Cubes and Cube Roots Test

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Cubes and Cube Roots Test - 9

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Weekly Quiz Competition

Question 1

1 / -0

Which of the following numbers is a perfect cube?

20250

21296

21952

36000

Solution

We start by factorising 21952.

2	21952
2	10976
2	5488
2	2744
2	1372
2	686
7	343
7	49
	7

∴ 21952 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7

It can be seen that all prime factors occur in groups of three.

Thus, 21952 is a perfect cube.

The correct answer is C.

Question 2

1 / -0

What is the least number by which 2592 should be multiplied so that the product is a perfect cube as well as a perfect square?

2

6

18

36 Solution

The number 2592 can be factorized as:

2	2592
2	1296
2	648
2	324
2	162
3	81
3	27
3	9
	3

∴ 2592 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

It can be seen that if we multiply 2592 with 2 × 3 × 3 = 18, then the number can be grouped into two’s as well as three’s.

∴2592 × 18 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

= 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Thus, the required number is 18.

Hence, the correct answer is option C.

Question 3
1 / -0

Use the following information to answer the next question.

A pattern is given as:

1 = 1 = 1³

3 + 5 = 8 = 2³

7 + 9 + 11 = 27 = 3³

13 + 15 + 17 + 19 = 64 = 4³

According to the given pattern, 5³ can be expressed as

A
21 + 23 + 25 + 27 + 29 = 5³

B
20 + 22 + 24 + 28 + 31 = 5³

C
28 + 30 + 32 + 35 = 5³

D
15 + 19 + 23 + 31 + 37 = 5³

Solution

The given pattern can be extended as:

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

Thus, 5³ can be expressed as 21 + 23 + 25 + 27 + 29.

The correct answer is A.
Question 4
1 / -0

A pattern is given as:

2³ − 1³ = 1 + 2 × 1 × 3

3³ − 2³ = 1 + 3 × 2 × 3

4³ − 3³ = 1 + 4 × 3 × 3

According to the given pattern, 41³ − 40³ can be expressed as

A
41³ − 40³ = 1 + 41 × 3 × 3

B
41³ − 40³ = 1 + 41 × 40 × 3

C
41³ − 40³ = 1 + 41 × 40 × 39

D
41³ − 40³ = 1 + 41 × 40 × 40

Solution

The given pattern of difference of cubes of consecutive numbers can be generalised as:

(a + 1)³ − a³ = 1 + (a + 1) × a × 3, where a is a natural number.

In particular for a = 40, we get:

(40 + 1)³ − 40³ = 1 + (40 + 1) × 40 × 3

⇒ 41³ − 40³ = 1 + 41 × 40 × 3

Hence, the correct answer is option B.
Question 5
1 / -0

If a natural number N can be expressed as N = 4x²y × 54xy², where x and y are natural numbers, then what is the cube root of N?

A
6xy

B
12xy

C
9xy²

D
27x²y

Solution

N = 4x²y × 54xy²

= 2 × 2 × x × x × y × 2 × 3 × 3 × 3 × x × y × y

= 2 × 2 × 2 × 3 × 3 × 3 × x × x × x × y × y × y

∴ = 2 × 3 × x × y = 6xy

Thus, the cube root of N is 6xy.

The correct answer is A.

Question 6

1 / -0

What is the least number by which 27440 can be divided to make it a perfect cube?

5

10

15

20 Solution

The number 27440 can be factorized as:

2	27440
2	13720
2	6860
7	3430
7	490
7	70
2	10
	5

∴ 27440 = 2 × 2 × 2 × 2 × 5 × 7 × 7 × 7

It can be seen that the prime factors 2 and 5 do not occur in a group of three. Therefore, when we divide 27440 by (2 × 5) = 10, we get 2 × 2 × 2 × 7 × 7 × 7 = 2744, which is a perfect cube.

Thus, the required number is 10.

The correct answer is B.

Question 7

1 / -0

What is the smallest number by which 24696 is to be multiplied with so that the product becomes a cube number?

2

3

6

7 Solution

The number 24696 can be factorized as:

2	24696
2	12348
2	6174
3	3087
3	1029
7	343
7	49
7	7
	1

∴ 24696 = 2 × 2 × 2 × 3 × 3 × 7 × 7 × 7

It can be seen that 3 does not occur in a group of three. Therefore, if we multiply 24696 with 3, we get a perfect cube.

∴24696 × 3 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 7

Thus, the required number is 3.

The correct answer is B.

Question 8
1 / -0

What is the smallest perfect cube divisible by 54, 132 and 242?

A
245684

B
256874

C
268716

D
287496

Solution

We start by finding the L.C.M of 54, 132 and 242.

2

54, 132, 242

3

27, 66, 121

11

9, 22, 121

9, 2, 11

∴ L.C.M (54, 132, 242) = 2 × 3 × 11 × 9 × 2 × 11

= 2 × 2 × 3 × 3 × 3 × 11 × 11

= 13068

It can be seen that in the prime factorisation of 13068, 2 and 11 do not occur in groups of three.

Therefore, if we multiply 13068 with 2 × 11 = 22, then the product will be a perfect cube.

∴ 13068 × 22 = 287496 = 2 × 2 × 2 × 3 × 3 × 3 × 11 × 11 × 11

Thus, the smallest perfect cube which is divisible by 54, 132 and 242 is 287496.

Hence, the correct answer is option D.
Question 9
1 / -0

What is the value of the expression?

A
22

B
25

C
35

D
44

Solution

The number 15625 can be factorized as:

5

15625

5

3125

5

625

5

125

5

25

5

∴15625 = 5 × 5 × 5 × 5 × 5 × 5

∴

Thus, the value of the expressionis 25.

The correct answer is B.

Question 10

1 / -0

Use the following information to answer the next question.

Some statements are given as:

I. If a perfect cube ends with 7, then its cube root ends with 3.

II. If a perfect cube ends with 1, then its cube root ends with 1 or 9.

III. If a number ends with 6, then its cube ends with 6.

IV. If a number ends with 5, then its cube ends with 0.

Which of the given statements are correct?

I and III

II and III

II and IV

I and IV

Solution

Statement I

If a perfect cube ends with 7, then its cube root ends with 3. For example, 3 × 3 × 3 = 27

Statement III

If a number ends with 6, then its cube ends with 6. For example, 6 × 6 × 6 = 216

Thus, statements I and III are correct.

The correct answer is A.

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2	54, 132, 242
3	27, 66, 121
11	9, 22, 121
	9, 2, 11

5	15625
5	3125
5	625
5	125
5	25
	5

Cubes and Cube Roots Test - 9

Result

Cubes and Cube Roots Test - 9

Score

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Question 1

20250

21296

21952

36000

Solution

Question 2

2

6

18

36

Solution

Question 3

21 + 23 + 25 + 27 + 29 = 53

20 + 22 + 24 + 28 + 31 = 53

28 + 30 + 32 + 35 = 53

15 + 19 + 23 + 31 + 37 = 53

Solution

Question 4

413 − 403 = 1 + 41 × 3 × 3

413 − 403 = 1 + 41 × 40 × 3

413 − 403 = 1 + 41 × 40 × 39

413 − 403 = 1 + 41 × 40 × 40

Solution

Question 5

6xy

12xy

9xy2

27x2y

Solution

Question 6

5

10

15

20

Solution

Question 7

2

3

6

7

Solution

Question 8

245684

256874

268716

287496

Solution

Question 9

22

25

35

44

Solution

Question 10

I and III

II and III

II and IV

I and IV

Solution

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21 + 23 + 25 + 27 + 29 = 5³

20 + 22 + 24 + 28 + 31 = 5³

28 + 30 + 32 + 35 = 5³

15 + 19 + 23 + 31 + 37 = 5³

41³ − 40³ = 1 + 41 × 3 × 3

41³ − 40³ = 1 + 41 × 40 × 3

41³ − 40³ = 1 + 41 × 40 × 39

41³ − 40³ = 1 + 41 × 40 × 40

9xy²

27x²y