Cubes and Cube Roots Test

Result

Cubes and Cube Roots Test - 21

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Find the cube root of each of the following numbers by prime factorisation method: $110592$

A
$38$

B
$48$

C
$58$

D
none of these

Solution

On prime factorising, we get,
$110592 = \underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{2\times3}\times\underline{2\times3}\times\underline{2\times3}$
$= 8\times8\times8\times6\times6\times6$
$= 8^3\times6^3$ .
Then, cube root of $110592$ is:
$\sqrt[3]{110592}=\sqrt[3]{8^3 \times 6^3}= 8\times6=48$ .
Therefore, $48$ is the required solution.
Hence, option $B$ is correct.
Question 2
1 / -0

The cube root of $\displaystyle -729$ is:

A
$-19$

B
$-9$

C
$-39$

D
$-3$

Solution

Prime factorising, we get,

$729=(9)\times(9)\times (9)$

But $-729=(-9)\times(-9)\times(-9)$

Then,

$\sqrt[3]{-729}$ $=\sqrt[3]{-9\times -9\times -9}=-9$

Therefore, cube root of $-729$ is $-9$

Hence, option $B$ is correct.
Question 3
1 / -0

The cube root of $35937$ is:

A
$33$

B
$45$

C
$27$

D
$31$

Solution

On prime factorising, we get,
$35937=(3\times3\times3)\times(11\times11\times11)$
$= 3^3 \times 11^3$ .
Then, cube root of $35937$ is:
$\sqrt[3]{35937}=\sqrt[3]{3^3 \times 11^3}$
$= 3\times 11$
$=33$ .
Therefore, option $A$ is correct.
Question 4
1 / -0

Find the value of: $\displaystyle \sqrt[3]{4096}$ .

A
$46$

B
$36$

C
$26$

D
$16$

Solution

On prime factorising, we get,
$4096=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}$
$=2^3\times 2^3\times 2^3\times 2^3$ .
Then, value of $\sqrt[3]{4096}$ is:
$\sqrt[3]{4096}=\sqrt[3]{2^3 \times 2^3 \times 2^3 \times 2^3}= 2\times2\times2\times2=16$ .

Therefore, option $D$ is correct.
Question 5
1 / -0

Evaluate: $\displaystyle \sqrt[3]{1331}$ .

A
$11$

B
$121$

C
$51$

D
$21$

Solution

On prime factorising, we get,
$1331=\underline{11\times 11\times 11}$ $=11^3$ .
Therefore, value of $\sqrt[3]{1331}$ is:
$\sqrt[3]{1331}=\sqrt[3]{(11)^3}= 11$ .
Therefore, option $A$ is correct.
Question 6
1 / -0

$\displaystyle \sqrt[3]{-1331}=$ ?

A
$36$

B
$-11$

C
$-21$

D
$16$

Solution

On prime factorising, we get,
$1331=\underline{11\times 11\times 11}$ $=11^3$ .
Then, $-1331$ $=(-11)^3$ .
Therefore, value of $\sqrt[3]{-1331}$ is:
$\sqrt[3]{-1331}=\sqrt[3]{(-11)^3}= -11$ .
Therefore, option $B$ is correct.
Question 7
1 / -0

Find the cube root of $-8000$ .

A
$80$

B
$-20$

C
$40$

D
Does not exist

Solution

On prime factorising, we get,
$8000=\underline{2\times 2\times 2}\times \underline{2\times2\times2} \times \underline{5\times5\times5}$ $=2^3\times 2^3 \times 5^3=20^3$ .
Then, $-8000$ $=(-20)^3$ .
Therefore, value of $\sqrt[3]{-8000}$ is:
$\sqrt[3]{-8000}=\sqrt[3]{(-20)^3}= -20$ .
Therefore, option $B$ is correct.
Question 8
1 / -0

$\sqrt[3]{-27000}=$ ?

A
$-90$

B
$-30$

C
$-60$

D
None of these

Solution

On prime factorising, we get,
$27000=\underline{3\times 3\times 3}\times \underline{2\times2\times2}\times \underline{5\times5\times5}$ $=2^3\times3^3\times5^3=30^3$ .
Then, $-27000$ $=(-30)^3$ .
Therefore, value of $\sqrt[3]{-27000}$ is:
$\sqrt[3]{-27000}=\sqrt[3]{(-30)^3}= -30$ .
Therefore, option $B$ is correct.
Question 9
1 / -0

What number must be multiplied to $6912$ , so that the product becomes a perfect cube?

A
$2$

B
$3$

C
$4$

D
$6$

E
$10$

Solution

Prime factorising $6912$ , we get,
$6912 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
$= 2 ^8 \times 3 ^3$ .
We know, a perfect cube has multiples of $3$ as powers of prime factors.
Here, number of $2$ 's is $8$ and number of $3$ 's is $3$ .
So we need to multiply another $2$ to the factorization to make $6912$ a perfect cube.
Hence, the smallest number by which $6912$ must be multiplied to obtain a perfect cube is $2$ .
Hence, option $A$ is correct.
Question 10
1 / -0

What is the value of $\displaystyle \sqrt[3]{\left (64\times 729 \right )}=$ ?

A
$9$

B
$6$

C
$36$

D
$52$

Solution

On prime factorisation of the numbers individually, we get,
$64=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }$ $=2^3 \times 2^3=4^3$ .

$729=\underline { 3 \times 3 \times 3 } \times \underline { 3 \times 3\times 3 }$ $=3^3 \times 3^3=9^3$ .

Therefore,
$\sqrt [ 3 ]{ 64 \times 729 }$ $= \sqrt [ 3 ]{4^3 \times 9^3 }$ $=4\times9=36$ .

Hence, option $C$ is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Cubes and Cube Roots Test - 21

Result

Cubes and Cube Roots Test - 21

Score

-

Rank

-

SHARING IS CARING

Question 1

383838

484848

585858

none of these

Solution

Question 2

−19-19−19

−9-9−9

−39-39−39

−3-3−3

Solution

Question 3

333333

454545

272727

313131

Solution

Question 4

464646

363636

262626

161616

Solution

Question 5

111111

121121121

515151

212121

Solution

Question 6

363636

−11-11−11

−21-21−21

161616

Solution

Question 7

808080

−20-20−20

404040

Does not exist

Solution

Question 8

−90-90−90

−30-30−30

−60-60−60

None of these

Solution

Question 9

222

333

444

666

101010

Solution

Question 10

999

666

363636

525252

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$38$

$48$

$58$

$-19$

$-9$

$-39$

$-3$

$33$

$45$

$27$

$31$

$46$

$36$

$26$

$16$

$11$

$121$

$51$

$21$

$36$

$-11$

$-21$

$16$

$80$

$-20$

$40$

$-90$

$-30$

$-60$

$2$

$3$

$4$

$6$

$10$

$9$

$6$

$36$

$52$