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**Q.1 What could be the possible â€˜oneâ€™sâ€™ digits of the square root of each of the following numbers?
**

Hence, the oneâ€™s digit of square root of 9801 is either 1 or 9.

(ii) 99856 - The digit 99856 ends with â€˜6â€™ then the oneâ€™s digit of the square root of that number may be 4 or 6.

Hence, the oneâ€™s digit of square root of 99856 is either 4 or 6.

(iii) 998001 - The digit 998001 ends with â€˜1â€™ then the oneâ€™s digit of the square root of that number may be 1 or 9.

Hence, the oneâ€™s digit of square root of 998001 is either 1 or 9.

(iv) 657666025 - The digit 657666025 ends with â€˜5â€™ then the oneâ€™s digit of the square root of that number will be 5.

Hence, the oneâ€™s digit of square root of 657666025 is 5.

**Q.2 Without doing any calculation, find the numbers which are surely not perfect squares.
**

(i) 153

Since, 153 ends with â€˜3â€™ it cannot be a perfect square number.

(ii) 257

Since, 257 ends with â€˜7â€™ it cannot be a perfect square number.

(iii) 408

Since, 408 ends with â€˜8â€™ it cannot be a perfect square number.

(iv) 441

Since, 441 ends with â€˜1â€™ it can be a perfect square number.

**Q.3 Find the square roots of 100 and 169 by the method of repeated subtraction.
**

Finding square root of 100 by using repeated subtraction:

(i) 100 â€“ 1 = 99 (ii) 99 â€“ 3 = 96 (iii) 96 â€“ 5 = 91 (iv) 91 â€“ 7 = 84

(v) 84 â€“ 9 = 75 (vi) 75 â€“ 11 = 64 (vii) 64 â€“ 13 = 51 (viii) 51 â€“ 15 = 36

(ix) 36 â€“ 17 = 19 (x) 19 â€“ 19 = 0

We obtained zero at the 10

Finding square root of 169 by using repeated subtraction:

(i) 169 â€“ 1 = 168 (ii) 168 â€“ 3 = 165 (iii) 165 â€“ 5 = 160 (iv) 160 â€“ 7 = 153

(v) 153 â€“ 9 = 144 (vi) 144 â€“ 11 = 133 (vii) 133 â€“ 13 = 120 (viii) 120 â€“ 15 = 105

(ix) 105 â€“ 17 = 88 (x) 88 â€“ 19 = 69 (xi) 69 â€“ 21 = 48 (xii) 48 â€“ 23 = 25

(xii) 25 â€“ 25 = 0

We obtained zero at the 13^{th} step. Hence, = 13.

**Q.4 Find the square roots of the following numbers by the Prime Factorisation Method.
**

3 | 729 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

Â | 1 |

Here, 729 = 3 x 3 x 3 x 3 x 3 x 3

Therefore, = 3 x 3 x 3 = 27.

(ii) 400

2 | 400 |

2 | 200 |

2 | 100 |

2 | 50 |

5 | 25 |

5 | 5 |

Â | 1 |

Here, 400 = 2 x 2 x 2 x 2 x 5 x 5

Therefore, = 2 x 2 x 5 = 20

(iii) 1764

2 | 1764 |

2 | 882 |

3 | 441 |

3 | 147 |

7 | 49 |

7 | 7 |

Â | 1 |

Here, 1764 = 2 x 2 x 3 x 3 x 7 x 7

Therefore, = 2 x 3 x 7 = 42

(iv) 4096

2 | 4096 |

2 | 2048 |

2 | 1024 |

2 | 512 |

2 | 256 |

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

Â | 1 |

Here, 4096 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Therefore, = 2 x 2 x 2 x 2 x 2 x 2 = 64

(v) 7744

2 | 7744 |

2 | 3872 |

2 | 1936 |

2 | 968 |

2 | 484 |

2 | 242 |

11 | 121 |

11 | 11 |

Â | 1 |

Here, 7744 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11

Therefore, = 2 x 2 x 2 x 11 = 88

(vi) 9604

2 | 9604 |

2 | 4802 |

7 | 2401 |

7 | 343 |

7 | 49 |

7 | 7 |

Â | 1 |

Here, 9604 = 2 x 2 x 7 x 7 x 7 x 7

Therefore, = 2 x 7 x 7 = 98

(vii) 5929

7 | 5929 |

7 | 847 |

11 | 121 |

11 | 11 |

Â | 1 |

Here, 5929 = 7 x 7 x 11 x 11

Therefore, = 7 x 11 = 77

(viii) 9216

2 | 9216 |

2 | 4608 |

2 | 2304 |

2 | 1152 |

2 | 576 |

2 | 288 |

2 | 144 |

2 | 72 |

2 | 36 |

2 | 18 |

3 | 9 |

3 | 3 |

Â | 1 |

Here, 9216 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

Therefore, = 2 x 2 x 2 x 2 x 2 x 3 = 96

(ix) 529

23 | 529 |

23 | 23 |

Â | 1 |

Here, 529 = 23 x 23

Therefore, = 23

(x) 8100

2 | 8100 |

2 | 4050 |

3 | 2025 |

3 | 675 |

3 | 225 |

3 | 75 |

5 | 25 |

5 | 5 |

Â | 1 |

Here, 8100 = 2 x 2 x 3 x 3 x 3 x 3 x 5 x 5

Therefore, = 2 x 3 x 3 x 5 = 90

**Q.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.
**

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 | 7 |

Â | 1 |

Here, 252 = 2 x 2 x 3 x 3 x 7

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

Therefore, 252 x 7 = 1764

Therefore, = 2 x 3 x 7 = 42

(ii) 180

2 | 180 |

2 | 90 |

3 | 45 |

3 | 15 |

5 | 5 |

Â | 1 |

Here, 180 = 2 x 2 x 3 x 3 x 5

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

Therefore, 180 x 5 = 900

Therefore, = 2 x 3 x 5 = 30

(iii) 1008

2 | 1008 |

2 | 504 |

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 | 7 |

Â | 1 |

Here, 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7

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

Therefore, 1008 x7 = 7056

Therefore, = 2 x 2 x 3 x 7 = 84

(iv) 2028

2 | 2028 |

2 | 1014 |

3 | 507 |

13 | 169 |

13 | 13 |

Â | 1 |

Here, 2028 = 2 x 2 x 3 x 13 x 13

Here, prime factor 3 has no pair. Therefore 2028 must be multiplied by 7 to make it a perfect square.

Therefore, 2028 x3 = 6084

Therefore, = 2 x 3 x 13 = 78

(v) 1458

2 | 1458 |

3 | 729 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

Â | 1 |

Here, 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3

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

Therefore, 1458 x2 = 2916

Therefore, = 2 x 3 x 3 x 3 = 54

(vi) 768

2 | 768 |

2 | 384 |

2 | 192 |

2 | 96 |

2 | 48 |

2 | 24 |

2 | 12 |

2 | 6 |

3 | 3 |

Â | 1 |

Here, 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3

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

Therefore, 768 x3 = 2304

Therefore, = 2 x 2 x 2 x 2 x 3 = 48

**Q.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.
**

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 | 7 |

Â | 1 |

Here, 252 = 2 x 2 x 3 x 3 x 7

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

Therefore, 252 Ã·7 = 36

Therefore, = 2 x 3 = 6

(ii) 2925

3 | 2925 |

3 | 975 |

5 | 325 |

5 | 65 |

13 | 13 |

Â | 1 |

Here, 2925 = 3 x 3 x 5 x 5 x 13

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

Therefore, 2925 Ã·13 = 225

Therefore, = 3 x 5 = 15

(iii) 396

2 | 396 |

2 | 198 |

3 | 99 |

3 | 33 |

11 | 11 |

Â | 1 |

Here, 396 = 2 x 2 x 3 x 3 x 11

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

Therefore, 396 Ã·11 = 36

Therefore, = 2 x 3 = 6

(vi) 2645

5 | 2645 |

23 | 198 |

23 | 99 |

Â | 1 |

Here, 2645 = 5 x 23 x 23

Here, prime factor 5 has no pair. Therefore 396 must be multiplied by 5 to make it a perfect square.

Therefore, 2645 Ã·5 = 529

Therefore, = 23 x 23 = 23

(v) 2800

2 | 2800 |

2 | 1400 |

2 | 700 |

2 | 350 |

5 | 175 |

5 | 35 |

1 | 7 |

Â | 1 |

Here, 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7

Here, prime factor 7 has no pair. Therefore 396 must be multiplied by 7 to make it a perfect square.

Therefore, 2800 Ã·7 = 400

Therefore, = 2 x 2 x 5 = 20

(vi) 1620

2 | 1620 |

2 | 810 |

3 | 405 |

3 | 135 |

3 | 45 |

3 | 15 |

5 | 5 |

Â | 1 |

Here, 1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5

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

Therefore, 1620 Ã·5 = 324

Therefore, = 2 x 3 x 3 = 18

**Q.7 The students of Class VIII of a school donated Rs 2401 in all, for Prime Ministerâ€™s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
**

Now, total amount donated is Rs 2401.

Hence, number of students in class =

Here, 2401 = 7 x 7 x 7 x 7

= 7 x 7 = 49.

Therefore, number of students in the class is 49.

**Q.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.
**

Therefore, number of rows = number of plants in each row

Here, Total number of plants = number of rows x number of plants in each row

Number of rows x number of plants in each row = 2025

(Number of rows)

Number of rows =

Now, 2025 = 5 x 5 x 3 x 3 x 3 x 3

Therefore, = 5 x 3 x 3 = 45

Hence, the number of rows and the number of plants in each row is 45.

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

2 | 180 |

2 | 90 |

3 | 45 |

3 | 15 |

5 | 5 |

Â | 1 |

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

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

Therefore, 180 x 5 = 900

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

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

** Sol.** The LCM of 8, 15 and 20 is 120.

2 | 120 |

2 | 60 |

2 | 30 |

3 | 15 |

5 | 5 |

Â | 1 |

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

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

Therefore, 120 x 2 x 3 x 5 = 3600

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

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very nice brother and thank you

all questions helps me to under stand problems

its very big answer

Nice site . All que are correct but one mistake in que 5 you write multipied instead of divide and the roman number in same question point v is uncorrect u write 6th