Question:

In a bag there are some gold coins.In another bag there are \(\frac 13^{\text {rd}}\) extra gold coins as compared to first bag.If the difference in the number of gold coins in first and second bag is 5, then how many coins are there in the first bag?

Updated On: Jul 15, 2026
  • 7
  • 9
  • 13
  • 15
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The Correct Option is D

Approach Solution - 1

The correct option is (D): 15.
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Approach Solution -2

Let the first bag have x coins. The second bag has 1/3rd extra coins compared to the first, that is \( x + \frac{x}{3} = \frac{4x}{3} \) coins, and the difference between the two bags is given as 5, that is \( \frac{4x}{3} - x = \frac{x}{3} = 5 \). Let's test each option directly against this condition:

  1. Option A (7): one-third of 7 is \( \frac{7}{3} \approx 2.33 \), not 5, so this does not satisfy the difference condition.
  2. Option B (9): one-third of 9 is 3, not 5, so this does not satisfy the condition either.
  3. Option C (13): one-third of 13 is \( \frac{13}{3} \approx 4.33 \), close to but not exactly 5, so it fails the exact condition.
  4. Option D (15): one-third of 15 is exactly 5, which matches the given difference of 5 coins perfectly.

Therefore, the correct answer is 15.

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Approach Solution -3

Let the first bag have \( x \) coins. The second bag has one-third extra, so it holds \( x + \frac{x}{3} = \frac{4x}{3} \) coins. Comparing the two bags as a ratio, first bag to second bag is \( 3 : 4 \). So the first bag can be written as \( 3k \) and the second bag as \( 4k \) for some whole number \( k \), and the difference between them is \( 4k - 3k = k \), which is given as 5. That fixes \( k = 5 \), so the first bag, being \( 3k \), holds \( 3 \times 5 = 15 \) coins. Checking each option against this ratio confirms it:

  1. Option A (7): 7 is not a multiple of 3, so it cannot represent \( 3k \) for a whole-number \( k \), it does not fit the 3:4 ratio structure at all.
  2. Option B (9): 9 is \( 3 \times 3 \), so \( k = 3 \), giving a difference of \( k = 3 \), not the required 5.
  3. Option C (13): 13 is not a multiple of 3 either, so it cannot fit the 3:4 ratio in whole coins.
  4. Option D (15): 15 is \( 3 \times 5 \), so \( k = 5 \), giving a difference of exactly \( k = 5 \), matching the given condition precisely.

Only 15 fits the 3:4 ratio with a difference of exactly 5 coins.

Therefore, the correct answer is 15.

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