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For the same principal amount, the compound interest for two years at 5%5\% per annum exceeds the simple interest for three years at 3%3\% per annum by Rs. 11251125. Then the principal amount in rupees is

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Solution

✅ Correct Answer: 90000

We solve this using the relationship between compound interest and simple interest.

We need to find a principal amount where compound interest for 2 years at 5% per annum exceeds simple interest for 3 years at 3% per annum by Rs. 1125.


We let the principal amount = PP

Simple Interest = Principal × Rate × Time ÷ 100

SI for 3 years at 3% = P×3×3÷100=9P100P \times 3 \times 3 \div 100 = \tfrac{9P}{100}


Compound Interest = P[(1+R100)n−1]P[(1 + \tfrac{R}{100})^n - 1]

CI for 2 years at 5% = P[(1+5100)2−1]P[(1 + \tfrac{5}{100})^2 - 1]

=P[(1.05)2−1]=P[1.1025−1]=10.25P100= P[(1.05)^2 - 1] = P[1.1025 - 1] = \tfrac{10.25P}{100}


Given: CI - SI = 1125

10.25P100−9P100=1125\tfrac{10.25P}{100} - \tfrac{9P}{100} = 1125

1.25P100=1125\tfrac{1.25P}{100} = 1125

P=1125×1001.25=90000P = 1125 \times \tfrac{100}{1.25} = 90000


Alternative approach using test value:

We test with P = Rs. 8000 (chosen to make calculations easy)


SI for 3 years at 3% = 8000×3×3÷100=Rs.7208000 \times 3 \times 3 \div 100 = Rs. 720

CI for 2 years at 5% = 8000×[(1.05)2−1]=8000×0.1025=Rs.8208000 \times [(1.05)^2 - 1] = 8000 \times 0.1025 = Rs. 820


Difference = 820 - 720 = Rs. 100

When difference is Rs. 100, principal is Rs. 8000

When difference is Rs. 1125, principal is: 8000×1125100=Rs.900008000 \times \tfrac{1125}{100} = Rs. 90000


The test value method works because the difference between CI and SI is directly proportional to the principal amount. This means if we know the difference for any principal, we can find the required principal using simple proportion.

In competitive exams, choosing a convenient test value (like 8000) makes mental calculations faster than solving the complete algebraic equation.

Therefore, the principal amount is Rs. 90000

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