Q.1 What could be the possible â€˜oneâ€™sâ€™ digits of the square root of each of the following numbers?
(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025
Sol. (i) 9801 - The digit 9801 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 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 (ii) 257 (iii) 408 (iv) 441
Sol. We know that the perfect square of a number can end only with 0, 1, 4, 5, 6, 9 or even number of zeros.
(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.
Sol. We know that the sum of first n odd natural numbers is n^{2}.
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^{th} step. Hence, = 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.
(i) 729 (ii) 400 (iii) 1764 (iv) 4096
(v) 7744 (vi) 9604 (vii) 5929 (viii) 9216
(ix) 529 (x) 8100
Sol. (i) 729
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.
(i) 252 (ii) 180 (iii) 1008 (iv) 2028Â Â (v) 1458 (vi) 768
Sol. (i) 252
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.
(i) 252 (ii) 2925 (iii) 396 (iv) 2645Â Â (v) 2800 (vi) 1620
Sol. (i) 252
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.
Sol. Given, each student donated as many rupees as the number of students in the class. Therefore, number of students in class will be the square root of the amount donated by the students of 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.
Sol. Given, each row contains as many plants as the number of rows.
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)^{2} = 2025
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.
Sol. The LCM of 4, 9 and 10 is 180.
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.
Get CBSE Exam Ready in 1-Month |
Avail 90% Off |
Good site
It's help me
GoÃ²dddd
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