# NCERT Solutions for Class 8 Maths

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1.    (i) The number 81 contains its unit’s place digit 1. So, square of 1 is 1.

Hence, unit’s digit of square of 81 is 1.

(ii) The number 272 contains its unit’s place digit 2. So, square of 2 is 4.

Hence, unit’s digit of square of 272 is 4.

(iii) The number 799 contains its unit’s place digit 9. So, square of 9 is 81.

Hence, unit’s digit of square of 799 is 1.

(iv) The number 3853 contains its unit’s place digit 3. So, square of 3 is 9.

Hence, unit’s digit of square of 3853 is 9.

(v) The number 1234 contains its unit’s place digit 4. So, square of 4 is 16.

Hence, unit’s digit of square of 1234 is 6.

(vi) The number 26387 contains its unit’s place digit 7. So, square of 7 is 49.

Hence, unit’s digit of square of 26387 is 9.

(vii) The number 52698 contains its unit’s place digit 8. So, square of 8 is 64.

Hence, unit’s digit of square of 52698 is 4.

(viii) The number 99880 contains its unit’s place digit 0. So, square of 0 is 0.

Hence, unit’s digit of square of 99880 is 0.

(ix) The number 12796 contains its unit’s place digit 6. So, square of 6 is 36.

Hence, unit’s digit of square of 12796 is 6.

(x) The number 55555 contains its unit’s place digit 5. So, square of 5 is 25.

Hence, unit’s digit of square of 55555 is 5.

2.   (i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0.

Therefore 1057 is not a perfect square because its unit’s place digit is 7.

(ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 23453 is not a perfect square because its unit’s place digit is 3.

(iii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 7928 is not a perfect square because its unit’s place digit is 8.

(iv) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 222222 is not a perfect square because its unit’s place digit is

(v) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 64000 is not a perfect square because its unit’s place digit is single 0.

(vi) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 89722 is not a perfect square because its unit’s place digit is 2.

(vii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 222000 is not a perfect square because its unit’s place digit is triple 0.

(viii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.

Therefore 505050 is not a perfect square because its unit’s place digit is 0.

3.    (i) 431 – Unit’s digit of given number is 1 and square of 1 is 1. Therefore, square of 431 would be an odd number.

(ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36. Therefore, square of 2826 would not be an odd number.

(iii) 7779 – Unit’s digit of given number is 9 and square of 9 is 81. Therefore, square of 7779 would be an odd number.

(iv) 82004 – Unit’s digit of given number is 4 and square of 4 is 16. Therefore, square of 82004 would not be an odd number.

4.   Sol.

11= 121

101= 10201

1001= 1002001

100001= 10000200001

1000000= 100000020000001

5.    Sol.

11= 121

101= 10201

10101= 102030201

1010101= 1020304030201

101010101= 10203040504030201

6.    Sol.

12 + 22 + 22 = 32

22 + 32 + 62 = 72

32 + 42 + 122 = 132

42 + 52 + 202 = 212

52 + 62 + 302 = 312

62 + 72 + 422 = 432

7.    (i) Here, there are five odd numbers. Therefore square of 5 is 25.

∴ 1 + 3 + 5 + 7 + 9 = 52 = 25

(ii) Here, there are ten odd numbers. Therefore square of 10 is 100.

∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 102 = 100

(iii) Here, there are twelve odd numbers. Therefore square of 12 is 144.

∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 122 = 144

8.    (i) 49 is the square of 7. Therefore it is the sum of 7 odd numbers.

49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) 121 is the square of 11. Therefore it is the sum of 11 odd numbers.

121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9.    (i) Since, non-perfect square numbers between n2 and (n +1)2 are

Here, n = 12

Therefore, non-perfect square numbers between 12 and 13 = 2n = 2 × 12 = 24

(ii) Since, non-perfect square numbers between n2 and (n + 1)2 are 2n.

Here, n = 25

Therefore, non-perfect square numbers between 25 and 26 = 2n = 2 × 25 = 50

(iii) Since, non-perfect square numbers between n2 and (n+ 1)2 are 2n.

Here, n = 99

Therefore, non-perfect square numbers between 99 and 100 = 2n = 2 × 99 = 198.

10.  (i) (32)2 = (30 + 2)2 = (30)2 + 2 × 30 × 2 + (2)2         [∵ (a + b)2 = a2 + 2ab + b2]

= 900 + 120 + 4 = 1024

(ii) (35 )2 = (30 + 5)2 = (30)2 + 2 × 30 × 5 + (5)2          [∵ (a + 13)2 = a2 + 2ab + b2]

= 900 + 300 + 25 = 1225

(iii) (86)2 = (80 + 6)2 = (80)2 + 2 × 80 × 6 +(6)2         [∵ (a + 17)2 = a2 + 2ab + b2]

= 1600 + 960 + 36 = 7386

(iv) (93)2 = (90 + 3)2 = (90)2 + 2 × 90 × 3 + (3)2        [∵ (a+02 = a2 +2ab+b2]

= 8100 + 540 + 9 = 8649

(v) (71)2 = (70+1)2 = (70)2 +2 × 70 × 1+ (1)2                 [∵ (a+0)2 = a2 +2ab – Fb2]

= 4900 + 140 + 1 = 5041

(vi) (46)2 =(40 + 6)2 = (40)2 + 2 × 40 × 6 + (6)2        [∵ (a + 0)2 = a2 + 2ab – Fb2]

= 1600 + 480 + 36 = 2116

11.  There are three numbers 2rn, m2 – 1 and m+ 1 in a Pythagorean Triplet.

Here, 2m = 6           ⇒

Therefore,         Second number (m2 – 1) = (3)– 1 = 9 – 1 = 8

Third number m+ 1 = (3)+ 1 = 9 + 1 = 10

Hence, Pythagorean triplet is (6, 8, 10).

(ii) There are three numbers 2m, m2 – 1 and m2 + 1 in a Pythagorean Triplet.

Here, 2m = 14                   ⇒

Therefore,           Second number (m2 – 1) = (7)– 1 = 49 – 1 = 48

Third number m2 + 1 = (7)+ 1 = 49 + 1 = 50

Hence, Pythagorean triplet is (14, 48, 50).

(iii) There are three numbers 2m, m2 – 1 and m2 + 1 in a Pythagorean Triplet.

Here, 2m = 16            ⇒

Therefore,       Second number (m2 – 1) = (8)2 -1 = 64 – 1 = 63

Third number m2 + 1 = (8)2 + 1 = 64 + 1 = 65

Hence, Pythagorean triplet is (16, 63, 65).

(iv) There are three numbers 2m, m2 – 1 and m2 + 1 in a Pythagorean Triplet.

Here, 2m = 18            ⇒

Therefore,         Second number (m2 – 1) = (9)2 – 1 = 81 – 1 = 80

Third number m2 + 1 = (9)2 + 1 = 81 + 1 = 82

Hence, Pythagorean triplet is (18, 80, 82).

12.  Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are:

(i) 1          (ii) 6                (iii) 1                      (iv) 5

13.  Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.

(i) But given number 153 has its unit digit 3. So it is not a perfect square number.

(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.

(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.

(iv) Given number 441 has its unit digit 1. So it would be a perfect square number

14.  By successive subtracting odd natural numbers from 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

This successive subtraction is completed in 10 steps.

Therefore √100 =10

By successive subtracting odd natural numbers from 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

This successive subtraction is completed in 13 steps.

Therefore √169 = 13.

15.  (i) 729

(ii) 400

(iii) 1764

(iv) 4096

(v) 7744

(vi) 9604

(vii) 5929

(vii) 9216

(ix) 529

(x) 8100

16.  (i) 252 = 2 × 2 × 3 × 3 × 7

Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square.

∴ 252 × 7 = 1764

And √1764 = 2 × 3 × 7 = 42

(ii) 180 = 2 × 2 × 3 × 3 × 5

Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.

∴ 180 × 5 = 900

And √900 = 2 × 3 × 5 = 30

(iii) 1008 =2 × 2 × 2 × 2 × 3 × 3 × 7

Here, prime factor 7 has no pair. Therefore 1008 must be multiplied by 7 to make it a perfect square.

∴ 1008 × 7 = 7056

And √7056 = 2 × 2 × 3 × 7 = 84

(iv) 2028 = 2 × 2 × 3 × 13 × 13

Here, prime factor 3 has no pair. Therefore 2028 must be multiplied by 3 to make it a perfect square.

∴ 2028 × 3 = 6084

And    √6084 = 2 × 2 × 3 × 3 × 13 × 13 = 78

(v) 1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, prime factor 2 has no pair. Therefore 1458 must be multiplied by 2 to make it a perfect square.

∴ 1458 × 2 = 2916

And √2916 = 2 × 3 × 3 × 3 = 54

(vi) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, prime factor 3 has no pair. Therefore 768 must be multiplied by to make it a perfect square.

∴ 768 × 3 = 2304

And √2304 = 2 × 2 × 2 × 2 × 3 = 48

17.  (i) 252 = 2 × 2 × 3 × 3 × 7

Here, prime factor 7 has no pair. Therefore 252 must be divided by 7 to make it a perfect square.

∴ 252 ÷ 7 = 36

And √36 = 2 × 3 = 6

(ii) 2925 = 3 × 3 × 5 × 5 × 13

Here, prime factor 13 has no pair. Therefore 2925 must be divided by 13 to make it a perfect square.

∴ 2925 ÷ 13 = 225

And     √225 = 3 × 5 = 15

(iii) 396 = 2 × 2 × 3 × 3 × 11

Here, prime factor 11 has no pair. Therefore 396 must be divided by 11 to make it a perfect square.

∴ 396 ÷ 11 = 36

And √36 = 2 × 3 = 6

(iv) 2645 = 5 × 23 × 23

Here, prime factor 5 has no pair. Therefore 2645 must be divided by 5 to make it a perfect square.

∴ 2645 ÷ 5 = 529

And    √529 = 23 × 23 = 23

(v) 2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7

Here, prime factor 7 has no pair. Therefore 2800 must be divided by 7 to make it a perfect square.

∴ 2800 ÷ 7 = 400

And     √400 = 2 × 2 × 5 = 20

(vi) 1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5

Here, prime factor 5 has no pair. Therefore 1620 must be divided by 5 to make it a perfect square.

∴ 1620 ÷ 5 = 324

And     √324 = 2 × 3 × 3 = 18

18.  Here, Donated money = ` 2401

Let the number of students be x.

Therefore donated money = x × x

According to question,

Hence, the number of students is 49.

19.  Here, Number of plants = 2025

Let the number of rows of planted plants be x.

And each row contains number of plants = x

According to question,

Hence, each row contains 45 plants.

20.  L.C.M. of 4, 9 and 10 is 180.

Prime factors of 180 = 2 × 2 × 3 × 3 × 5

Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.

∴ 180 × 5 = 900

Hence, the smallest square number which is divisible by 4, 9 and 10 is 900.

21.  L.C.M. of 8, 15 and 20 is 120.

Prime factors of 120 = 2 × 2 × 2 × 3 × 5

Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be multiplied by 2 × 3 × 5 to make it a perfect square.

∴ 120 × 2 × 3 × 5 = 3600

Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600.

22.  (i) 2304

Hence, the square root of 2304 is 48.

(ii) 4489

Hence, the square root of 4489 is 67.

(iii) 3481

Hence, the square root of 3481 is 59.

(iv) 529

Hence, the square root of 529 is 23.

(v) 3249

Hence, the square root of 3249 is 57.

(vi) 1369

Hence, the square root of 1369 is 37.

(vii) 5776

Hence, the square root of 5776 is 76.

(viii) 7921

Hence, the square root of 7921 is 89.

(ix) 576

Hence, the square root of 576 is 24.

(x) 1024

Hence, the square root of 1024 is 32.

(xi) 3136

Hence, the square root of 3136 is 56.

(xii) 900

Hence, the square root of 900 is 30.

23.  (i) Here, 64 contains two digits which is even.

Therefore, number of digits in square root =

(ii) Here, 144 contains three digits which is odd.

Therefore, number of digits in square root =

(iv) Here, 4489 contains four digits which is even.

Therefore, number of digits in square root =

(v) Here, 390625 contains six digits which is even.

Therefore, number of digits in square root =

24.  (i) 2.56

Hence, the square root of 2.56 is 1.6.

(ii) 7.29

Hence, the square root of 7.29 is 2.7.

(iii) 51.84

Hence, the square root of 51.84 is 7.2.

(iv) 42.25

Hence, the square root of 42.25 is 6.5.

(v) 31.36

Hence, the square root of 31.36 is 5.6

25.  (i) 402

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 2. Therefore 2 must be subtracted from 402 to get a perfect square.

∴ 402 – 2 = 400

Hence, the square root of 400 is 20.

(ii) 1989

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 53. Therefore 53 must be subtracted from 1989 to get a perfect square.

∴ 1989 – 53 = 1936

Hence, the square root of 1936 is 44.

(iii) 3250

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 1. Therefore 1 must be subtracted from 3250 to get a perfect square.

∴ 3250 – 1 = 3249

Hence, the square root of 3249 is 57.

(iv) 825

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 41. Therefore 41 must be subtracted from 825 to get a perfect square.

∴ 825 – 41 = 784

Hence, the square root of 784 is 28.

(v) 4000

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 31. Therefore 31 must be subtracted from 4000 to get a perfect square.

∴ 4000 – 31 = 3969

Hence, the square root of 3969 is 63.

26.  (i) 525

Since remainder is 41. Therefore 222<525

Next perfect square number 232 = 529

Hence, number to be added = 529 – 525 = 4

∴ 525 + 4 = 529

Hence, the square root of 529 is 23.

(ii) 1750

Since remainder is 69. Therefore 412<1750

Next perfect square number 422 = 1764

Hence, number to be added = 1764 – 1750 = 14

∴ 1750 + 14 = 1764

Hence, the square root of 1764 is 42.

(iii) 252

Since remainder is 27. Therefore 152<252

Next perfect square number 162 = 256

Hence, number to be added = 256 – 252 = 4

∴ 252 + 4 = 256

Hence, the square root of 256 is 16.

(iv) 1825

Since remainder is 61. Therefore 422<1825

Next perfect square number 432 = 1849

Hence, number to be added = 1849 – 1825 = 24

∴ 1825 + 24 = 1849

Hence, the square root of 1849 is 43.

(v) 6412

Since remainder is 12. Therefore 802<6412

Next perfect square number 812 = 6561

Hence, number to be added = 6561 – 6412 = 149

∴ 6412 + 149 = 6561

Hence, the square root of 6561 is 81.

27.  Let the length of side of a square be x meter.

Area of square = (side)2 = x2

According to question,

28.  (a) Using Pythagoras theorem,

AC2 = AB2 + BC2

⇒ AC2 = (6)2 + (8)2

⇒ AC2 = 36 + 84 = 100

⇒ AC2 = 10 cm

(b) Using Pythagoras theorem,

AC2 = AB2 + BC2

⇒ (13)2 = AB2 + (5)2

⇒ 169 = AB2 + 25

⇒ AB = 169 – 25

⇒ AB= 144

⇒ AB = 12 cm.

29.  Here, plants = 1000

Since remainder is 39. Therefore 31 < 1000

Next perfect square number 32 = 1024

Hence, number to be added = 1024 – 1000 = 24

1000 + 24 = 1024

Hence, the gardener required 24 more plants.

30.  Here, Number of children = 500

By getting the square root of this number, we get,

In each row, the number of children is 22.

And left out children are 16.

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