9
SQUARE ROOTS AND
CUBE ROOTS
1. Find the smallest natural number by which the following numbers
must be multiplied to make them a perfect square:
(i) 1682
(ii) 2178
2 1682
Ans. (i) 1682 = 2 × 29 × 29
29 841
Since the factor 2 does not have its pair.
So, the given number must be multi29 29
plied by 2.
1
(ii) 2178 = 2 × 3 × 3 × 11 × 11
2 2178
Since the factor 2 does not have its pair,
3 1089
So, the given number must be multi3 363
plied by 2.
11 121
11 11
1
2. Find the smallest natural number by which the following number
must be divided to make them a perfect square:
(i) 2592
(ii) 16224
2 2592
Ans. (i) 2592 = 2 × 2 × 2 × 2 × 2 × 9 × 9
2 1296
Since the factor 2 does not have its pair,
so, the given number must be divided
2 648
by 2.
2 324
2 162
9
81
9
9
1
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2 16224
2 8112
2 4056
2 2028
2 1014
3
507
13 169
13
13
1
3. What least number must be subtracted from the following numbers
to make them a perfect square?
(i) 8934
(ii) 11021
Ans. (i) The square root of 8934 can be calculated as:
(ii) 16224 = 2 × 2 × 2 × 2 × 2 × 3 × 13 × 13
Since the factor 2 × 3 i.e. 6 does not have its
pair, so, the given number must be
divided by 6.
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94
9 8934
81
184 834
736
98
188
Hence, we will have to subtract 98.
(ii) The square root of 11021 can be calculated as:
104
1 11021
1
204 1021
816
205
Hence, we will have to subtract 205.
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4. What least number must be added to the following numbers to make
them perfect squares?
(i) 5678
(ii) 8000
Ans. (i) The square root of 5678 can be calcu75
lated as:
7 5678
Hence, we will have to add
49
2
= (76) – 5678
145 778
= 5776 – 5678 = 98
725
53
(ii) The square root of 8000 can be calcu89
lated as:
8 8000
Hence, we will have to add
64
2
= (90) – 8000
169 1600
= 8100 – 8000 = 100
1521
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79
5. Find the square root of:
(i) 5184
67
121
(v) 0.018769
(ii) 9
(iv) 67.0761
Ans. (i) 5184
= 2 × 2 × 2 × 2 × 2 × 2 × 3× 3× 3× 3
= 2 × 2 × 2 × 3 × 3 = 72
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(iii) 25
544
729
2
2
2
2
2
2
3
3
3
3
5184
2592
1296
648
324
162
81
27
9
3
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(ii)
9
67
9 × 121 + 67
=
121
121
1156
2 × 2 × 17 × 17
=
121
11 × 11
2 × 17 34
1
=
=3
=
11
11
11
=
(iii)
25
544
25 × 729 + 544
=
729
729
2
2
17
17
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1156
578
289
17
1
11 121
11 11
1
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18769 137
2
18769
18769
=
=
=5
=
27
27
729
729
729
137
27
1 18769
2 729
1
4
23 87
47 329
69
329
267 1869
×
1869
×
(iv) 67.0761
=
8.19
8 67.0761
64
161 307
161
1629 14661
14661
×
67.0761 = 8.19
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(v) 0.018769
0.137
1 0.018769
1
87
23
69
267
1869
1869
×
0.018469 = 0.137
6. Find the value of the following:
(i)
(iii)
Ans. (i)
(ii)
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(ii)
85 − 16
1.3
2
− 1.2
2
85 − 16 = 85 − 4 = 81 = 9 × 9 = 9
2
14
21
256
121
− 1
=
−
121
100
121
100
=
=
(iii)
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14
21
2
− 1
121
100
(1.3)2 − (1.2 )2
Math Class VIII
16 × 16
11× 11
−
11× 11
10 × 10
16 11 160 − 121
39
− =
=
11 10
110
110
=
(1.3 + 1.2 )(1.3 − 1.2 )
=
( 2.5)( 0.1) =
=
5× 5
5
=
= 0⋅5
10 × 10 10
5
0.25 =
25
100
Question Bank
7. Find the value of 15625 and hence evaluate 156.25 + 1.5628.
125
Ans. 15625 = 125
1 15625
15625
1
= 12.5
Thus 156.25 =
100
22 56
44
15625 125
245 1225
=
= 1.25
and 1.5625 =
10000 100
1225
×
Hence, 156.25 + 1.5625 = 12.5 + 1.25 = 13.75
8. Evaluate :
(i) 99 × 396
(ii)
147 × 243
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25 13
× 2 × 0.25
32 18
(iii)
Ans. (i)
99 × 396
= 99 × 396
2
2
3
3
11
396
198
99
33
11
1
3
7
7
147
49
7
1
= 3 × 3 × 11 × 11 × 2 × 2 × 3 × 3
32 × 112 × 22 × 32
=
3
3
3
3
3
= 3 × 11 × 2 × 3 = 198
(ii)
147 × 243
= 147 × 243
=
3× 7 × 7 × 3× 3× 3× 3× 3
243
81
27
9
3
1
32 × 32 × 32 × 7 2
= 3 × 3 × 3 × 7 = 189
=
(iii)
25 13
25 49
× 2 × 0.25 =
× × 0.25
32 18
32 18
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=
25 49 25
25 × 49 × 25
× ×
=
32 18 100
32 × 18 × 100
=
25 × 49
32 × 18 × 4
=
5× 5× 7 × 7
( 2 × 2 × 2 × 2 × 2 ) × ( 2 × 3 × 3) × ( 2 × 2 )
=
5× 5× 7 × 7
2 × 2 × 2 × 2 × 2 × 2 × 3× 3× 2 × 2
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5× 7
35
=
2 × 2 × 2 × 3 × 2 48
9. A man, after a tour, finds that he had spent every day as many rupees
as the number of days he had been on tour. How long did his tour
last, if he had spent in all Rs 1,296?
Ans. Let the number of days he had spent be x
∴ Number of rupees spent in each day = x
∴ Total money spent = x × x = x2 = 1,296
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=
Thus
x = 1296 ⇒ x = 4 × 4 × 9 × 9
⇒ x=4×9
⇒ x = 36
10. A society collected Rs 9,216, each member contributed as many
rupees as there were members in the society. Find the number of
members.
Ans.Let the number of members be x.
∴ Money collected by each member be x.
Thus total money collected = x × x = x2 = 9216
x = 4× 4× 4× 4× 6× 6
⇒ x = 9216
⇒
x = 96
⇒ x=4×4×6 ⇒
Hence, number of members are 96.
11. A basket contains 125 flowers. A man goes for worship and puts
as many flowers in each temple as there are temples in the city.
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If the man needs 20 baskets of flowers, find the number of flowers
that he puts in each temple.
Ans.Let the number of flowers put in each temple be x.
Thus total number of flowers used = x × x = x2 = 125 × 20
x2 = 2500
⇒
x = 50 × 50
x = 50
∴
Hence, the number of flowers put it each temple is 50.
12. Find the square root of the following numbers by division method :
(i) 213444
(ii) 18.4041
(iii) 5.774409
⇒
Ans. (i)
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(ii)
213444
= 462 × 462 = 462
462
4 213444
16
86 534
516
922
1844
1844
×
(iii)
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18.4041
= 4.29 × 4.29 = 4.29
4.29
4 18.4041
16
82 240
164
849
7641
7641
×
5.774409
= 2.403 × 2.403 = 2.403
2.403
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44 177
176
4803 14409
14409
×
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13. Find the square root of 3 correct to 3 significant figures. Hence
2+ 3
correct to 2 decimal places.
2− 3
find the value of
Ans. 3 = 1.73
2+ 3
2+ 3 2+ 3
=
×
2− 3
2− 3 2+ 3
Now
2
(2 + 3) = (2 + 3)
2
4−3
( 2 )2 − ( 3 )
1
2
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27
1.732
3.000000
1
200
189
1100
1029
7100
6924
176
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=
=
2+ 3
343
3462
2
= 2+ 3
1
= 2 + 1.73 = 3.73
2+ 3
= 3.73
2− 3
14. Find the least number which must be subtracted from 7581 to obtain
a perfect square. Find this perfect square and its square root.
87
Ans.
8 7581
64
167 1181
1169
12
Hence,
∴ The number which is to be subtracted = 12 and
perfect square
= 7581 – 12 = 7569 and square root = 87
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15. Find the least number which must be subtracted from 43379 to obain
a perfect square. Find the perfect square and its square root.
208
Ans.
2 43379
4
408 3379
3264
115
Hence, the number which must be subtracted = 115
and perfect square = 43379 – 115 = 43264
and square root = 43264 = 208
16. Find the least number which must be added to 6203 to obtain a
perfect square. Find the perfect square and its square root.
79
Ans.
7 6203
49
149 1303
1341
–38
1341 – 1303 = 38 must be added in order to get a perfect square.
Thus the number to be added = 38
Perfect square = 6203 + 38 = 6241 and square root = 79
17. Find the least number which must be added to 506900 to make it
a perfect square. Find this perfect square and its square root.
Ans.
712
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141 169
141
2800
1422
2844
–44
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2844 – 2800 = 44 must be added to get a perfect square
Thus number to be added = 44
Perfect square = 506900 + 44 = 506944 and square root = 712
18. Find the greatest number of six-digit which is a perfect square. Find
the square root of the number.
Ans. The greatest 6-digit number = 999999
999
81
189 1899
1701
1989 19899
17901
1998
Thus the required 6-digits number which is a perfect square
= 999999 – 1998 =998001.
Hence, square root of the number is 999.
19. The students of a class arranged a picnic. Each student contributed
as many rupees as the number of students in the class. If the total
contribution is Rs 2601, find the strength of the class.
51
Ans.Total contribution = Rs 2601
5 2601
Let number of students be x
25
∴ Contribution of each students = Rs x
101 101
Thus Rs x × x = x2 = 2601
101
x = 51
⇒ x = 2601
⇒
×
Hence, number of students are 51
20. Test whether the given number is a perfect cube or not : 2 3380
(i) 3380
(ii) 10584
2 1690
Ans. (i) 3380 = 2 × 2 × 5 × 13 × 13
5 845
Making a groups of 3, its factors are
13 169
not in triplet.
13 13
Hence, it is not a perfect cube
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2
2
2
3
3
3
7
7
(ii) 10584 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7
= 23 × 33 × 7 × 7
Thus its factors, 7 is not in triplet.
Hence, it is not a perfect cube.
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B
10584
5292
2646
1323
441
147
49
7
1
21. Find the smallest number by which 11979 must be multiplied so
that the product is a perfect cube.
3 11979
Ans.Resolving 11979 into prime factors, we get
3 3993
11979 = 3 × 3 × 11 × 11 × 11
11 1331
To make the above number a perfect cube
11 121
we have to multiply it by 3.
11 11
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22. Find the smallest number by wihch 8788 must be divided so that
the quotient is a perfect cube.
2 8788
Ans.Writing 8788 as a product of prime factors, we get
2 4394
8788 = 2 × 2 × 13 × 13 × 13
13 2197
from above it is clear that the smallest
13 169
number to divide the given number to make
13 13
it perfect cube is = 2 × 2 = 4
1
Hence, the required number is 4.
23 . Find the cube root of :
(i) 74088
(ii)
5
1182
2197
(iii) 4
508
1331
(iv) 42.875
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Ans. (i) 74088 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7
3
74088 = 2 × 3 × 7 = 42
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2
2
3
3
3
7
7
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1182 12167 ( 23 × 23 × 23)
(ii) 5
=
=
2197 2197 (13 × 13 × 13)
74088
37044
18522
9261
3087
1029
343
49
7
1
23 12167
23 529
23 23
1182 3 ( 23 × 23 × 23)
1
Thus , 3 5
=
13 2197
2197
13 × 13 × 13
13 169
23 10
13 13
=
=1
13
13
1
2 5832
508 3 508 3 4 × 1331 + 508
(iii) 4
= 4
=
2 2916
1331
1331
1331
2 1458
3
5832
5832
3 729
=3
= 3
1331
3 243
1331
3 81
2
×
2
×
2
×
3
×
3
×
3
×
3
×
3
×
3
3 27
=3
11× 11 × 11
3 9
3 3
23 × 33 × 33 2 × 3 × 3
3
11 1331
=
=
1
3
11
11
11 121
11 11
18
7
=1
=
11
11
1
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(iv)
3
5
5
5
7
7
7
3
3
5
×
7
42.875 = 3
103
5× 5× 5× 7 × 7 × 7
=3
10 × 10 × 10
42875
8575
1715
343
49
7
1
3
3
5
7
5 × 7 35
×
=
=
= 3.5
=3
3
10
10
10
24. Multiply 6561 by the smallest number so that the product is a perfect
3 6561
cube. Also find the cube root of the product.
3 2187
Ans. Splitting 6561 into prime factors, we have
3 729
6561 = 3 × 3 × 3 ×3 × 3 × 3 × 3 × 3
3 243
The required number is 3 which is to be
3 81
multiplied to 6561, so that the product is a
perfect cube.
3 27
3 9
Hence the cube root of the product
3
= 3 × 3 × 3 = 27.
25. Divide 137592 by the smallest number so that the quotient is a
perfect cube. Also find the cube root of the quotient.
Ans.First of all, we find the prime factorisation of 137592
137592 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 13
2 137592
We note that the prime factors 2 and 3 occur
2 68796
thrice while prime factors 7 and 13 occur
2 34398
twice and once. Therefore, the smallest
3 17199
number by which the given number must
3 5733
be divided so that the quotient is a perfect
3 1911
cube is 7 × 7 × 13 i.e. 637.
7 637
137592
Also the quotient =
7 91
637
Now 216 = 2 × 2 × 2 × 3 × 3 × 3
13 13
= (2 × 2 × 2) × (3 × 3 × 3)
1
Hence, 3 216 = 2 × 3 = 6.
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