CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 4 - MCQExams.com

Prime factorising $$25$$ , we get,

$$25=5\times 5=5^2$$ .

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$5$$ 's is $$2$$ .

So we need to multiply another $$5$$ in the factorization to make $$25$$ a perfect cube.

Hence, the smallest number by which $$25$$ must be multiplied to obtain a perfect cube is $$5$$ .

Prime factorising $$1296$$ , we get,

$$1296= 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$$

           $$= 2^4 \times 3 ^4$$ .

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$2$$ 's is $$4$$ and number of $$3$$ 's is $$4$$ .

So we need to divide $$2$$ and $$3$$ from the factorization to make $$1296$$ a perfect cube.

Hence, the smallest number by which $$1296$$ must be divided to obtain a perfect cube is $$2 \times 3=6$$ .

Prime factorising $$4096$$ , we get,

$$4096 = 2\times2\times2\times2\times2\times2\times$$ $$2\times2\times2\times2\times2\times2$$   $$= 2^{12}$$ .

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$2$$ 's is $$12$$ , which is a multiple of $$3$$ .

Therefore, $$4096$$ is a perfect cube.

On prime factorisation of $$175616$$ , we get,

$$175616= 2\times2\times2\times2\times2\times2\times2\times2\times2\times7\times7\times7$$

              $$= 8\times8\times8\times7\times7\times7$$

              $$= 8^3 \times 7^3$$ .

Then, cube root of $$175616$$ is:

$$\sqrt[3]{175616}=\sqrt[3]{8^3 \times 7^3}$$

                  $$= 8\times 7\times$$

                  $$=56$$

Therefore, option $$A$$ is correct.

Prime factorising $$1188$$ , we get,

$$1188 = 2 \times 2 \times 3 \times 3 \times 3 \times 11$$

           $$= 2^2 \times 3^3 \times 11$$ .

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$2$$ 's is $$2$$ , number of $$3$$ 's is $$3$$ and number of $$11$$ 's is $$1$$ .

So we need to divide $$2^2$$ and $$11$$ from the factorization to make $$1188$$ a perfect cube.

Hence, the smallest number by which $$1188$$ must be divided to obtain a perfect cube is $$2^2 \times 11 = 44$$ .

Prime factorising $$4320$$ , we get,

$$4320 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$$

           $$= 2 ^5 \times 3 ^3 \times 5^1$$ .

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$2$$ 's is $$5$$ , number of $$3$$ 's is $$3$$ and number of $$5$$ 's is $$1$$ .

So we need to multiply another $$2$$ , and $$5^2$$ in the factorization to make $$4320$$ a perfect cube.

Hence, the smallest number by which $$4320$$ must be multiplied to obtain a perfect cube is $$2\times 5^2=50$$ .

Hence, option $$C$$ is correct.

On prime factorising, we get,

$$27000 = 2\times2\times2\times5\times5\times5\times3\times3\times3$$

            $$= 2^3\times5^3\times3^3$$ .

Then, cube root of $$27000$$ is:

$$\sqrt[3]{27000}=\sqrt[3]{3^3 \times 5^3 \times 2^3}= 3\times5\times2=30$$ .

Therefore, $$30$$ is the required solution.

Hence, option $$A$$ is correct.

On prime factorising, we get,

$$91125=(5 \times 5\times 5)\times (3\times3\times3)\times(3\times3\times3)$$

            $$= 5^3 \times 3^3 \times 3^3$$ .

Then, cube root of $$91125$$ is:

$$\sqrt[3]{91125}=\sqrt[3]{5^3 \times 3^3 \times 3^3 }=5\times 3\times 3=45$$ .

Therefore, option $$A$$ is correct.

On prime factorisation of the numbers individually, we get,

$$125=\underline { 5 \times 5 \times 5 }=5^3$$ .

$$1331=\underline { 11 \times 11 \times 11 }=11^3$$ .

Then, cube root of $$-\dfrac{125}{1331}$$ is:

$$\displaystyle \sqrt [ 3 ]{-\dfrac{125}{1331}}$$ $$=\displaystyle \sqrt [ 3 ]{-\dfrac{5^3}{11^3}}$$ $$=\displaystyle \sqrt [ 3 ]{(-\dfrac{5}{11})^3}$$ $$= \displaystyle {-\dfrac{5}{11}}$$ .

Thus, option $$B$$ is correct.

On prime factorisation of the numbers individually, we get,

$$27=\underline { 3 \times 3 \times 3 }=3^3$$ .

$$125=\underline { 5 \times 5 \times 5 }=5^3$$ .

Therefore, cube root of $$\dfrac{27}{125}$$ is:

$$\displaystyle {\sqrt [ 3 ]{ \dfrac{27}{125}}=\sqrt [ 3 ]{ \dfrac{3^3}{5^3}}=\dfrac{3}{5}}$$ .

Therefore, option $$A$$ is correct.

On prime factorising, we get,

$$216=\underline{6\times 6\times 6}$$ $$=6^3$$ .

$$343=\underline{7\times 7\times 7}$$ $$=7^3$$ .

Then, $$-343$$ $$=(-7)^3$$ .

Therefore, value of $$\sqrt[3]{216\times(-343)}$$ is:

$$\sqrt[3]{6^3\times(-7)^3}=6\times(-7)=-42$$ .

Therefore, option $$B$$ is correct.

On prime factorising, we get,

$$13824 = 3\times3\times3\times2\times2\times2\times2\times2\times2\times2\times2\times2$$

            $$= 2^3\times2^3\times2^3\times3^3=24^3$$ .

Then, cube root of $$13824$$ is:

$$\sqrt[3]{13824}=\sqrt[3]{24^3}= 24$$ .

Therefore, option $$A$$ is correct.

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.

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.

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.

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.

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.

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.