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Published on: 04/09/2019
Cubes and Cube Roots
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Questions + Answers key
Take MCQ Mathematics Test
1.
Which of the following are perfect cubes?
400
2.
1 = 1 = 13
3 + 5 = 8 = 23
7 + 9 + 11 = 27 = 33
13 + 15 + 17 + 19 = 64 = 43
21 + 23 + 25 + 27 + 29 = 125 = 53
Express the following numbers as the sum of odd numbers using the above pattern.
83
3.
Find the one's digit of the cube of each of the following numbers.
1005
4.
Find the one's digit of the cube of each of the following numbers.
3331
5.
Cube of an odd number is odd.
6.
If a2 ends in 9, then a3 ends in 7.
7.
If a2 ends in 5, then a3 ends in 25.
8.
There are five perfect cubes between 1 to 100
9.
The cube of 0.4 is 0.064.
10.
The cube of a two digits number may have seven or more digits.
11.
There is no perfect cube which ends with 8.
12.
If square of a number ends with 5, then its cube ends with 25.
13.
A perfect cube does not end with two zeroes.
14.
Cube of any odd number is even.
15.
Using prime factorisation, find the cube root of 5832.
16.
Find the cube root of 614125 through estimation.
17.
Is 68600 a perfect cube? If not, find the smallest number by which 68600 must be multiplied to get a perfect cube.
18.
Check whether 21600 is a perfect cube or not.
19.
The value of \(\sqrt [ 3 ]{ 512 } \times \sqrt [ 3 ]{ -27 } \) is
24
-24
48
-48
20.
Which of the following is the cube of an even natural number?
1331
4913
3375
1728
21.
The cube root of \(\frac { -64 }{ 125 } \) is
\(\frac { 8 }{ 5 } \)
\(\frac { -8 }{ 5 } \)
\(\frac { -4 }{ 5 } \)
\(\frac { 5 }{ -8 } \)
22.
Which of the following number is a perfect cube?
243
512
392
23.
The one's digit of the cube of 23 is
3
6
7
24.
Cube of a number ending in 7 will end in the digit ______
25.
The least number by which 72 is divided to make it a perfect cube is _____
26.
The least number by which 72 is multiplied to make it a perfect cube is _____
27.
The least number by which 250 is multiplied to make it a perfect cube is _____
28.
One's digit in the cube of 27 is _____
29.
Three numbers are in ratio to one another 2: 3: 4. The sum of their cubes is 33957. Find the numbers.
30.
What is the smallest number by which 392 may be divided so that the quotient is a perfect cube?
1.
We have, 400
Resolving 400 into prime factors, we get
400 = 2 x 2 x 2 x 2 x 5 x 5
Clearly, the prime factors 2 and 5 do not appear in group of three (triples).
So, 400 is not a perfect cube.
2.
From the given pattern, we observe that it follows the relation
n3 = [n (n -1) + 1]+ [n (n -1) + 3] + [n (n -1) + 5] + ...+n terms
83 = [8(8 -1) + 1][8(8 -1) + 3] + [8(8 -1) + 5]++ [8(8 -1) + 7] + [8(8 -1) + 9] + [8(8 -1) + 11]+[8(8 -1) + 13] + [8(8 - 1)+ 15]
= [8 x 7 + 1]+ [8 x 7 + 3] + [8 x 7 + 5]+(8 x 7 + 7] + (8 x 7 + 9]+ [8 x 7 + 11]+[8 x 7 + 13]+[8 x 7 + 15]
= (56 + 1)+ (56 + 3) + (56 + 5) + (56 + 7) + (56 + 9) + (56 + 11)+ (56 + 13) + (56 + 15)
= 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512
3.
We have, 1005
One's digit of 1005 = 5
Now, cube of one's digit of 1005 = (5)3 = 5 x 5 x 5 = 125
Hence,one's digit in the cube of 1005 is 5.
4.
We have, 3331 = 1
One's digit of 3331 = 1
Now, cube of the one's digit of 3331 = (1)3= 1 x 1 x 1=1
Hence, one's digit in the cube of 3331 is 1.
5.
(a)
6.
(b)
7.
(b)
8.
(b)
9.
(a)
10.
(b)
11.
(b)
12.
(b)
13.
(a)
14.
(b)
15.
The prime factorisation of 5832
5832 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Therefore, \(\sqrt [ 3 ]{ 5832 } \)
= \(\sqrt [ 3 ]{ \underline { 2\times 2\times 2 } \underline { 3\times 3\times 3 } \underline { 3\times 3\times 3 } } \)
= 2 x 3 x 3
=18
16.
Given, number is 614125.
So, groups of 614125 are
In the first group, the number 125 ends with 5. We know that, 5 comes at the unit's place of a number only when its cube ends in 5.
So, 5 will come at unit's place.
Now, in the second group, the number is 614
∵ (8)3 = 512 and (9)3=729
and 512 < 614 < 729
So, ten's place of required cube root is 8.
Hence, \(\sqrt [ 3 ]{ 614125 } \) = 85
17.
Given, number is 68600.
Now, prime factorisation of 68600
68600 = 2 x 2 x 2 x 5 x 5 x 7 x 7 x 7
The prime factor 5 does not have triples.
∴ 68600 is not a perfect cube
So, if we multiply the number 68600 by 5, then we get a perfect cube.
i.e., 68600 x 5 = 343000
= 2 x 2 x 2 x 5 x 5 x 5 x 7 x 7 x 7
= 2 x 5 x 7 =70
18.
Given, number is 21600.
Now, prime factorisation of 21600
21600 = 2 x 2 x 2 x 3 x 3 x 3 x 2 x 2 x 5 x 5
Prime factors 2 and 5 do not have triples.
So, 21600 is not a perfect cube.
19.
(b)
-24
20.
(d)
1728
21.
(c)
\(\frac { -4 }{ 5 } \)
22.
8640
23.
9
24.
∵ 7 x 7 x 7 = 343
So, cube of a number ending in 7 will end in the digit 3.
25.
∵ 72 = 2 x 2 x 2 x 3 x 3
So, dividing by 3 x 3 = 9, we get a perfect cube.
26.
72 = 2 x 2 x 2 x 3 x 3
So, multiplying by 3, we get a perfect cube.
27.
∵ 250 = 5 x 5 x 5 x 2 ∴ Least number = 2 x 2 =4
28.
(27)3 = 27 x 27 x 27 = 19683
29.
14, 21, 28
30.
49
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