Exercise 6.1

Q.1 What will be the unit digit of the squares of the following numbers?
(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880
(ix) 12796 (x) 55555
Sol. (i) 81
The digit 81 ends with ‘1’. And square of 1 is 1.
Hence, the unit digit of square of 81 is 1.

(ii) 272
The digit 272 ends with ‘2’. And square of 2 is 4.
Hence, the unit digit of square of 272 is 4.

(iii) 799
The digit 799 ends with ‘9’. And square of 9 is 81.
Hence, the unit digit of square of 799 is 1.

(iv) 3853
The digit 3853 ends with ‘3’. And square of 3 is 9.
Hence, the unit digit of square of 3583 is 9.

(v) 1234
The digit 1234 ends with ‘4’. And square of 4 is 16.
Hence, the unit digit of square of 1234 is 6.

(vi) 26387
The digit 26387 ends with ‘7’. And square of 7 is 49.
Hence, the unit digit of square of 26387 is 9.

(vii) 52698
The digit 52698 ends with ‘8’. And square of 8 is 64.
Hence, the unit digit of square of 52698 is 4.

(viii) 99880
The digit 99880 ends with ‘0’. And square of 0 is 0.
Hence, the unit digit of square of 99880 is 0.

(ix) 12796
The digit 12796 ends with ‘6’. And square of 6 is 36.
Hence, the unit digit of square of 12796 is 6.

(x) 55555
The digit 55555 ends with ‘5’. And square of 5 is 25.
Hence, the unit digit of square of 55555 is 5.

Q.2 The following numbers are obviously not perfect squares. Give reason.
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222
(v) 64000 (vi) 89722 (vii) 222000 (viii) 505050
Sol. (i) 1057
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(ii) 23453
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(iii) 7928
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(iv) 222222
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(v) 64000
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(vi) 89722
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(vii) 222000
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(viii) 505050
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

Q.3 The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Sol. (i) 431
The digit 431 ends with ‘1’. And square of 1 is 1.
Hence, the square of 431 would be an odd number.

(ii)2826
The digit 2826 ends with ‘6’. And square of 6 is 36.
Hence, the square of 2826 would not be an odd number.

(iii)7779
The digit 7779 ends with ‘9’. And square of 9 is 81.
Hence, the square of 7779 would be an odd number.

(iv)82004
The digit 82004 ends with ‘4’. And square of 4 is 16.
Hence, the square of 82004 would not be an odd number.

Q.4 Observe the following pattern and find the missing digits.
11² = 121
101² = 10201
1001² = 1002001
100001² = 1 ……… 2 ……… 1
10000001² = ………………………
Sol.
100001² = 10000200001
10000001² = 100000020000001

Q.5 Observe the following pattern and supply the missing numbers.
11² = 1 2 1
101² = 1 0 2 0 1
10101² = 102030201
1010101² = ………………………
………… ² = 10203040504030201
Sol.
1010101² = 1020304030201
101010101² = 10203040504030201

Q.6 Using the given pattern, find the missing numbers.
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + _² = 21²
5² + _² + 30² = 31²
6² + 7² + _² = __²
Sol.
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² +42² = 43²

Q.7 Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Sol. We know that the sum of first n odd natural numbers is n².
(i) 1 + 3 + 5 + 7 + 9
Here, there are five odd numbers.
Therefore, 1 + 3 + 5 + 7 + 9 = 5² = 25.

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
Here, there are ten odd numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 = 10² = 100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Here, there are twelve odd numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23= 12² = 144

Q.8 (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Sol. We know that the sum of first n odd natural numbers is n².

(i) 49 = 7². Therefore, it is the sum of 7 odd numbers.
Hence, 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) 121 = 11². Therefore, it is the sum of 11 odd numbers.
Hence, 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Q.9 How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
Sol. We know that there will be 2n numbers in between the squares of the numbers n and (n+1).
(i) 12 and 13
Here, n = 12. Therefore, there will 2n = 2 x 12 = 24 square numbers between 12 and 13.

(ii) 25 and 26
Here, n = 25. Therefore, there will 2n = 2 x 25 = 50 square numbers between 25 and 26.

(iii) 99 and 100
Here, n = 99. Therefore, there will 2n = 2 x 99 = 198 square numbers between 99 and 100.

 

Exercise 6.2

 

Q.1 Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93  (v) 71 (vi) 46
Sol. (i) 32
32² = (30 + 2)²
30(30 + 2) + 2(30 + 2)
30² + 30 x 2 + 2 x 30 + 2²
900 + 60 +60 + 4
1024

(ii) 35
35² = (30 + 5)²
30(30 + 5) + 5(30 + 5)
30² + 30 x 5 + 5 x 30 + 5²
900 + 150 +150 + 25
1225

(iii) 86
86² = (80 + 6)²
80(80 + 6) + 6(80 + 6)
80² + 80 x 6 + 6 x 80 + 6²
6400 + 480 +480 + 36
7396

(iv) 93
93² = (90 + 3)²
90(90 + 3) + 3(90 + 3)
90² + 90 x 3 + 3 x 90 + 3²
8100 + 270 +270 + 9
8649

(v) 71
71² = (70 + 1)²
70(70 + 1) + 1(70 + 1)
70² + 70 x 1 + 1 x 70 + 1²
4900 + 70 +70 + 1
5041

(vi) 46
46² = (40 + 6)²
40(40 + 6) + 6(40 + 6)
40² + 40 x 6 + 6 x 40 + 6²
1600 + 240 +240 + 36
2116

Q.2 Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv)18
Sol. We know that, for any natural number m > 1, we have (2m) ² + (m² – 1)² = (m² + 1)² where, 2m, m² – 1 and m² + 1 forms a Pythagorean triplet.

(i) 6
For 6 to be member of Pythagorean triplet we select 2m = 6.
Therefore, m = 3.
Second number = (m² – 1) = (3² – 1) = 9 – 1 = 8
Third number = (m² + 1) = (3² + 1)² = 9 + 1 = 10
Thus, the Pythagorean triplet is (6, 8, 10).

(ii) 14
For 14 to be member of Pythagorean triplet we select 2m = 14.
Therefore, m = 7.
Second number = (m² – 1) = (7² – 1) = 49 – 1 = 48
Third number = (m² + 1) = (7² + 1) = 49 + 1 = 50
Thus, the Pythagorean triplet is (14, 48, 50).

(iii) 16
For 16 to be member of Pythagorean triplet we select 2m = 16.
Therefore, m = 8.
Second number = (m² – 1) = (8² – 1) = 64 – 1 = 63
Third number = (m² + 1) = (8² + 1) = 64 + 1 = 65
Thus, the Pythagorean triplet is (16, 63, 65).

(iv)18
For 18 to be member of Pythagorean triplet we select 2m = 18.
Therefore, m = 9.
Second number = (m² – 1) = (9² – 1) = 81 – 1 = 80
Third number = (m² + 1) = (9² + 1) = 81 + 1 = 82
Thus, the Pythagorean triplet is (18, 80, 82).

 

Exercise 6.3

Q.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
Sol. (i) 9801 – The digit 9801 ends with ‘1’ then the one’s digit of the square root of that number may be 1 or 9.
Hence, the one’s digit of square root of 9801 is either 1 or 9.

(ii) 99856 – The digit 99856 ends with ‘6’ then the one’s digit of the square root of that number may be 4 or 6.
Hence, the one’s digit of square root of 99856 is either 4 or 6.

(iii) 998001 – The digit 998001 ends with ‘1’ then the one’s digit of the square root of that number may be 1 or 9.
Hence, the one’s digit of square root of 998001 is either 1 or 9.

(iv) 657666025 – The digit 657666025 ends with ‘5’ then the one’s digit of the square root of that number will be 5.
Hence, the one’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 ‘3’ it cannot be a perfect square number.
(ii) 257
Since, 257 ends with ‘7’ it cannot be a perfect square number.

(iii) 408
Since, 408 ends with ‘8’ it cannot be a perfect square number.

(iv) 441
Since, 441 ends with ‘1’ 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 n².
Finding square root of 100 by using repeated subtraction:
(i) 100 – 1 = 99 (ii) 99 – 3 = 96 (iii) 96 – 5 = 91 (iv) 91 – 7 = 84
(v) 84 – 9 = 75 (vi) 75 – 11 = 64 (vii) 64 – 13 = 51 (viii) 51 – 15 = 36
(ix) 36 – 17 = 19 (x) 19 – 19 = 0
We obtained zero at the 10^th step. Hence, = 10.

Finding square root of 169 by using repeated subtraction:
(i) 169 – 1 = 168 (ii) 168 – 3 = 165 (iii) 165 – 5 = 160 (iv) 160 – 7 = 153
(v) 153 – 9 = 144 (vi) 144 – 11 = 133 (vii) 133 – 13 = 120 (viii) 120 – 15 = 105
(ix) 105 – 17 = 88 (x) 88 – 19 = 69 (xi) 69 – 21 = 48 (xii) 48 – 23 = 25
(xii) 25 – 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
	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
	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
	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
	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
	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
	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
	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
	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
	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
	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) 2028  (v) 1458 (vi) 768
Sol. (i) 252

2	252
2	126
3	63
3	21
7	7
	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
	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
	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
	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
	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
	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) 2645  (v) 2800 (vi) 1620
Sol. (i) 252

2	252
2	126
3	63
3	21
7	7
	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 ÷7 = 36
Therefore, = 2 x 3 = 6

(ii) 2925

3	2925
3	975
5	325
5	65
13	13
	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 ÷13 = 225
Therefore, = 3 x 5 = 15

(iii) 396

2	396
2	198
3	99
3	33
11	11
	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 ÷11 = 36
Therefore, = 2 x 3 = 6

(vi) 2645

5	2645
23	198
23	99
	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 ÷5 = 529
Therefore, = 23 x 23 = 23

(v) 2800

2	2800
2	1400
2	700
2	350
5	175
5	35
1	7
	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 ÷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
	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 ÷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 Minister’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
	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
	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.