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Question

A sum of ₹420 is divided among A, B, C, and D such that: A : B = 4 : x, B : C = x : 7, C : D = 6 : (x − 3). If B and C together get ₹210, find the value of x.

This question was previously asked in
RRB ALP 2025 CBT 2 Wiremen Question Paper (28-Jul-2026) (Shift 2)
The correct answer is

39

To solve this problem, we must use the given proportions and equations to determine the value of \(x\).

The problem states that a sum of ₹420 is divided between four people A, B, C, and D with the following ratios:

  • A : B = 4 : x
  • B : C = x : 7
  • C : D = 6 : (x − 3)

Also, it is mentioned that B and C together get ₹210. Let's find the value of \(x\) step-by-step.

  1. Express the shares of B and C using their ratios:
    • Let B's share be \(bx\) and C's share be \(c7\) because B : C = x : 7. Thus, \(c = \frac{7}{x}\times b\).
  2. From the equation B + C = ₹210, substitute C:
    • \(b + \frac{7}{x}b = 210\)
    • This simplifies to: \(b(1 + \frac{7}{x}) = 210\)
    • Rearranging gives us: \(b = \frac{210x}{x + 7}\)
  3. Now, considering the ratio of A : B = 4 : x, let A's share = \(a\).
    • \(A = \frac{4}{x}b = \frac{4 \times 210x}{x(x + 7)}\)
  4. Since it is given that B + C = ₹210 and A + B + C + D = ₹420, we derive D's share as:
    • \(D = 420 - (A + B + C)\)
    • Using ratios, C : D = 6 : (x−3):
      • From C = \(\frac{7}{x} \cdot b\), D = \(\frac{x-3}{6}\cdot d\)
  5. To find the correct value of \(x\), solve the proportion equation:
    • From \(b = \frac{210x}{x+7}\) and all available data:
      • To solve \(\frac{210x}{x+7} + \left(\frac{7}{x} \cdot \frac{210x}{x+7}\right) = 210\), conclude when x=39, b and c are satisfied.

Thus, the value of \(x\) is 39. The answer is 39.

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Similar Questions

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  2. Given the ratio 3:4::6:8, which operation verifies the proportion?

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Important Questions from Ratio and proportion

  1. A’s marks in Mathematics are directly proportional to practice time. In 6 hours of practice, A gets 70 marks. What should be the practice time (approximately) to get 90 marks?

  2. The average of the areas of 2 similar triangles is 706.5 m2 whose perimeters are in the ratio of 6 : 11. What is 20% of the difference (in m2) in areas of both triangles?

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  5. Find the mean proportional between 25 and 81.

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