Question:

In a bag there are some gold coins. In another bag there are 1/3rd extra gold coins as compared to the first bag. If the difference in the number of gold coins in the first and second bag is 5, then how many coins are there in the first bag?

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When “fraction extra” is mentioned, add it to the original quantity, then subtract to find the difference and solve for the original.
Updated On: Jul 15, 2026
  • 7
  • 9
  • 13
  • 15
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The Correct Option is D

Approach Solution - 1

Step 1: Let the number of coins in the first bag be $x$
The second bag has $\frac{1}{3}$ extra coins than the first bag. So coins in second bag = $x + \frac{1}{3}x = \frac{4x}{3}$.
Step 2: Difference between the coins in second and first bag
Difference = $\frac{4x}{3} - x = \frac{4x - 3x}{3} = \frac{x}{3}$.
Step 3: Use given difference = 5
$\frac{x}{3} = 5$
$x = 15$
Step 4: Conclusion
The first bag contains 15 coins.
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Approach Solution -2

Since the second bag has one-third extra coins compared to the first, the ratio of coins in the first bag to the second bag is 3 : 4 (three parts becomes four parts once a third more is added). The gap between the two bags is therefore 4 - 3 = 1 part, and that 1 part is given to equal 5 coins. So each part equals 5 coins, meaning the first bag, at 3 parts, holds \(3 \times 5 = 15\) coins.

  1. 7: If the first bag held 7 coins, one-third extra would be \(7 + \frac{7}{3} \approx 9.33\) coins in the second bag, which is not even a whole number of coins, so this cannot be right.
  2. 9: If the first bag held 9 coins, the second bag would hold \(9 + \frac{9}{3} = 12\) coins, a difference of 3 coins, not the 5 coins the question states.
  3. 13: If the first bag held 13 coins, the second bag would hold \(13 + \frac{13}{3} \approx 17.33\) coins, again not a whole number, so this does not fit either.
  4. 15: If the first bag holds 15 coins, the second bag holds \(15 + \frac{15}{3} = 15 + 5 = 20\) coins, and the difference is \(20 - 15 = 5\) coins, exactly matching the given difference.

Only 15 coins in the first bag produces a whole-number second bag and the exact difference of 5 coins stated in the question.

Therefore, the correct answer is 15.

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Approach Solution -3

Since the second bag has one-third extra coins compared to the first, the first bag's coin count must be divisible by 3 for that extra third to come out to a whole number of coins, since you cannot have a fractional coin. This divisibility requirement is a quick way to narrow down the options before even computing the difference.

  1. 7: 7 is not divisible by 3, so one-third of 7 is not a whole number of coins, meaning this option cannot represent a real number of coins in a bag that gains a whole-number "extra third." This fails the divisibility check immediately.
  2. 9: 9 is divisible by 3, giving an extra third of 3 coins, so the second bag would hold 12 coins. But the difference between 12 and 9 is 3 coins, not the 5 coins stated in the question, so this fails once the actual difference is checked.
  3. 13: 13 is not divisible by 3, so it fails the same divisibility requirement as 7, since one-third of 13 is not a whole number of coins.
  4. 15: 15 is divisible by 3, giving an extra third of 5 coins, so the second bag would hold 20 coins. The difference between 20 and 15 is exactly 5 coins, matching the question precisely.

Filtering out the options that are not divisible by 3 leaves only 9 and 15, and checking the actual difference for each confirms that only 15 coins in the first bag produces the stated difference of 5.

Therefore, the correct answer is 15.

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