A varies directly as the positive square root of B, and inversely as the cube of C. If A = 15, when B = 27 and C = 2, then find B when A = 9 and C = 2.

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 05 Dec 2022 Shift 2)

Answer 1

Given:

A varies directly as the positive square root of B.

A = 15, when B = 27 and C = 2.

Concept used:

If P ∝ Q and P ∝ R, then P ∝ QR and P = H × QR. (Here, H is a constant)

Calculation:

According to the question,

A ∝ \(\sqrt {B}\) and A ∝ 1/C³

So, A ∝ \(\frac {\sqrt B}{C^3}\)

⇒ A = k × \(\frac {\sqrt B}{C^3}\) ....(1) (Here, k is a constant)

Now putting A = 15, B = 27 and C = 2 in (1),

A = k × \(\frac {\sqrt B}{C^3}\)

⇒ 15 = k × \(\frac {\sqrt {27}}{2^3}\)

⇒ k = \(\frac {120}{3\sqrt {3}}\)

⇒ k = \(\frac {40}{\sqrt {3}}\)

Now when A = 9 and C = 2,

A = k × \(\frac {\sqrt B}{C^3}\)

⇒ 9 = \(\frac {40}{\sqrt {3}}\) × \(\frac {\sqrt B}{2^3}\)

⇒ \(\sqrt {\frac {B}{3}}\) = 9/5

⇒ B/3 = 81/25

⇒ B = 243/25

∴ The value of B is \(\frac{243}{25}\).