When x is added to each of 26, 40, 22 and 34, then the numbers so obtained, in this order, are in proportion. Then, if 4x : y :: y : (7x-6), and y > 0, what is the value of y?

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RRB NTPC Graduate Level CBT-I Official Paper (Held On: 05 Jun, 2025 Shift 2)
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  1. 9
  2. 2
  3. 3
  4. 8

Answer (Detailed Solution Below)

Option 4 : 8
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Detailed Solution

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Given:

When x is added to each of 26, 40, 22, and 34, the numbers obtained are in proportion.

If 4x : y :: y : (7x - 6), and y > 0, find the value of y.

Formula used:

For numbers in proportion: \(\frac{a}{b} = \frac{c}{d}\)

For ratio: \(\frac{4x}{y} = \frac{y}{7x-6}\)

Calculation:

For the first condition:

The numbers after adding x: (26 + x), (40 + x), (22 + x), (34 + x).

Given these are in proportion:

\(\frac{26+x}{40+x} = \frac{22+x}{34+x}\)

⇒ (26 + x)(34 + x) = (40 + x)(22 + x)

⇒ 884 + 26x + 34x + x2 = 880 + 40x + 22x + x2

⇒ 884 + 60x = 880 + 62x

⇒ 4 = 2x

⇒ x = 2

For the second condition:

Given: \(\frac{4x}{y} = \frac{y}{7x-6}\)

Substitute x = 2:

\(\frac{4(2)}{y} = \frac{y}{7(2)-6}\)

\(\frac{8}{y} = \frac{y}{14-6}\)

\(\frac{8}{y} = \frac{y}{8}\)

⇒ 8 × 8 = y × y

⇒ y2 = 64

⇒ y = √64 = 8 (since y > 0)

∴ The correct answer is option (4).

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