Cubes and Cube Roots — Statement I-II Type Questions

Statement I: The cube of 231 is odd.

Statement II: If a number has 1 in its unit's place, then its cube ends in 1.

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

231 is an odd number and cubes of odd numbers are always odd. So, the cube of 231 is odd.

∴ Statement I is true.

If a number has 1 in its unit's place, then its cube also ends in 1, as 1 × 1 × 1 = 1.

∴ Statement II is true.

Hence, Option 3 is the correct option.

Statement I: If $a$ is a positive integer, then $(-a)^3$ > $a^2$

Statement II: If $a$ is a negative integer, then $(-a)^3$ > $a^2$

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

If $a$ is a positive integer, then $(-a)^3 = -a^3$, which is negative, while $a^2$ is positive.

So, $(-a)^3 \lt a^2$.

∴ Statement I is false.

If $a$ is a negative integer, then $(-a)$ is positive, so $(-a)^3$ is positive and $a^2$ is also positive.

Taking $a$ = -1, we get $(-a)^3$ = 1³ = 1 and $a^2$ = (-1)² = 1.

So, $(-a)^3 = a^2$, which means $(-a)^3 \gt a^2$ is not true for every negative integer.

∴ Statement II is false.

Hence, Option 4 is the correct option.

Statement I: $(ab)^3 = a^3b^3$

Statement II: $\sqrt[3]{ab} = \sqrt[3]{a}\ \sqrt[3]{b}$

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

$(ab)^3 = (ab) × (ab) × (ab) = (a × a × a) × (b × b × b) = a^3b^3$

∴ Statement I is true.

We know that the cube root of the product of two numbers is equal to the product of their cube roots.

$\Rightarrow \sqrt[3]{ab}\\[1em] \Rightarrow (ab)^{\dfrac{1}{3}}\\[1em] \Rightarrow a^{\dfrac{1}{3}} \times b^{\dfrac{1}{3}}\\[1em] \Rightarrow \sqrt[3]{a}\ \sqrt[3]{b}$

∴ Statement II is true.

Hence, Option 3 is the correct option.