NCERT Solutions for Class 8 Maths

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.2.

• Cubes and Cube Roots Class 8 Ex 7.1

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.2

Question 1.
Find the cube root of each of the following numbers by prime factorisation method:
(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125
Solution.
(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

Question 2.
State true or false:
(i) Cube of any odd number is even,
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.
Solution.
(i) False
(ii) True
(iii) False ⇒ \({ 15 }^{ 2 }\) = 225, \({ 15 }^{ 3 }\) = 3375
(iv) False ⇒ \({ 12 }^{ 3 }\) = 1728
(v) False ⇒ \({ 10 }^{ 3 }\) = 1000, \({ 99 }^{ 3 }\) = 970299
(vi) False ⇒ \({ 10 }^{ 3 }\) = 1000, \({ 99 }^{ 3 }\) = 970299
(vii) True ⇒ \({ 1 }^{ 3 }\) = 1; \({ 2 }^{ 3 }\) = 8

Question 3.
You are told that 1,331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution.
By guess,
Cube root of 1331 =11
Similarly,
Cube root of 4913 = 17
Cube root of 12167 = 23
Cube root of 32768 = 32
EXPLANATIONS
(i)
Cube root of 1331
The given number is 1331.

Step 1. Form groups of three starting from the rightmost digit of 1331. 1 331
In this case, one group i.e., 331 has three digits whereas 1 has only 1 digit.
Step 2. Take 331.
The digit 1 is at one’s place. We take the one’s place of the required cube root as 1.
Step 3. Take the other group, i.e., 1. Cube of 1 is 1.
Take 1 as ten’s place of the cube root of 1331.
Thus, \(\sqrt [ 3 ]{ 1331 } =11\)

(ii)
Cube root of 4913
The given number is 4913.

Step 1. Form groups of three starting from the rightmost digit of 4913.
In this case one group, i.e., 913 has three digits whereas 4 has only one digit.
Step 2. Take 913.
The digit 3 is at its one’s place. We take the one’s place of the required cube root as 7.
Step 3. Take the other group, i.e., 4. Cube of 1 is 1 and cube of 2 is 8. 4 lies between 1 and 8.
The smaller number among 1 and 2 is 1.
The one’s place of 1 is 1 itself. Take 1 as ten’s place of the cube root of 4913.
Thus, \(\sqrt [ 3 ]{ 4913 } =17\)

(iii)
Cube root of 12167
The given number is 12167.

Step 1. Form groups of three starting from the rightmost digit of 12167.
12 167. In this case, one group, i. e., 167 has three digits whereas 12 has only two digits.
Step 2. Take 167.
The digit 7 is at its one’s place. We take the one’s place of the required cube root as 3.
Step 3. Take the other group, i.e., 12. Cube of 2 is 8 and cube of 3 is 27. 12 lies between 8 and 27. The smaller among 2 and 3 is 2.
The one’s place of 2 is 2 itself. Take 2 as ten’s place of the cube root of 12167.
Thus, A/12167 = 23.
Thus, \(\sqrt [ 3 ]{ 12167 } =23\).

(iv)
Cube root of 32768
The given number is 32768.

Step 1. Form groups of three starting from the rightmost digit of 32768.
32 768. In this case one group,
i. e., 768 has three digits whereas 32 has only two digits.
Step 2. Take 768.
The digit 8 is at its one’s place. We take the one’s place of the required cube root as 2.
Step 3. Take the other group, i.e., 32.
Cube of 3 is 27 and cube of 4 is 64.
32 lies between 27 and 64.
The smaller number between 3 and 4 is 3.
The ones place of 3 is 3 itself. Take 3 as ten’s place of the cube root of 32768.
Thus, \(\sqrt [ 3 ]{ 32768 } =32\).

We hope the NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4.

• Squares and Square Roots Class 8 Ex 6.1
• Squares and Square Roots Class 8 Ex 6.2
• Squares and Square Roots Class 8 Ex 6.3

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4

Question 1.
Find the square root of each of the following numbers by Division method:

Solution.
(i) 2304

(ii) 4489

(iii) 3481

(iv) 529

(v) 3249

(vi) 1369
(vii) 5776

(viii) 7921

(ix) 576

(x) 1024

(xi) 3136

(xii) 900

Question 2.
Find the number of digits in the square root of each of the following numbers (without any calculation):
(i) 64
(ii) 144
(iii) 4489
(iv) 27225
(v) 390625.
Solution.
(i) 64
Number (n) of digits in 64 = 2 which is even.
∴ Number of digits in the square root of 64 \(\frac { n }{ 2 } =\frac { 2 }{ 2 } =1\)

(ii) 144
Number (n) of digits in 144 = 3 which is
∴ Number of digits in the square root of 144 \(\frac { n+1 }{ 2 } =\frac { 3+1 }{ 2 } =\frac { 4 }{ 2 } =2\)

(iii) 4489
Number (n) of digits in 4489 = 4 which is even.
∴ Number of digits in the square root of 4489 \(\frac { n }{ 2 } =\frac { 4 }{ 2 } =2\)

(iv) 27225
Number (n) of digits in 27225 = 5 which is odd.
∴ Number of digits in the square root of 27225 \(\frac { n+1 }{ 2 } =\frac { 5+1 }{ 2 } =\frac { 6 }{ 2 } =3\)

(v) 390625
Number (n) of digits in 390625 = 6 which is even.
∴ Number of digits in the square root of 390625 \(\frac { n }{ 2 } =\frac { 6 }{ 2 } =3\)

Question 3.
Find the square root of the following decimal numbers:
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 31.36
Solution.
(i) 2.56

(ii) 7.29

(iii) 51.84

(iv) 42.25

(v) 31.36

Question 4.
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402
(ii) 1989
(iii) 3250
(iv) 825
(v) 4000
Solution.
(i)402

(ii) 1989

(iii) 3250

(iv) 825

(v) 4000

Question 5.
Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 525
(ii) 1750
(iii) 252
(iv) 1825
(v) 6412
Solution.
(i) 525

(ii) 1750

(iii) 252

(iv) 1825

(v) 6412

Question 6.
Find the length of the side of a square whose area is 441 \({ m }^{ 2 }\).
Solution.
Area of the square = 441 \({ m }^{ 2 }\)
∴ Length of the side of the square

Question 7.
In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC
(b) If AC 13 cm, BC = 5 cm, find AB.
Solution.
(a) In the right triangle ABC,
∠B = 90°
Given

Question 8.
A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Solution.
Let the number of rows be x.
Then the number of columns is x.

Question 9.
There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?
Solution.
Let the number of rows be x.
Then the number of columns is x.

We hope the NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4 are part help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4 are part, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3.

• Squares and Square Roots Class 8 Ex 6.1
• Squares and Square Roots Class 8 Ex 6.2
• Squares and Square Roots Class 8 Ex 6.4

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3

Question 1.
What could be the possible ‘one’s’ digits of the square root of each of the following numbers ?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025.
Solution.
(i) 9801
∵ 1 x 1 = 1 and 9 x 9 = 81
∵ The possible one’s digit of the square root of the number 9801 could be 1 or 9.

(ii) 99856
∵ 4 x 4 = 16 and 6 x 6 = 36
∵ The possible one’s digit of the square root of the number 99856 could be 4 or 6.

(iii) 998001
∵ 1×1 = 1 and 9 x 9 = 81
∵ The possible one’s digit of the square root of the number 998001 could be 1 or 9.

(iv) 657666025
∵ 5 x 5 = 25
∵ The possible one’s digit of the square root of the number 657666025 could be 5.

Question 2.
Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153
(ii) 257
(iii) 408
Solution.
(i) 153
The number 153 is surely not a perfect square because it ends in 3 whereas the square numbers end with 0, 1, 4, 5, 6 or 9.

(ii) 257
The number 257 is surely not a perfect square because it ends in 7 whereas the square numbers end with 0, 1, 4, 5, 6 or 9.

(iii) 408
The number 408 is surely not a perfect square because it ends in 8 whereas the square numbers end with 0, 1, 4, 5, 6 or 9.

(iv) 441
The number may be a perfect square as the square numbers end wTith 0, 1, 4, 5, 6 or 9.

Question 3.
Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution.
(i) 100

• 100 – 1 = 99
• 99 – 3 = 96
• 96 – 5 = 91
• 91 – 7 = 84
• 84 – 9 = 75
• 75 – 11 = 64
• 64 – 13 = 51
• 51 – 15 = 36
• 36 – 17 = 19
• 19 – 19 = 0

Since from 100, we subtracted successive odd numbers starting from 1 and obtained 0 at the 10th step, therefore,
\(\sqrt { 100 } =10\)

(ii) 169

• 169 – 1 = 168
• 168 – 3 = 165
• 165 – 5 = 160
• 160 – 7 = 153
• 153 – 9 = 144
• 144-11 = 133
• 133 – 13 = 120
• 120 – 15 = 105
• 105 – 17 = 88
• 88 – 19 = 69
• 69 – 21 = 48
• 48 – 23 = 25
• 25 – 25 = 0

Since rom 169, we subtracted successive odd numbers starting from 1 and obtained 0 at the 13th step, therefore,
\(\sqrt { 169 } =13\)

Question 4.
Find the square roots of the following numbers by the Prime Factorisation Method:
(i) 729
(ii) 400
(iii) 1764
(iv) 4095
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100
Solution.
(i) 729
The prime factorisation of 729 is
729 = 3 x 3 x 3 x 3 x 3 x 3.

(ii) 400
The prime factorisation of 400 is
400 = 2 x 2 x 2 x 2 x 5 x 5.

(iii) 1764

(iv) 4096

(v) 7744

(vi) 9604

(vii) 5929

(viii) 9216

(ix) 529

(x) 8100

Question 5.
For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.
(i) 252
(ii) 180
(iii) 1008
(iv) 2028
(v) 1458
(vi) 768
Solution.
(i) 252

(ii) 180

(iii) 1008

(iv) 2028

(v) 1458

(vi) 768

Question 6.
For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.
(i) 252
(ii) 2925
(iii) 396
(iv) 2645
(v) 2800
(vi) 1620
Solution.
(i) 252

(ii) 2925

(iii) 396

(iv) 2645

(v) 2800

(vi) 1620

Question 7.
The students of Class VIII of a school donated? 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as rr’iny rupees as the number of students in the class. Find the number of students in the class.
Solution.
Let the number of students in the class be x.

Question 8.
2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution.
Let the number of rows be x.
Then, number of plants in each row = x.

Question 9.
Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution.

In order to get a perfect square, each factor of 180 must be paired. So, we need to make pair of 5.
Therefore, 180 should be multiplied by 5.
Hence, the required smallest square number is 180 x 5 = 900.

Question 10.
Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3 are part help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3 are part, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2.

• Squares and Square Roots Class 8 Ex 6.1
• Squares and Square Roots Class 8 Ex 6.3
• Squares and Square Roots Class 8 Ex 6.4

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2

Question 1.
Find the square of the following numbers:
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Solution.
(i) 32
32 = 30 + 2
Therefore, \({ 32 }^{ 2 }\) = \({ \left( 30+2 \right) }^{ 2 }\)
= 30 (30 + 2) + 2 (30 + 2)
= 900 + 60 + 60 + 4
= 1024

(ii) 35
35 = 30 + 5
Therefore, \({ 35 }^{ 2 }\) = \({ \left( 30+5 \right) }^{ 2 }\)
= 30 (30 + 5) + 5 (30 + 5)
= 900 + 150 + 150 + 25
= 1225

(iii) 86
86 = 80 + 6
Therefore, \({ 86 }^{ 2 }\)= \({ \left( 80+6 \right) }^{ 2 }\)
= 80 (80 + 6) + 6 (80 + 6)
= 6400 + 480 + 480 + 36
= 7396

(iv) 93
93 = 90 + 3
Therefore, \({ 90 }^{ 2 }\)= \({ \left( 90+3\right) }^{ 2 }\)
= 90 (90 + 3) + 3 (90 + 3)
= 8100 + 270 + 270 + 9
= 8649

(v) 71
71 = 70 + 1
Therefore, \({ 70 }^{ 2 }\)= \({ \left( 70+1\right) }^{ 2 }\)
= 70 (70 + 1) + 1 (70 + 1)
= 4900 + 70 + 70 + 1
= 5041

(vi) 46
46 = 40 + 6
Therefore, \({ 40 }^{ 2 }\)= \({ \left( 40+6\right) }^{ 2 }\)
= 40 (40 + 6) + 6 (40 + 6)
= 1600 + 240 + 240 + 36
= 2116

Question 2.
Write a Pythagorean triplet whose one number is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution.
(i) 6
Let 2m = 6
⇒ \(m=\frac { 6 }{ 2 } =3\)

(ii) 14
Let 2m = 14

(iii) 16
Let 2m = 16

(iv) 18
Let 2m = 18

We hope the NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2 are part help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2 are part, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3.

• Data Handling Class 8 Ex 5.1
• Data Handling Class 8 Ex 5.2

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3

Question 1.
List the outcomes you can see in these experiments.
(a) Spinning a wheel

(b) Tossing two coins together
Solution.
(a) Outcomes in spinning the given wheel are A, B, C and D.
(b) Outcomes in tossing two coins together are HT, HH, TH, TT (Here HT means Head on first coin and Tail on the second coin and so on).

Question 2.
When a die is thrown, list the outcomes of an event of getting
(i)
(a) a prime number
(b) not a prime number

(ii)
(a) a number greater than 5
(b) a number not greater than 5.
Solution.
Possible outcomes are:
1, 2, 3, 4, 5, and 6.
Out of these, prime numbers are
2, 3 and 5.

(i)
(a) Outcomes of an event of getting a prime number are: 2, 3 and 5
(b) Outcomes of an event of not getting a prime number are 1, 4 and 6.

(ii)
(a) Outcomes of an event of getting a number greater than 5 are 6
(b) Outcomes of an event of getting a; number not greater than 5 are 1, 2, 3, 4 and 5.

Question 3.
Find the
(a) Probability of the pointer stopping on D in (Question l-(a))?
(b) Probability of getting an ace from a j well shuffled deck of 52 playing cards?
(c) Probability of getting a red apple. (See figure)

Solution.
(a) There are in all 5 outcomes of the event. These are A, B, C and D. The pointer stopping on D has only 1 outcome, i.e., D
∴ Probability of the pointer stopping on D =\(\frac { 1 }{ 5 } \)

(b) Total number of playing cards = 52 Number of possible outcomes = 52
Number of aces in a deck of playing cards = 4
cards = 4
∴ Probability of getting an ace from a well shuffled deck of 52
playing cards = \(\frac { 4 }{ 52 } \) = \(\frac { 1 }{ 13 } \)

(c) Total number of apples = 7
Number of red apples = 4
∴ Probability of getting a red apple = \(\frac { 4 }{7} \)

Question 4.
Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of?
(i) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv) getting a 1-digit number?
Solution.
Total number of outcomes of the event (1, 2, 3, 4, 5, 6, 7, 8, 9 and 10) = 10
(i)
∵ Number of outcomes of getting a number 6=1
∴ Probability of getting a number \(\frac { 1 }{10} \)

(ii)
∵ There are 5 numbers (1, 2, 3, 4 and 5) less than 6.
∴ Number of outcomes of getting a number less than 6 = 5
∴ Probability of getting a number less than \(6=\frac { 5 }{ 10 } =\frac { 1 }{ 2 } \)

(iii)
∵ There are 4 numbers (7, 8, 9 and 10) greater than 6
∴ Number of outcomes of getting a number greater than 6 = 4
∴ Probability of getting a number greater than \(6=\frac { 4 }{ 10 } =\frac { 2}{ 15 } \)

(iv)
∵ There are 9 1-digit numbers (1, 2, 3, 4, 5, 6, 7, 8 and 9)
∴ Number of outcomes of getting a 1-digit number = 9
∴ Probability of getting a 1-digit number \(=\frac { 9 }{ 10 } \)

Question 5.
If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a nonblue sector?
Solution.
Number of green sectors = 3
Number of blue sectors = 1
Number of red sectors = 1
∴ Total number of sectors = 3 + 1 + 1=5
∴ Total number of outcomes of the event = 5
Number of outcomes of getting a green sector = 3
∴ Probability of getting a green sector = \(\frac { 3 }{ 5 } \)
Number of outcomes of getting a non-blue sector = Number of green sectors + Number of red sectors
=3+1=4
∴ Probability of getting a non-blue sector = \(\frac { 4 }{ 5 } \).

Question 6.
Find the probabilities of the events given in Question 2.
Solution.
Total number of outcomes of the event (1, 2, 3, 4, 5 and 6) = 6
(i)
(a) Number of prime numbers (2, 3 and 5) = 3
∴ Number of outcomes of getting a prime number = 3
∴ Probability of getting a prime number = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \).
(b) Number of non-prime numbers (1, 4 and 6) = 3
∴ Number of outcomes of getting a non-prime number = 3
∴ Probability of getting a non-prime number = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \).

(ii)
(a) Number greater than 5 = 6, i.e., only one.
∴ Number of outcomes of getting a number greater than 5 = 1
∴ Probability of getting a number greater than 5 = \(\frac { 1 }{ 6 } \).
(b) Number of numbers not greater than 5 (1, 2, 3, 4 and 5) = 5
∴ Number of outcomes of getting a number not greater than 5 = 5
∴ Probability of getting a number not greater than 5 = \(\frac { 5 }{ 6 } \).

We hope the NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3, drop a comment below and we will get back to you at the earliest.