles.
Solution:
(i) Let no. of sides = n
Sum of angles of polygon = 900°
(n – 2) x 180° = 900°
n – 2 = \(\frac { 900 }{ 180 }\)
n – 2 = 5
n = 5 + 2
n = 7
(ii) Let no. of sides = n
Sum of angles of polygon = 1620°
(n – 2) x 180° = 1620°
n – 2 = \(\frac { 1620 }{ 180 }\)
n – 2 = 9
n = 9 + 2
n = 11
(iii) Let no. of sides = n
Sum of angles of polygon = 16 right angles = 16 x 90 = 1440°
(n – 2) x 180° = 1440°
n – 2 = \(\frac { 1440 }{ 180 }\)
n – 2 = 8
n = 8 + 2
n = 10
(iv) Let no. of sides = n
Sum of angles of polygon = 32 right angles = 32 x 90 = 2880°
(n – 2) x 180° = 2880
n – 2 = \(\frac { 2880 }{ 180 }\)
n – 2 = 16
n = 16 + 2
n = 18

Question 4.
Is it possible to have a polygon ; whose sum of interior angles is :
(i) 870°
(ii) 2340°
(iii) 7 right-angles
(iv) 4500°
Solution:
(i) Let no. of sides = n
Sum of angles = 870°
(n – 2) x 180° = 870°
n – 2 = \(\frac { 870 }{ 180 }\)
n – 2 = \(\frac { 29 }{ 6 }\)
n = \(\frac { 29 }{ 6 }\) + 2
n = \(\frac { 41 }{ 6 }\)
Which is not a whole number.
Hence it is not possible to have a polygon, the sum of whose interior angles is 870°
(ii) Let no. of sides = n
Sum of angles = 2340°
(n – 2) x 180° = 2340°
n – 2 = \(\frac { 2340 }{ 180 }\)
n – 2 = 13
n = 13 + 2 = 15
Which is a whole number.
Hence it is possible to have a polygon, the sum of whose interior angles is 2340°.
(iii) Let no. of sides = n
Sum of angles = 7 right angles = 7 x 90 = 630°
(n – 2) x 180° = 630°
n – 2 = \(\frac { 630 }{ 180 }\)
n – 2 = \(\frac { 7 }{ 2 }\)
n = \(\frac { 7 }{ 2 }\) + 2
n = \(\frac { 11 }{ 2 }\)
Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles.
(iv) Let no. of sides = n
(n – 2) x 180° = 4500°
n – 2 = \(\frac { 4500 }{ 180 }\)
n – 2 = 25
n = 25 + 2
n = 27
Which is a whole number.
Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.

Question 5.
(i) If all the angles of a hexagon are equal ; find the measure of each angle.
(ii) If all the angles of a 14-sided figure are equal ; find the measure of each angle.
Solution:
(i) No. of sides of hexagon, n = 6
Let each angle be = x°
Sum of angles = 6x°
(n – 2) x 180° = Sum of angles
(6 – 2) x 180° = 6x°
4 x 180 = 6x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 2

Question 6.
Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with :
(i) 7 sides
(ii) 10 sides
(iii) 250 sides.
Solution:
(i) No. of sides n = 7
Sum of interior & exterior angles at one vertex = 180°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 3

Question 7.
The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are (6x – 1)°, (10x + 2)°, (8x + 2)° (9x – 3)°, (5x + 4)° and (12x + 6)° ; find each exterior angle.
Solution:
Sum of exterior angles of hexagon formed by producing sides of order = 360°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 5
i.e. 41° ; 72°, 58° ; 60° ; 39° and 90°

Question 8.
The interior angles of a pentagon are in the ratio 4 : 5 : 6 : 7 : 5. Find each angle of the pentagon.
Solution:
Let the interior angles of the pentagon be 4x, 5x, 6x, 7x, 5x.
Their sum = 4x + 5x + 6x + 7x + 5x = 21x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 6

Question 9.
Two angles of a hexagon are 120° and 160°. If the remaining four angles are equal, find each equal angle.
Solution:
Two angles of a hexagon are 120°, 160°
Let remaining four angles be x, x, x and x.
Their sum = 4x + 280°
But sum of all the interior angles of a hexagon
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 7

Question 10.
The figure, given below, shows a pentagon ABCDE with sides AB and ED parallel to each other, and ∠B : ∠C : ∠D = 5 : 6 : 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 8
(i) Using formula, find the sum of interior angles of the pentagon.
(ii) Write the value of ∠A + ∠E
(iii) Find angles B, C and D.
Solution:
(i) Sum of interior angles of the pentagon

Question 11.
Two angles of a polygon are right angles and the remaining are 120° each. Find the number of sides in it.
Solution:
Let number of sides = n
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 10
n = \(\frac { 300 }{ 60 }\)
n = 5

Question 12.
In a hexagon ABCDEF, side AB is parallel to side FE and ∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3. Find ∠B and ∠D.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 11

Question 13.
the angles of a hexagon are x + 10°, 2x + 20°, 2x – 20°, 3x – 50°, x + 40° and x + 20°. Find x.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 12

Question 14.
In a pentagon, two angles are 40° and 60°, and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.
Solution:
In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles = 3 x 180°
= 540° – 40° – 60° = 540° – 100° = 440°
Ratio in these 3 angles =1 : 3 : 7
Sum of ratios =1 + 3 + 7 = 11
Biggest angle = \(\frac { 440\times 7 }{ 11 }\) = 280°

Understanding Shapes Exercise 16B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Fill in the blanks :
In case of regular polygon, with :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 14
Solution:

Question 2.
Find the number of sides in a regular polygon, if its each interior angle is :
(i) 160°
(ii) 135°
(iii) \(1\frac { 1 }{ 5 }\) of a right-angle
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 17

Question 3.
Find the number of sides in a regular polygon, if its each exterior angle is :
(i) \(\frac { 1 }{ 3 }\) of a right angle
(ii) two-fifth of a right-angle.
Solution:
(i) Each exterior angle = \(\frac { 1 }{ 3 }\) of a right angle
= \(\frac { 1 }{ 3 }\) x 90
= 30°
Let number of sides = n
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 19

Question 4.
Is it possible to have a regular polygon whose each interior angle is :
(i) 170°
(ii) 138°
Solution:
(i) No. of sides = n
each interior angle = 170°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 20

Which is not a whole number.
Hence it is not possible to have a regular polygon having interior angle of 138°.

Question 5.
Is it possible to have a regular polygon whose each exterior angle is :
(i) 80°
(ii) 40% of a right angle.
Solution:
(i) Let no. of sides = n each exterior angle = 80°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 22
Which is not a whole number.
Hence it is not possible to have a regular polygon whose each exterior angle is of 80°.
(ii) Let number of sides = n
Each exterior angle = 40% of a right angle

Which is a whole number.
Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.

Question 6.
Find the number of sides in a regular polygon, if its interior angle is equal to its exterior angle.
Solution:
Let each exterior angle or interior angle be = x°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 24

Question 7.
The exterior angle of a regular polygon is one-third of its interior angle. Find the number of sides in the polygon.
Solution:
Let interior angle = x°
Exterior angle = \(\frac { 1 }{ 3 }\) x°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 25

Question 8.
The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find :
(i) measure of each interior angle ;
(ii) measure of each exterior angle and
(iii) number of sides in the polygon.
Solution:
Let exterior angle = x°
Interior angle = 5x°
x + 5x = 180°
6x = 180°
x = 30°
Each exterior angle = 30°
Each interior angle = 5 x 30° = 150°
Let no. of sides = n
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 27

Question 9.
The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :
(i) each exterior angle of the polygon ;
(ii) number of sides in the polygon
Solution:
Interior angle : exterior angle = 2 : 1
Let interior angle = 2x° & exterior angle = x°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 28

Question 10.
The ratio between the exterior angle and the interior angle of a regular polygon is 1 : 4. Find the number of sides in the polygon.
Solution:
Let exterior angle = x° & interior angle = 4x°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 29

Question 11.
The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.
Solution:
Let number of sides = n
Sum of exterior angles = 360°
Sum of interior angles = 360° x 2 = 720°
Sum of interior angles = (n – 2) x 180°
720° = (n – 2) x 180°
n – 2 = \(\frac { 720 }{ 180 }\)
n – 2 = 4
n = 4 + 2
n = 6

Question 12.
AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = 20° ; find :
(i) its each interior angle,
(ii) its each exterior angle
(iii) the number of sides in the polygon.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 30

Question 13.
Two alternate sides of a regular polygon, when produced, meet at the right angle. Calculate the number of sides in the polygon.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 31

Question 14.
In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of:
(i) ∠BAE
(ii) ∠ABE
(iii) ∠BED
Solution:
(i) Since number of sides in the pentagon = 5
Each exterior angle = \(\frac { 360 }{ 5 }\) = 72°
∠BAE = 180° – 72°= 108°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 32

Question 15.
The difference between the exterior angles of two regular polygons, having the sides equal to (n – 1) and (n + 1) is 9°. Find the value of n.
Solution:
We know that sum of exterior angles of a polynomial is 360°
(i) If sides of a regular polygon = n – 1
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 33

Question 16.
If the difference between the exterior angle of a n sided regular polygon and an (n + 1) sided regular polygon is 12°, find the value of n.
Solution:
We know that sum of exterior angles of a polygon = 360°
Each exterior angle of a regular polygon of 360°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 35

Question 17.
The ratio between the number of sides of two regular polygons is 3 : 4 and the ratio between the sum of their interior angles is 2 : 3. Find the number of sides in each polygon.
Solution:
Ratio of sides of two regular polygons = 3 : 4
Let sides of first polygon = 3n
and sides of second polygon = 4n
Sum of interior angles of first polygon
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 37

Question 18.
Three of the exterior angles of a hexagon are 40°, 51 ° and 86°. If each of the remaining exterior angles is x°, find the value of x.
Solution:
Sum of exterior angles of a hexagon = 4 x 90° = 360°
Three angles are 40°, 51° and 86°
Sum of three angle = 40° + 51° + 86° = 177°
Sum of other three angles = 360° – 177° = 183°
Each angle is x°
3x = 183°
x = \(\frac { 183 }{ 3 }\)
Hence x = 61

Question 19.
Calculate the number of sides of a regular polygon, if:
(i) its interior angle is five times its exterior angle.
(ii) the ratio between its exterior angle and interior angle is 2 : 7.
(iii) its exterior angle exceeds its interior angle by 60°.
Solution:
Let number of sides of a regular polygon = n
(i) Let exterior angle = x
Then interior angle = 5x
x + 5x = 180°
=> 6x = 180°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 38

Question 20.
The sum of interior angles of a regular polygon is thrice the sum of its exterior angles. Find the number of sides in the polygon.
Solution:
Sum of interior angles = 3 x Sum of exterior angles
Let exterior angle = x
The interior angle = 3x
x + 3x=180°
=> 4x = 180°
=> x = \(\frac { 180 }{ 4 }\)
=> x = 45°
Number of sides = \(\frac { 360 }{ 45 }\) = 8

Understanding Shapes Exercise 16C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.
Solution:
Let the other angle = x°
According to given,
89° + 113° + x° + x° = 360°
2x° = 360° – 202°
2x° = 158°
x° = \(\frac { 158 }{ 2 }\)
other two angles = 79° each

Question 2.
Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5 : 7; find the measure of each of them.
Solution:
Two angles are 68° and 76°
Let other two angles be 5x and 7x
68° + 76°+ 5x + 7x = 360°
12x + 144° = 360°
12x = 360° – 144°
12x = 216°
x = 18°
angles are 5x and 7x
i.e. 5 x 18° and 7 x 18° i.e. 90° and 126°

Question 3.
Angles of a quadrilateral are (4x)°, 5(x+2)°, (7x – 20)° and 6(x+3)°. Find :
(i) the value of x.
(ii) each angle of the quadrilateral.
Solution:
Angles of quadrilateral are,
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 40

Question 4.
Use the information given in the following figure to find :
(i) x
(ii) ∠B and ∠C
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 42
Solution:

Question 5.
In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4 : 5
(i) Calculate each angle of the quadrilateral.
(ii) Assign a special name to quadrilateral ABCD
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 44

Question 6.
From the following figure find ;
(i) x
(ii) ∠ABC
(iii) ∠ACD
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 45
(i) In Quadrilateral ABCD,
x + 4x + 3x + 4x + 48° = 360°
12x = 360° – 48°
12x = 312

Question 7.
Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ; ∠A = 5(a+2)° and ∠B = 2(2a+7)°.
Calculate ∠A.
Solution:
∠C = 64° (Given)
∠D = ∠C – 8° = 64°- 8° = 56°
∠A = 5(a+2)°
∠B = 2(2a+7)°
Now ∠A + ∠B + ∠C + ∠D = 360°
5(a+2)° + 2(2a+7)° + 64° + 56° = 360°
5a + 10 + 4a + 14° + 64° + 56° = 360°
9a + 144° = 360°
9a = 360° – 144°
9a = 216°
a = 24°
∠A = 5 (a + 2) = 5(24+2) = 130°

Question 8.
In the given figure : ∠b = 2a + 15 and ∠c = 3a + 5; find the values of b and c.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 46
Solution:
Stun of angles of quadrilateral = 360°
70° + a + 2a + 15 + 3a + 5 = 360°
6a + 90° = 360°
6a = 270°
a = 45°
b = 2a + 15 = 2 x 45 + 15 = 105°
c = 3a + 5 = 3 x 45 + 5 = 140°
Hence ∠b and ∠c are 105° and 140°

Question 9.
Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.
Solution:
Let each equal angle be x°
x + x + x + 69° = 360°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 47
3x = 360°- 69
3x = 291
x = 97°
Each, equal angle = 97°

Question 10.
In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.
Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other
(i) Is PS also parallel to QR ?
(ii) Assign a special name to quadrilateral PQRS.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 48

Question 11.
Use the informations given in the following figure to find the value of x.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 49
Solution:
Take A, B, C, D as the vertices of Quadrilateral and BA is produced to E (say).
Since ∠EAD = 70°
∠DAB = 180° – 70°= 110°
[EAB is a straight line and AD stands on it ∠EAD+ ∠DAB = 180°]
110° + 80° + 56° + 3x – 6° = 360°
[sum of interior angles of a quadrilateral = 360°]
3x = 360° – 110° – 80° – 56° + 6°
3x = 360° – 240° = 120°
x = 40°

Question 12.
The following figure shows a quadrilateral in which sides AB and DC are parallel. If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 50
Solution:
Let ∠A = 4x
∠D = 5x
Since ∠A + ∠D = 180° [AB||DC]
4x + 5x = 180°
=> 9x = 180°
=> x = 20°
∠A = 4 (20) = 80°,
∠D = 5 (20) = 100°
Again ∠B + ∠C = 180° [ AB||DC]
3x – 15° + 4x + 20° = 180°
7x = 180° – 5°
=> 7x = 175°
=> x = 25°
∠B = 75° – 15° = 60°
and ∠C = 4 (25) + 20 = 100°+ 20°= 120°

Question 13.
Use the following figure to find the value of x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 51
Solution:
The sum of exterior angles of a quadrilateral

=> y + 80° + 60° + 90° = 360°
=> y + 230° = 360°
=> y = 360° – 230° = 130°
At vertex A,
∠y + ∠x = 180° (Linear pair)
x = 180° – 130°
=> x = 50°

Question 14.
ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 53
Given : ABCDE is a regular pentagon.
The bisector ∠A of the pentagon meets the side CD at point M.
To prove : ∠AMC = 90°
Proof: We know that, the measure of each interior angle of a regular pentagon is 108°.
∠BAM = \(\frac { 1 }{ 2 }\) x 108° = 54°
Since, we know that the sum of a quadrilateral is 360°
In quadrilateral ABCM, we have
∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°
54° + 108° + 108° + ∠AMC = 360°
∠AMC = 360° – 270°
∠AMC = 90°

Question 15.
In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively. Show that:
∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
Solution:
Given : AO and BO are the bisectors of ∠A and ∠B respectively.
∠1 = ∠4 and ∠3 = ∠5 ……..(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 54
To prove : ∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
Proof: In quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
\(\frac { 1 }{ 2 }\) (∠A + ∠B + ∠C + ∠D) = 180° …………(ii)
Now in ∆AOB
∠1 + ∠2 + ∠3 = 180° ………(iii)
Equating equation (ii) and equation (iii), we get
∠1 + ∠2 + ∠3 = ∠A + ∠B + \(\frac { 1 }{ 2 }\) (∠C + ∠D)
∠1 + ∠2 + ∠3 = ∠1 + ∠3 + \(\frac { 1 }{ 2 }\) (∠C + ∠D)
∠2 = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
Hence proved.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 12 Algebraic Identities. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Algebraic Identities Exercise 12A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Use direct method to evaluate the following products :
(i) (x + 8)(x + 3)
(ii) (y + 5)(y – 3)
(iii) (a – 8)(a + 2)
(iv) (b – 3)(b – 5)
(v) (3x – 2y)(2x + y)
(vi) (5a + 16)(3a – 7)
(vii) (8 – b) (3 + b)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -1

Question 2.
Use direct method to evaluate :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -3
Solution:

Question 3.
Evaluate :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -7
Solution:

Question 4.
Use the product (a + b) (a – b) = a² – b² to evaluate:
(i) 21 x 19
(ii) 33 x 27
(iii) 103 x 97
(iv) 9.8 x 10.2
(v) 7.7 x 8.3
(vi) 4.6 x 5.4
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -10

Question 5.
Evaluate :
(i) (6 – xy) (6 + xy)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -11
Solution:

Algebraic Identities Exercise 12B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Expand :
(i) (2a + b)²
(ii) (a – 2b)²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -15
Solution:

Question 2.
Find the square of :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -19
Solution:

Question 3.
Evaluate:
Using expansion of (a + b)² or (a – b)²
(i) (208)²
(ii) (92)²
(iii)(415)²
(iv) (188)²
(v) (9.4)²
(vi) (20.7)²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -23

Question 4.
Expand :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -25
Solution:

Question 5.
Find the cube of :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -28
Solution:

Algebraic Identities Exercise 12C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
If a+b=5 and ab = 6; find a² + b²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -31

Question 2.
If a – b = 6 and ab = 16; find a² + b²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -32

Question 3.
If a² + b² = 29 and ab = 10 ; find :
(i) a + b
(ii) a – b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -33

Question 4.
If a² + b²= 10 and ab = 3; find :
(i) a – b
(ii) a + b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -34

Question 5.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -35
Solution:

Question 6.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -38
Solution:

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -41
Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -42
Solution:

Question 9.
If a + b + c = 10 and a² + b² + c² = 38; find ab + bc + ca
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -46

Question 10.
Find a² + b² + c² ; if a + b + c = 9 and ab + bc + ca = 24
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -47

Question 11.
Find a + b + c; if a² + b² + c² = 83 and ab + bc + ca = 71
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -49

Question 12.
If a + b = 6 and ab=8; find a³ + b³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -50

Question 13.
If a – b=3 and ab = 10; find a³ – b³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -51

Question 14.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -53
Solution:

Question 15.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -55
Solution:

Question 16.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -58
Solution:

Question 17.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -62
Solution:

Question 18.
The sum of the squares of two numbers is 13 and their product is 6. Find:
(i) the sum of the two numbers.
(ii) the difference between them.
Solution:
Let x and y be the two numbers, then
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -65

Algebraic Identities Exercise 12D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -66
Solution:

Question 2.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -69
Solution:

Question 3.
Find the square of :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -75
Solution:

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -78
Solution:

Question 5.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -80
Solution:

Question 6.
If a² + b² = 41 and ab = 4, find :
(i) a – b
(ii) a + b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -82

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -83
Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -85.
Solution:

Question 9.
Expand :
(i) (3x – 4y + 5z)²
(ii) (2a – 5b – 4c)²
(iii) (5x + 3y)³
(iv) (6a – 7b)³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -87

Question 10.
If a + b + c = 9 and ab + bc + ca = 15, find: a² + b² + c².
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -89

Question 11.
If a + b + c = 11 and a² + b² + c² = 81, find ab + bc + ca.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -90

Question 12.
If 3x – 4y = 5 and xy = 3, find : 27x³ – 64y³.
Solution:
27x³ – 64x³ = (3x)³ – (4y)³
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -91

Question 13.
If a + b = 8 and ab = 15, find : a³ + b³.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -92

Question 14.
If 3x + 2y = 9 and xy = 3, find : 27x³ + 8y³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -93

Question 15.
If 5x – 4y = 7 and xy = 8, find : 125x³ – 64y³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -94

Question 16.
The difference between two numbers is 5 and their products is 14. Find the difference between their cubes.
Solution:
Let x and y be two numbers, then x – y = 5 and xy = 14
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -95

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 3 Squares and Square Roots. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square of :
(i) 59
(ii) 63
(iii) 15
Solution:
(i) Square of 59= 59 x 59 = 3481
(ii) Square of 6.3 = 6.3 x 6.3 = 39.69
(iii) Square of 15 = 15 x 15 = 225

Question 2.
By splitting into prime factors, find the square root of :
(i) 11025
(if) 396900
(iii) 194481
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -1

Question 3.
(i) Find the smallest number by which 2592 be multiplied so that the product is a perfect square.
(ii) Find the smallest number by which 12748 be mutliplied so that the product is a perfect square?
Solution:

On grouping the prime factors of 2592 as shown; on factor i.e. 2 is left which cannot be paired with equal factor.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -4
The given number should be multiplied by 2 to make the given number a perfect square.

On grouping the prime factors of 12748 as shown; one factor i.e. 3187 is left which cannot be paired with equal factor.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -6
The given number should be multiplied by 3187.

Question 4.
Find the smallest number by which 10368 be divided, so that the result is a perfect square. Also, find the square root of the resulting numbers.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -7

Question 5.
Find the square root of :
(i) 0.1764
(ii) \(96\frac { 1 }{ 25 }\)
(iii) 0.0169
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -8

Question 6.
Evaluate
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -9

Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -11

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -14
Solution:

Question 8.
A man, after a tour, finds that he had spent every day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all ₹ 1,296
Solution:
Let the number of days he had spent = x
Number of rupees spent in each day = x
Total money spent = x x x = x² = 1,296 (given)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -18

Question 9.
Out of 745 students, maximum are to be arranged in the school field for a P.T. display, such that the number of rows is equal to the number of columns. Find the number of rows if 16 students were left out after the arrangement.
Solution:
Total number of students = 745
Students left after standing in arrangement = 16
No. of students who were to be arranged = 745 – 16 = 729
The number of rows = no. of students in each row
No. of rows = √729
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -19

Question 10.
13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Find two more such pairs.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -20

Question 11.
Find the smallest perfect square divisible by 3, 4, 5 and 6.
Solution:
L.C.M. of 3, 4, 5, 6 = 2 x 2 x 3 x 5 = 60
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -21
in which 3 and 5 are not in pairs L.C.M. = 2 x 3 x 2 x 5 = 60
We should multiple it by 3 x 5 i.e. by 15
Required perfect square = 60 x 15 = 900

Question 12.
If √784 = 28, find the value of:
(i) √7.84 + √78400
(ii) √0.0784 + √0.000784
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -22

Squares and Square Roots Exercise 3B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square root of:
(i) 4761
(ii) 7744
(iii) 15129
(iv) 0.2916
(v) 0.001225
(vi) 0.023104
(vii) 27.3529
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -23

Question 2.
Find the square root of:
(i) 4.2025
(ii) 531.7636
(iii) 0.007225
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -26

Question 3.
Find the square root of:
(i) 245 correct to two places of decimal.
(ii) 496 correct to three places of decimal.
(iii) 82.6 correct to two places of decimal.
(iv) 0.065 correct to three places of decimal.
(v) 5.2005 correct to two places of decimal.
(vi) 0.602 correct to two places of decimal
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -28

Required square root = 0.78 upto two places of decimals.

Question 4.
Find the square root of each of the following correct to two decimal places:
(i) \(3\frac { 4 }{ 5 }\)
(ii) \(6\frac { 7 }{ 8 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -31

Question 5.
For each of the following, find the least number that must be subtracted so that the resulting number is a perfect square.
(i) 796
(ii) 1886
(iii) 23497
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -33

Question 6.
For each of the following, find the least number that must be added so that the resulting number is a perfect square.
(i) 511
(ii) 7172
(iii) 55078
Solution:
(i) 511
Taking square root of 511, we find that 27 has been left We see that 511 is greater than (22)²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -35
On adding the required number to 511, we get (23)² i.e., 529
So, the required number = 529 – 511 = 18
(ii) 7172
Taking square root of 7172, we find that 116 has been left
We see that 7172 is greater than (84)²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -36
Taking square root of 55078, we find that 322 has been left
We see that 55078 is greater than (234)²
On adding the required number to 55078, we get (235)² i.e., 55225
Required number = 55225 – 55078 = 147

Question 7.
Find the square root of 7 correct to two decimal places; then use it to find the value of \(\sqrt { \frac { 4+\sqrt { 7 } }{ 4-\sqrt { 7 } } }\) correct to three significant digits.
Solution:
√7 = 2.645 = 2.65
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -37

Question 8.
Find the value of √5 correct to 2 decimal places; then use it to find the square root of \(\sqrt { \frac { 3-\sqrt { 5 } }{ 3+\sqrt { 5 } } }\) correct to 2 significant digits.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -38

Question 9.
Find the square root of:
(i) \(\frac { 1764 }{ 2809 }\)
(ii) \(\frac { 507 }{ 4107 }\)
(iii) \(\sqrt { 108\times 2028 }\)
(iv) 0.01 + √0.0064
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -40

Question 10.
Find the square root of 7.832 correct to :
(i) 2 decimal places
(ii) 2 significant digits.
Solution:
Square root of 7.832
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -42
√7.832 = 2.80 upto two decimal places
= 2.8 upto two significant places

Question 11.
Find the least number which must be subtracted from 1205 so that the resulting number is a perfect square.
Solution:
Clearly, if 49 is subtracted from 1205, the number will be a perfect square.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -43

Question 12.
Find the least number which must be added to 1205 so that the resulting number is a perfect square.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -44

Question 13.
Find the least number which must be subtracted from 2037 so that the resulting number is a perfect square.
Solution:
Clearly; if 12 is subtracted from 2037, the remainder will be a perfect square.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -45

Question 14.
Find the least number which must be added to 5483 so that the resulting number is a perfect square.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -46

Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Seeing the value of the digit at unit’s place, state which of the following can be square of a number :
(i) 3051
(ii) 2332
(iii) 5684
(iv) 6908
(v) 50699
Solution:
We know that the ending digit (the digit at units place) of the square of a number is 0, 1, 4, 5, 6, or 9
So, the following numbers can be squares : 3051, 5684, and 50699 i.e., (i), (iii), and (v)

Question 2.
Squares of which of the following numbers will have 1 (one) at their unit’s place :
(i) 57
(ii) 81
(iii) 139
(iv) 73
(v) 64
Solution:
The square of the following numbers will have 1 at their units place as (1)² = 1, (9)² = 81
81 and 139 i.e., (i) and (iii)

Question 3.
Which of the following numbers will not have 1 (one) at their unit’s place :
(i) 322
(ii) 572
(iii) 692
(iv) 3212
(v) 2652
Solution:
The square of the following numbers will not have 1 at their units place : as only (1)² = 1, (9)² = 81 have 1 at then units place
322, 572, 2652 i.e., (i), (ii) and (v)

Question 4.
Square of which of the following numbers will not have 6 at their unit’s place :
(i) 35
(ii) 23
(iii) 64
(iv) 76
(v) 98
Solution:
The squares of the following numbers, Will not have 6 at their units place as only (4)² = 16, (6)² = 36 has but its units place 35, 23 and 98 i.e., (i), (ii), and (v)

Question 5.
Which of the following numbers will have 6 at their unit’s place :
(i) 262
(ii) 492
(iii) 342
(iv) 432
(v) 2442
Solution:
The following numbers have 6 at their units place as (4)² = 16, (6)² = 36 has 6 at their units place 262, 342, 2442 i.e., (i), (iii) and (v)

Question 6.
If a number ends with 3 zeroes, how many zeroes will its square have ?
Solution:
We know that if a number ends with n zeros, then its square will have 2n zeroes at their ends
A number ends with 3 zeroes, then its square will have 3 x 2 = 6 zeroes

Question 7.
If the square of a number ends with 10 zeroes, how many zeroes will the number have ?
Solution:
We know that if a number ends with n zeros Then its square will have 2n zeroes Conversely, if square of a number have 2n zeros at their ends then the number will have n zeroes
The square of a number ends 10 zeroes, then the number will have \(\frac { 10 }{ 2 }\) = 5 zeroes

Question 8.
Is it possible for the square of a number to end with 5 zeroes ? Give reason.
Solution:
No, it is not possible for the square of a number, to have 5 zeroes which is odd because the number of zeros of the square must be 2n zeroes i.e., even number of zeroes.

Question 9.
Give reason to show that none of the numbers, given below, is a perfect square.
(i) 2162
(ii) 6843
(iii) 9637
(iv) 6598
Solution:
A number having 2,3,7 or 8 at the unit place is never a perfect square.

Question 10.
State, whether the square of the following numbers is even or odd?
(i) 23
(ii) 54
(iii) 76
(iv) 75
Solution:
(i) 23 – odd
(ii) 54 – even
(iii) 76 – odd
(iv) 75 – even

Question 11.
Give reason to show that none of the numbers 640, 81000 and 3600000 is a perfect square.
Solution:
No, number has an even number of zeroes.

Question 12.
Evaluate:
(i) 37² – 36²
(ii) 85² – 84²
(iii) 101² – 100²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -47

Question 13.
Without doing the actual addition, find the sum of:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(ii) 1 + 3 + 5 + 7 + 9 + ……………… + 39 + 41
(iii) 1 + 3 + 5 + 7 + 9 + ………………… + 51 + 53
Solution:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23
= Sum of first 12 odd natural numbers = 122 = 144
(ii) 1+3 + 5 + 7 + 9 + ……….. + 39 + 41
= Sum of first 21 odd natural numbers = 212 = 441
(iii) 1 + 3 + 5 + 7 + 9 + ……………. + 51 + 53
= Sum of first 27 odd natural number = 272 = 729

Question 14.
Write three sets of Pythagorean triplets such that each set has numbers less than 30.
Solution:
The three sets of Pythagorean triplets such that each set has numbers less than 30 are 3, 4 and 5; 6, 8 and 10; 5, 12 and 13
Proof:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -49

Selina Concise Physics Class 8 ICSE Solutions – Force and Pressure

August 11, 2022January 27, 2019 by Kalyan

Selina Concise Physics Class 8 ICSE Solutions – Force and Pressure

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 3 Force and Pressure. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 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 8 Physics Chapter – 3 – Force and Pressure

FORCE : “Is the cause which changes the state of a body (rest or
state of motion) or it changes the size or shape of a body”.

Weight of a body → The force with which a body is attracted towards the centre of earth mg = force of gravity.

A force does not change the mass of the body that is why mass of a
body on earth and moon in same but weight → force exerted on
body is different.
Force cannot be seen but it is felt.
→ represents force, length of arrow gives magnitude and arrow points the direction.
S.I. unit of force is Newton (N).
Newton: “Is that much force, which when acting on a body of mass 1 kg produces in it (increases) a speed of 1 M s^-1 in the direction of its motion.
1 kgf = 9.8 N = 10 N (nearly)
RIGID body: “When a force is applied on a body and inter-spacing between its constituent particles do not change is called RIGID body” force can cause only the motion in it.
NON-RIGID body: “When force applied changes inter-spacing.” Force causes both change in its size (shape) and the motion in body.
TURNING EFFECT: “When force is applied on a pivoted (at a point) body, it can turn it and turning of body about point of rotation is called TURNING EFFECT.” or Moment of force.
This is measured as:
TURNING EFFECT = MOMENT OF FORCE
Force x perpendicular distance from point of rotation.

Moment of force = F x OP
S.I. unit of moment of force=N × m=Nm
THRUST: “Force acting normally on a surface.” Smaller the area of surface, larger is thrust.
PRESSURE : “Thrust per unit area.p = Thrust/area = F/A S.I unit of area A pressure is Nm^-2 or pascal (Pa)
If Thrust is measured in kgf and area in Cm², then pressure is expressed as kgf Cm^-2.
ATMOSPHERIC Pressure: 1 atm = 76 cm of mercury column 1 atm = 1.013 x 10⁵ Pa
FACTORS AFFECTING THE PRESSURE : P = F/A
(i) Area: Greater the area, lesser is pressure and lesser area, greater is pressure.
(ii) Magnitude of thrust acting: greater thrust, greater pressure.
Factors Affecting LIQUID PRESSURE = hdg
(i) High of liquid column: increases with height
(ii) Density of liquid: increases with density of liquid.
(iii) Gravity constant.
ATMOSPHERIC PRESSURE : “Pressure exerted by the air of atmosphere around us.”

STANDARD ATMOSPHERIC PRESSURE 1 Atm = 76 cm of Hg column = 1.013 x 10⁵Pa

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) The S.I. unit of force is kgf.
Answer. False.
The S.I. unit of force is newton.

(b) A force always produces both the linear and turning motions.
Answer. False.

(c) Moment of force = force × perpendicular distance of force – from the pivoted point.
Answer. True.

(d) Less force is needed when applied at a farther distance from the pivoted point.
Answer. True.

(e) For a given thrust, pressure is more on a surface of large j area.
Answer. False.
For a given thrust, pressure is less on a surface of large area.

(f) The pressure on a surface increases with an increase in the thrust on the surface.
Answer. True.

(g) A man exerts same pressure on the ground whether he is standing or he is lying.
Answer. False.
A man exerts different pressure on the ground whether he is standing or he is lying.

(h) It is easier to hammer a blunt nail into a piece of wood than a sharply pointed nail.
Answer. False.
It is not easier to hammer a blunt nail into a piece of wood than a sharply pointed nail.

(i) The S.I. unit of pressure is pascal.
Answer. True.

(j) Water in a lake exerts pressure only at its bottom.
Answer. False.

(k) A liquid exerts pressure in all directions.
Answer. True.

(l) Gases exert pressure in all directions.
Answer. True.

(m) The atmospheric pressure is nearly 10⁵ Pa.
Answer. True.

(n) Higher we go, greater is the air pressure.
Answer. False.

2. Fill in the blanks

(a) 1 kgf = 10 N (nearly).
(b) Moment of force = force × distance of force from the point of turning
(c) In a door, handle is provided farthest from the hinges.
(d) The unit of thrust is newton .
(e) Thrust is the normal force acting on a surface.
(f) Pressure is the thrust acting on a surface of unit area.
(g) The unit of pressure is pascal
(h) Pressure is reduced if area of surface increases.
(i) Pressure in a liquid increases with the depth.
(j) The atmospheric pressure on earth surface is nearly 10⁵ Pa.

3. Match the following

Selina Concise Physics Class 8 ICSE Solutions Chapter 3 Force and Pressure 2

4. Select the correct alternative

(a) SI. unit of moment of force ¡s

N
N cm
kgfm
N m

(b) To obtain a given moment of force for turning a body, the force needed can be decreased by

applying the force at the pivoted point
applying the force very close to the pivoted point
applying the force farthest from the pivoted point
none of the above

(c) The unit of thrust is

kgf
kg
g
m s^-1

(d) The unit of pressure is

N × m
kgf
N m^-2
kgf m²

(e) The pressure and thrust are related as

Pressure = Thrust
Pressure = Thrust x Area
Pressure = Thrust / Area,
Pressure = Area / Thrust

(f) A body weighing 5 kgf, placed on a surface of area 0.1 m2, exerts a thrust on the surface equal to

50 kgf
5 kgf
50 kgf m^-2
5 kgf m^-2

P.Q. A body weighing 5 kgf, placed on a surface of area 0.1 m2, exerts a pressure on the surface equal to

50 kgf
5 kgf
50 kgf m^-2
5 kgf m^-2

(g) The feet of lizards act like

moving pads
drilling pads
suction pads
none of the above

(h) Pressure exerted by a liquid is due to its

weight
mass
volume
area

(i) Pressure inside a liquid increases with :

increase in depth
decrease in depth
decrease in density
none of the above

(j) The atmospheric pressure at sea level is nearly

10 Pa
100,000 Pa
100 Pa
10,000 Pa

(k) Nose bleeding may occur at a high altitude because

the atmospheric pressure decreases
the oxygen content of atmosphere decreases
the atmospheric pressure increasess
there are strong air currents at the high altitude

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths Solutions Physics Chemistry Biology Geography History & Civics

Linear Equations in one Variable Exercise 14A – Selina Concise Mathematics Class 8 ICSE Solutions

Solve the following equations:
Question 1.
20 = 6 + 2x
Solution:
20 = 6 + 2x
20 – 6 = 2x
14 = 2x
7 = x
x = 7

Question 2.
15 + x = 5x + 3
Solution:
15 – 3 = 5x – x
12 = 4x
3 = x
x = 3

Question 3.
\(\frac { 3x+2 }{ x-6 }\) = -7
Solution:
3x + 2 = -7 (x – 6) (by cross multiplying)
3x + 2 = -7x + 42
3x + 7x = 42 – 2
10x = 40
x = 4

Question 4.
3a – 4 = 2(4 – a)
Solution:
3a – 4 = 8 – 2a
3a + 2a = 8+4
5a = 12
a = 2.4

Question 5.
3(b – 4) = 2(4 – b)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 1

Question 6.
\(\frac { x+2 }{ 9 } =\frac { x+4 }{ 11 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 2

Question 7.
\(\frac { x-8 }{ 5 } =\frac { x-12 }{ 9 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 3

Question 8.
5(8x + 3) = 9(4x + 7)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 4

Question 9.
3(x +1) = 12 + 4(x – 1)
Solution:
3(x + 1) = 12 + 4(x – 1)
3x + 3 = 12 + 4x – 4
3x – 4x = 12 – 4 – 3
-x = 5
x = -5

Question 10.
\(\frac { 3x }{ 4 } -\frac { 1 }{ 4 } \left( x-20 \right) =\frac { x }{ 4 } +32\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 5

Question 11.
\(3a-\frac { 1 }{ 5 } =\frac { a }{ 5 } +5\frac { 2 }{ 5 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 6

Question 12.
\(\frac { x }{ 3 } -2\frac { 1 }{ 2 } =\frac { 4x }{ 9 } -\frac { 2x }{ 3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 7

Question 13.
\(\frac { 4\left( y+2 \right) }{ 5 } =7+\frac { 5y }{ 13 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 9

Question 14.
\(\frac { a+5 }{ 6 } -\frac { a+1 }{ 9 } =\frac { a+3 }{ 4 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 8

Question 15.
\(\frac { 2x-13 }{ 5 } -\frac { x-3 }{ 11 } =\frac { x-9 }{ 5 } +1\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 13

Question 16.
6(6x – 5) – 5 (7x – 8) = 12 (4 – x) + 1
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 14

Question 17.
(x – 5) (x + 3) = (x – 7) (x + 4)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 15

Question 18.
(x – 5)2 – (x + 2)2 = -2
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 16

Question 19.
(x – 1) (x + 6) – (x – 2) (x – 3) = 3
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 18

Question 20.
\(\frac { 3x }{ x+6 } -\frac { x }{ x+5 } =2\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 19

Question 21.
\(\frac { 1 }{ x-1 } +\frac { 2 }{ x-2 } =\frac { 3 }{ x-3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 20

Question 22.
\(\frac { x-1 }{ 7x-14 } =\frac { x-3 }{ 7x-26 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 22

Question 23.
\(\frac { 1 }{ x-1 } -\frac { 1 }{ x } =\frac { 1 }{ x+3 } -\frac { 1 }{ x+4 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 23

Question 24.
Solve: \(\frac { 2x }{ 3 } -\frac { x-1 }{ 6 } +\frac { 7x-1 }{ 4 } =2\frac { 1 }{ 6 }\)
Hence, find the value of ‘a’, if \(\frac { 1 }{ a }\) + 5x = 8.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 24

Question 25.
Solve: \(\frac { 4-3x }{ 5 } +\frac { 7-x }{ 3 } +4\frac { 1 }{ 3 } =0\)
Hence find the value of ‘p’ if 2p – 2x + 1 = 0
Solution:

Hence x = 8
Now, 3p – 2x + 1=0
⇒ 3p – 2 x 8 + 1 = 0
⇒ 3p – 16 + 1 =0
⇒ 3p – 15 = 0.
⇒ 3p=15
⇒ p = 5

Question 26.
Solve: \(0.25+\frac { 1.95 }{ x } =0.9\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 28

Question 27.
Solve: \(5x-\left( 4x+\frac { 5x-4 }{ 7 } \right) =\frac { 4x-14 }{ 3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 29

Linear Equations in one Variable Exercise 14B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Fifteen less than 4 times a number is 9. Find the number.
Solution:
Let the required number be x
4 times the number = 4x
15 less than 4 times the number = 4x-15
According to the statement :
4x – 15 = 9
⇒ 4x = 9 + 15
⇒ 4x = 24
⇒ x = 6

Question 2.
If Megha’s age is increased by three times her age, the result is 60 years. Find her age
Solution:
Let Megha’s age = x years
Three times Megha’s age = 3x years
According to the statement :
x + 3x = 60
=> 4x = 60
=> x = 15
Megha’s age = 15 years

Question 3.
28 is 12 less than 4 times a number. Find the number.
Solution:
Let the required number be x
4 times the number = 4x
12 less than 4 times the number = 4x – 12
According to the statement
4x – 12 = 28
=> 4x = 28 + 12
=> 4x = 40
x = 10
Required number = 10

Question 4.
Five less than 3 times a number is -20. Find the number.
Solution:
Let the required number = x
3 times the number = 3x
5 less than 3 times the number = 3x – 5
According to statement :
3x – 5 = -20
=> 3x = -20 + 5
=> 3x = -15
=> x = -5
Required number = -5

Question 5.
Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she ?
Solution:
Let Neetu’s age = x years
3 times Neetu’s age = 3x years
Fifteen more than 3 times Neetu’s age = (3x + 15) years
4 times Neetu’s age = 4x
According to the statement :
4x = 3x + 15
=> 4x – 3x = 15
=> x = 15
Neetu’s age = 15 years

Question 6.
A number decreased by 30 is the same as 14 decreased by 3 times the number; Find the number.
Solution:
Let the required number = x
The number decreased by 30 = x – 30
14 decreased by 3 times the number = 14 – 3x
According to the statement :
x – 30 = 14 – 3x
=> x + 3x = 14 + 30
=> 4x = 44
x = 11
Required number =11

Question 7.
A’s salary is same as 4 times B’s salary. If together they earn Rs.3,750 a month, find the salary of each.
Solution:
Let B’s salary = Rs. x
A’s salary = Rs. 4x
According to the statement :
x + 4x = 3750
=> 5x = 3750
=> x = 750
4x = 750 x 4 = 3000
A’s salary = Rs. 3000
B’s salary = Rs. 750

Question 8.
Separate 178 into two parts so that the first part is 8 less than twice the second part.
Solution:
Let first part = x
Second part = 178 – x
According to the problem :
First Part = 8 less than twice the second part
x = 2(178 – x) – 8
=> x = 356 – 2x – 8
=> x+2x = 356 – 8
=> 3x = 348
=> x = 116
First Part = 116
=> Second Part = 178 – x = 178 – 116 = 62
First Part = 116
=> Second Part = 62

Alternative Method :
Let Second part = x
First part = 2x – 8
According to the problem :
x + 2x – 8 = 178
=> x + 2x = 178 + 8
=> 3x = 186
=> x = 62
First part = 2x – 8 = 2 x 62 – 8 = 124 – 8 = 116
First part = 116
Second part = 62

Question 9.
Six more than one-fourth of a number is two-fifth of the number. Find the number.
Solution:
Let the required number = x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 30
x = 40
Required number = 40

Question 10.
The length of a rectangle is twice its width. If its perimeter is 54 cm; find its length.
Solution:
Let width of the rectangle = x cm
Length of the rectangle = 2x cm
Perimeter of the rectangle = 2 [Length + Width] = 2 [2x + x] = 2 x 3x = 6x cm
Given perimeter = 54 cm
6x = 54
=> x = 9
Length = 2x = 2 x 9 = 18 cm

Question 11.
A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the origi¬nal rectangle.
Solution:
Let width of the original rectangle = x cm
Length of the original rectangle = (2x – 5)cm
Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm
New width of the rectangle = (x + 2) cm
New perimeter = 2[Length+Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm
Given; new perimeter = 74 cm
6x – 16 = 74
=> 6x = 74 + 16
=> 6x = 90
=>x = 15
Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm
Width of the original rectangle = x = 15 cm

Question 12.
The sum of three consecutive odd numbers is 57. Find the numbers.
Solution:
Let the three consecutive odd numbers be x, x+2, x+4.
According to the statement :
x + x + 2 + x + 4 = 57
=> x + x + x = 57 – 2 – 4
=> 3x = 51
=> x = 17
Three consecutive odd numbers are 17, 19, 21

Question 13.
A man’s age is three times that of his son, and in twelve years he will be twice as old as his son would be. What are their present ages.
Solution:
Let present age of the son = x years
present age of the man = 3x years
In 12 years :
Son’s age will be = (x + 12) years
The man’s age will be = (3x + 12) years
According to the statement :
3x + 12 = 2(x + 12)
=> 3x + 12 = 2x + 24
=> 3x – 2x = 24 – 12
=> x = 12
3x = 3 x 12 = 36
Hence, present age of the man = 36 years
Present age of the son = 12 years.

Question 14.
A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half the age of the man at that time?
Solution:
Man’s age = 42 years
Son’s age = 12 years
Let after x years the age of the son will be half the age of the man.
Man’s age after x years = (42 + x) years
Son’s age after x years = (12 + x) years
According to the statement :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 31
Hence after 18 years, the age of the son will be half the age of the man

Question 15.
A man completed a trip of 136 km in 8 hours. Some part of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 32

Question 16.
The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.
Solution:
Let one number = x
Second number = x + 3 [Difference of two numbers is 3]
According to the statement :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 34
One number = 10
Second number = x + 3 = 10 + 3 = 13

Question 17.
Two consecutive natural numbers are such that one-fourth of the smaller exceeds one-fifth of the greater by 1. Find the numbers.
Solution:
Let two consecutive natural numbers = x, x+1
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 35
Two consecutive numbers are 24 and 25

Question 18.
Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively; the sum of the quotients is 40. Find the numbers.
Solution:
Let the three consecutive whole numbers be x, x + 1 and x + 2
According to the statement:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 36
x = 50
x + 1 = 50+1 = 51
x + 2 = 50 + 2 = 52
Three consecutive whole numbers are 50, 51 and 52

Question 19.
If the same number be added to the numbers 5, 11, 15 and 31, the resulting numbers are in proportion. Find the number.
Solution:
Let x be added to each number, then the numbers will be 5 + x, 11 + x, 15 + x and 31 + x
According to the condition
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 37
1 should be added

Question 20.
The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.
Solution:
Let present age of son = x year
Then age of his father = 2x
8 years hence,
Age of son = (x + 8) years and age of father = (2x + 8) years
According to the condition,
\(\frac { 2x+8 }{ x+8 } =\frac { 7 }{ 4 }\)
=> 8x + 32 = 7x + 56
=> 8x – 7x = 56 – 32
=> x = 24
Present age of son = 24 years
and age of father = 2x = 2 x 24 = 48 years
Hence age of man = 48 years and age of his son = 24 years

Linear Equations in one Variable Exercise 14C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Solve:
(i) \(\frac { 1 }{ 3 } x-6=\frac { 5 }{ 2 }\)
(ii) \(\frac { 2x }{ 3 } -\frac { 3x }{ 8 } =\frac { 7 }{ 12 }\)
(iii) (x + 2)(x + 3) + (x – 3)(x – 2) – 2x(x + 1) = 0
(iv) \(\frac { 1 }{ 10 } -\frac { 7 }{ x } =35\)
(v) 13(x – 4) – 3(x – 9) – 5(x + 4) = 0
(vi) x + 7 – \(\frac { 8x }{ 3 } =\frac { 17x }{ 6 } -\frac { 5x }{ 8 }\)
(vii) \(\frac { 3x-2 }{ 4 } -\frac { 2x+3 }{ 3 } =\frac { 2 }{ 3 } -x\)
(viii) \(\frac { x+2 }{ 6 } -\left( \frac { 11-x }{ 3 } -\frac { 1 }{ 4 } \right) =\frac { 3x-4 }{ 12 }\)
(ix) \(\frac { 2 }{ 5x } -\frac { 5 }{ 3x } =\frac { 1 }{ 15 }\)
(x) \(\frac { x+2 }{ 3 } -\frac { x+1 }{ 5 } =\frac { x-3 }{ 4 } -1\)
(xi) \(\frac { 3x-2 }{ 3 } +\frac { 2x+3 }{ 2 } =x+\frac { 7 }{ 6 }\)
(xii) \(x-\frac { x-1 }{ 2 } =1-\frac { x-2 }{ 3 }\)
(xiii) \(\frac { 9x+7 }{ 2 } -\left( x-\frac { x-2 }{ 7 } \right) =36\)
(xiv) \(\frac { 6x+1 }{ 2 } +1=\frac { 7x-3 }{ 3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 38

Question 2.
After 12 years, I shall be 3 times as old as 1 was 4 years ago. Find my present age.
Solution:
Let present age = x years
According to question,
(x + 12) = 3(x – 4)
x + 12 = 3x – 12
2x = 24
=> x = 12 years
Present age = 12 years

Question 3.
A man sold an article for 7396 and gained 10% on it. Find the cost price of the article
Solution:
S.P. of article = ₹ 396
Gain = 10%
Let cost price = ₹ x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 52
Cost price of an article = ₹ 360

Question 4.
The sum of two numbers is 4500. If 10% of one number is 12.5% of the other, find the numbers.
Solution:
Let the first number = x
and the second number = y
According to question,
x + y = 4500 ……(i)
and 10% x = 12.5% y
i.e. 10x = 12.5y
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 53
x = 2500
Hence, the numbers are 2500 and 2000

Question 5.
The sum of two numbers is 405 and their ratio is 8 : 7. Find the numbers.
Solution:
Let the first number = x
and the second number = 7
According to the question, x + y = 405 ……..(i)
and the numbers are in the ratio 8 : 7
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 54

x = 189
Hence, the numbers are 189 and 216

Question 6.
The ages of A and B are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.
Solution:
Ratio in the present ages of A and B = 7 : 5
Let age of A = 7x years
Let age of B = 5x years
10 years hence,
Then age of A = 7x + 10 years
and age of B = 5x + 10 years
According to the condition,
\(\frac { 7x+10 }{ 5x+10 } =\frac { 9 }{ 7 }\)
By crossing multiplication
7(7x + 10) = 9(5x + 10)
=> 49x + 70 = 45x + 90
=> 49x – 45x = 90 – 70
=> 4x = 20
=> x = 5
Present age of A = 7x = 7 x 5 = 35 years
and present age of B = 5x = 5 x 5 = 25 years

Question 7.
Find the number whose double is 45 greater than its half.
Solution:
Let the required number = x
Double of it = 2x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 56
Required number = 30

Question 8.
The difference between the squares of two consecutive numbers is 31. Find the numbers.
Solution:
Let first number = x
and The second number = x + 1
According to the condition,
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 57
First number = 15
and second number = 15 + 1 = 16
Hence, the numbers are 15, 16

Question 9.
Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.
Solution:
Let the required number = x
5 times of it = 5x
Twice of it = 2x
According to the condition,
5x – 5 = 2x + 4
=> 5x – 2x = 4 + 5
=> 3x = 9
=> x = 3
Required number = 3

Question 10.
The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator, and denominator both, the fraction becomes \(\frac { 2 }{ 3 }\). Find the original fraction.
Solution:
Let denominator of the original fraction = x
Then numerator = x – 5
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 58 .

Selina Concise Physics Class 8 ICSE Solutions – Light Energy

August 11, 2022January 27, 2019 by Kalyan

Selina Concise Physics Class 8 ICSE Solutions – Light Energy

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 5 Light Energy. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Physics ICSE Solutions Chemistry Biology Maths Geography History & Civics

Selina Concise ICSE Solutions for Class 8 Physics Chapter 5 Light Energy

Light is a form of energy which produces in us the sensation of sight i.e. we can see objects only when light falls on them and then reflected into our eye.
Velocity of light in air or in vacuum is 300000 km per second.
Or
3 x 10⁸ms^-1
As light passes into different mediums its speed changes and depends upon the density of medium i.e. it decreases with increase in density i.e. it is 2.25 × 10⁸ m/s in water and 2 x 10⁸ ms^-1 in glass as water is
denser than air ( \(_{ w }^{ a }{ \mu }\) = 1.33 ) and glass is still optically denser than water
( \(_{ g }^{ a }{ \mu }\) =1.5 ) i.e. slower in water and still slower in glass.
Light travels in a straight line.
As light travels from one transparent medium to other transparent medium and falls oblique at another medium, its path changes and this change in path is called REFRACTION OF LIGHT.
When ray of light travels from RARER (less-denser) to DENSER medium, it bends TOWARD the normal AND when it travels from a DENSER to a RARER medium it bends away from NORMAL
ANGLE of INCIDENCE : “The angle which incident ray makes with normal”. “∠i”
ANGLE OF REFRACTION: “The angle which refracted ray makes with normal” “ ∠r ”
∠i is not equal to ∠r
LAWS OF REFRACTION or SNELL’S LAWS OF REFRACTION:
(i) Incident ray, normal at the point of incidence and Refracted ray all lie in the same plane.
(ii) Ratio of sine of angle of incidence to the sine of angle of refraction is constant.
EFFECTS OF REFRACTION :
(i) A coin placed in water appears to be raised.
(ii) Swimming pool seen from above appears SHADOW.
(iii) A pencil in water appears to be bent.
(iv) MIRAGE in desert, EARLY Sunrise, LATE SUN set are all due to REFRACTION of light.
White light is a band of seven colours-VIBGYOR. Speed of all colours of the white light in AIR or VACUUM is same, but different different transparent mediums.
In glass or water Speed of VIOLET colour is MINIMUM and speed of RED light is MAXIMUM
Refractive index of medium is minimum for VIOLET light and R.I. of medium is maximum for red light.
DISPERSION: “The splitting (breaking) of white light into seven colours is called DISPERSION OF LIGHT.
CAUSE OF DISPERSION: Speed of different colours is different in glass or water and different colours get separated from each other on refraction at second surface of glass prism.

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) Water is optically denser than glass.
Answer. False.
Water is optically denser than air.

(b) A ray of light when passes from glass to air, bends towards the normal.
Answer. False.

(c) The speed of light is more in glass than in water.
Answer. False.

(d) The depth of a pond when seen from above appears to be less.
Answer. True.

(e) Light travels at a lower speed in water than in air.
Answer. True.

(f) Light travels in the same straight line path while passing through different media.
Answer. False.

(g) The angle formed between the normal and the refracted ray is known as the angle of incidence.
Answer. False.

(h) At the point of incidence, a line drawn at right angles to the surface, separating the two media, is called the normal.
Answer. True.

(i) Image is formed by a mirror due to refraction of light.
Answer. False.

(j) Rays of light incident parallel to the principal axis pass through the focus after reflection from a concave mirror.
Answer. True.

(k) A convex mirror is used as a shaving mirror.
Answer. False.

(l) The focal length of a convex mirror is equal to its radius of curvature.
Answer. False.

(m) A concave mirror converges the light-rays, but a convex mirror diverges them.
Answer. True.

(n) A virtual image formed by a spherical mirror is always erect and situated behind the mirror.
Answer. True.

2. Fill in the blanks

(a) Water is optically denser than air.
(b) Air is optically rarer than glass.
(c) When a ray of light travels from water to air, it bends away from the normal.
(d) When a ray of light travels from air to glass, it bends towards the normal.
(e) When white light passes through a prism, it disperses
(f) The splitting of white light into its constituent colours is called dispersion.
(g) A concave mirror is obtained on silvering the outer surface of a part of a hollow glass sphere.
(h) Radius of curvature of a spherical mirror is two times its focal length.
(i) The angle of incidence for a ray of light passing through the centre of curvature of a spherical mirror is 0°
(j) A convex mirror always forms a virtual image.
(k) A concave mirror forms a virtual image for an object placed between pole and focus.

3. Match the following

4. Select the correct alternative

(a) The speed of light in air or vacuum is

3 × 10⁸M s^-1
2.25 × 10⁸ m s^-1
332 ms^-1
2.0 × 10⁸ ms^-1

(b) A ray of light moving from an optically rarer to a denser medium

bends away from the normal
bends towards the normal
remains undeviated
none of the above

(c) The angle between the normal and refracted ray is called

angle of deviation
angle of incidence
angle of refraction
angle of emergence.

(d) The property of splitting of white light into its seven constituent colours is known as

rectilinear propagation
refraction
reflection
dispersion

(e) The seven colours in the spectrum of sunlight in order, are represented as :

VIBGYOR
VIGYBOR
BIVGYOR
RYOBIVG

(f) A ray of light passing through centre of curvature of a spherical mirror, after reflection

passes through the focus
passes through the pole
becomes parallel to the principal axis
retraces its own path.

(g) If the radius of curvature of a concave mirror is 20 cm, its focal length is:

10 cm
20 cm
40 cm
80 cm

(h) The image formed by a convex mirror is

erect and diminished
erect and enlarged
inverted and diminished
inverted and enlarged.

(i) The image formed by a concave mirror is of the same size as the object, if the object is placed

at the focus
between the pole and focus
between the focus and centre of curvature
at the centre of curvature.

(j) A convex mirror is used

as a shaving mirror
as a head mirror by a dentist
as a rear view mirror by a driver
as a reflector in torch.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 13 Factorisation. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths Solutions Physics Chemistry Biology Geography History & Civics

Factorisation Exercise 13A – Selina Concise Mathematics Class 8 ICSE Solutions

Factorise :
Question 1.
15x + 5
Solution:
15x + 5 = 5(3x + 1)

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 16

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 18

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 20

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 22

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 25

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 27

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 29

Question 16.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 30
Solution:

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 33

Factorisation Exercise 13B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Factorise : a² + ax + ab + bx
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 34

Question 2.
Factorise : a² – ab – ca + bc
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 35

Question 3.
Factorise : ab – 2b + a² – 2a
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 36

Question 4.
Factorise : a³ – a² + a – 1
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 37

Question 5.
Factorise : 2a – 4b – xa + 2bx
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 38

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 40

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 42

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 44

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 46

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 48

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 50

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 52

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 54

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 56

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 58

Question 16.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 60

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 62

Question 18.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 64

Question 19.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 66

Question 20.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 68

Factorisation Exercise 13C – Selina Concise Mathematics Class 8 ICSE Solutions

Note : a² – b² = (a + b) (a – b)
Question 1.
Factorise : 16 – 9x²
Solution:
16 – 9x² = (4)² – (3x)² = (4 + 3x) (4 – 3x)

Question 2.
Factorise : 1 – 100a²
Solution:
1 – 100a² = (1)² – (10a)² = (1 + 10a) (1 – 10a)

Question 3.
Factorise : 4x² – 81y²
Solution:
4x² – 81y² = (2x)² – (9y)² = (2x + 9y) (2x – 9y)

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 69
Solution:

Question 5.
Factorise : (a+2b)² – a²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 71

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 73

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 75

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 77

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 79

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 81

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 83

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 85

Question 13.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 86
Solution:

Question 14.
Evaluate :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 88
Solution:

Question 15.
Evaluate :
(0.7)² – (0.3)²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 90

Question 16.
Evaluate :
(4.5)² – (1.5)²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 91

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 93

Question 18.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 95

Question 19.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 97

Question 20.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 99

Factorisation Exercise 13D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 100
= (x+4) (x+2)

Question 2.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 102

Question 3.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 104

Question 4.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 106

Question 5.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 108

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 110

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 112

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 114

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 116

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 118

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 119
= (a – 1)(3a – 2)

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 121

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 123

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 125

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 127

Question 16.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 129

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 132

Question 18.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 134

Question 19.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 136

Question 20.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 138

Question 21.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 140

Question 22.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 142

Question 23.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 144

Question 24.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 146

Question 25.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 149

Factorisation Exercise 13E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
In each case find whether the trinomial is a perfect square or not:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 150
Solution:

Question 2.
Factorise completely 2 – 8x².
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 153

Question 3.
Factorise completely : 8x²y – 18y³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 154

Question 4.
Factorise completely : ax² – ay²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 155

Question 5.
Factorise completely : 25x³ – x
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 156

Question 6.
Factorise completely : a⁴ – b⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 157

Question 7.
Factorise completely : 16x⁴ – 81y⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 158

Question 8.
Factorise completely : 625 – x⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 159

Question 9.
Factorise completely : x² – y² – 3x – 3y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 160

Question 10.
Factorise completely : x² – y² – 2x + 2y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 162

Question 11.
Factorise completely : 3x² + 15x – 72
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 163

Question 12.
Factorise completely : 2a² – 8a – 64
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 164

Question 13.
Factorise completely : 5b² + 45b + 90
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 165

Question 14.
Factorise completely : 3x²y + 11xy + 6y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 166

Question 15.
Factorise completely : 5ap² + 11ap + 2a
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 167

Question 16.
Factorise completely : a² + 2ab + b² – c²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 168

Question 17.
Factorise completely : x² + 6xy + 9y² + x + 3y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 169

Question 18.
Factorise completely : 4a² – 12ab + 9b² + 4a – 6b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 170

Question 19.
Factorise completely : 2a²b² – 98b⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 171

Question 20.
Factorise completely : a² – 16b² – 2a – 8b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 172

Factorisation Exercise 13F – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 173
Solution:

Question 2.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 175
Solution:

Question 3.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 178
Solution:

Question 4.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 182
Solution:

Question 5.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 186
Solution:

Question 6.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 188

Solution:

Question 7.
Factorise xy² – xz², Hence, find the value of:

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 192

Question 8.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 193
Solution:

Question 9.
Factorise a²b – b³ Using this result, find the value of 101² x 100 – 100³.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 197

Question 10.
Evaluate (using factors): 301² x 300 – 300³.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 198