CAT 2025 Lesson : Surds & Indices - Past Questions

Observation/Strategy
1) The powers are in fractions and the denominators are the same.
2) We can multiply each of the powers by the LCM of the denominators and compare.

LCM$(2, 3, 4, 6, 12) = 12$

$2^{\frac{1}{12} \times 12} = 2^6 = 64$

$3^{\frac{1}{3} \times 12} = 3^4 = 81$

$4^{\frac{1}{4} \times 12} = 4^3 = 64$

$6^{\frac{1}{6} \times 12} = 6^2 = 36$

$12^{\frac{1}{12} \times 12} = 12$

The largest number here is $81$.

∴$3^{\frac{1}{3}}$ is the largest.

Answer: (2) $3^{1/3}$

Observation/Strategy
1) There are 2 equations with 2 variables. So, we need to substitute one in the other.

Substituting $q = 9p ⇒ p^{9p} = (9p)^{p}$

⇒ $p^{8p} \times p^{p} = 9^{p} \times p^{p}$

⇒ $(p^{8})^{p} = 9^{p}$

⇒ $p^{8} = 9$

⇒ $p = 9^{\frac{1}{8}} = \sqrt[8]{9}$

Answer: (4) $\sqrt[8]{9}$

$\left( \dfrac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{-p}}} \right)^{\frac{1}{p}} = \left( \dfrac{2^{2 \times (p + \frac{1}{4})} \times (2^{p + 1})^{\frac{1}{2}}}{2 \times 2^{\frac{-p}{2}}} \right)^{\frac{1}{p}}$

$= \left( 2^{2p + \frac{1}{2}} \times 2^{\frac{p + 1}{2}} \times 2^{\frac{p}{2} - 1} \right)^{\frac{1}{p}}$

$= (2^{3p})^{\frac{1}{p}} = 2^3 = 8$

Answer: (2) $8$