When x is added to each of 30, 15, 21 and 11, then the numbers so obtained, in this order, are in proportion. Then, if 6x : y :: y : (3x-9), and y > 0, what is the value of y?

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This question was previously asked in

RRB NTPC Graduate Level CBT-I Official Paper (Held On: 05 Jun, 2025 Shift 1)

Answer 1

Given:

Numbers 30, 15, 21, and 11 are added to x and are in proportion.

6x : y :: y : (3x - 9), and y > 0.

Formula used:

If numbers are in proportion, then:

(a + x)/(b + x) = (c + x)/(d + x)

For ratio problems: a:b::b:c ⇒ b² = a × c

Calculation:

⇒ (30 + x)/(15 + x) = (21 + x)/(11 + x)

⇒ (30 + x) × (11 + x) = (15 + x) × (21 + x)

⇒ 330 + 41x + x² = 315 + 36x + x²

⇒ 330 - 315 = 36x - 41x

⇒ x = -3

Solve the second ratio condition.

6x : y :: y : (3x - 9)

⇒ y² = 6x × (3x - 9)

Substitute x = -3:

⇒ y² = 6 × (-3) × (3 × (-3) - 9)

⇒ y² = 6 × (-3) × (-9 - 9)

⇒ y² = 324

⇒ y = √324

⇒ y = 18 (since y > 0)

∴ The correct answer is option (3).

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