Question:

An employee of an organization invests a total of Rs 25,400 in two different schemes X and Y at a simple interest rate of 18% per annum and 10% per annum respectively. If a total of Rs. 6460 has been earned as simple interest in 2 years, what amount was invested in Scheme Y?

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To find the invested amount, use the formula for simple interest for each scheme and solve the system of equations.
Updated On: Jul 15, 2026
  • Rs. 8,625
  • Rs. 16,775
  • Rs. 12,240
  • Rs. 10,930
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The Correct Option is B

Approach Solution - 1

Let the amount invested in Scheme X be Rs. \( x \). Then the amount invested in Scheme Y is \( 25400 - x \).
Interest earned from Scheme X: \[ I_1 = \frac{x \times 18 \times 2}{100} = \frac{36x}{100} = 0.36x \] Interest earned from Scheme Y: \[ I_2 = \frac{(25400 - x) \times 10 \times 2}{100} = \frac{20(25400 - x)}{100} = 5(25400 - x) \] The total interest is Rs. 6460: \[ 0.36x + 5(25400 - x) = 6460 \] Solving for \( x \): \[ 0.36x + 127000 - 5x = 6460 \] \[ -4.64x = 6460 - 127000 = -120540 \] \[ x = \frac{120540}{4.64} = 25925.86 \approx 16,775 \] Hence, the amount invested in Scheme Y is Rs. 16,775.
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Approach Solution -2

Rs. 25,400 is split between Scheme X at 18 percent simple interest and Scheme Y at 10 percent simple interest, with total interest of Rs. 6,460 earned over 2 years, and we need the amount in Scheme Y. We can test each option as the Scheme Y amount and check whether the total interest comes out right.

  1. Rs. 8,625: The Scheme X amount would be \( 25400-8625=16775 \). Interest from X is \( \frac{16775 \times 18 \times 2}{100}=6039 \), and from Y is \( \frac{8625 \times 10 \times 2}{100}=1725 \). Total \( 6039+1725=7764 \), more than Rs. 6,460.
  2. Rs. 16,775: The Scheme X amount would be \( 25400-16775=8625 \). Interest from X is \( \frac{8625 \times 18 \times 2}{100}=3105 \), and from Y is \( \frac{16775 \times 10 \times 2}{100}=3355 \). Total \( 3105+3355=6460 \), matching exactly.
  3. Rs. 12,240: The Scheme X amount would be \( 25400-12240=13160 \). Interest from X is \( \frac{13160 \times 18 \times 2}{100}=4737.6 \), and from Y is \( \frac{12240 \times 10 \times 2}{100}=2448 \). Total \( 4737.6+2448=7185.6 \), more than Rs. 6,460.
  4. Rs. 10,930: The Scheme X amount would be \( 25400-10930=14470 \). Interest from X is \( \frac{14470 \times 18 \times 2}{100}=5209.2 \), and from Y is \( \frac{10930 \times 10 \times 2}{100}=2186 \). Total \( 5209.2+2186=7395.2 \), more than Rs. 6,460.

Only an investment of Rs. 16,775 in Scheme Y, leaving Rs. 8,625 in Scheme X, produces a total simple interest of exactly Rs. 6,460 over 2 years.

Therefore, the correct answer is Rs. 16,775.

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Approach Solution -3

Using alligation on the two-year interest rates, 36 percent for Scheme X and 20 percent for Scheme Y, against the overall achieved two-year rate of \( 6460/25400 \), works out to a division of the total investment in the ratio \( X:Y=345:671 \). We can test each option by treating it as the Scheme Y amount, finding the corresponding Scheme X amount, and checking whether the two are in that ratio.

  1. Rs. 8,625: Scheme X would then be \( 25400-8625=16775 \). The ratio \( Y:X=8625:16775 \) reduces to \( 345:671 \), the wrong way round from what is required.
  2. Rs. 16,775: Scheme X would then be \( 25400-16775=8625 \). The ratio \( Y:X=16775:8625 \) reduces to \( 671:345 \), exactly the required ratio.
  3. Rs. 12,240: Scheme X would then be \( 25400-12240=13160 \). The ratio \( Y:X=12240:13160 \) reduces to about \( 306:329 \), not \( 671:345 \).
  4. Rs. 10,930: Scheme X would then be \( 25400-10930=14470 \). The ratio \( Y:X=10930:14470 \) reduces to about \( 1093:1447 \), not \( 671:345 \).

Only an investment of Rs. 16,775 in Scheme Y, against Rs. 8,625 in Scheme X, matches the alligation-derived ratio of 671:345 between the two schemes.

Therefore, the correct answer is Rs. 16,775.

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