The S.I. on a certain sum of money for 4 years at 4 percent per annum exceeds the C.I. on the same sum for 3 years at 5 percent per annum by Rs. 57. Find the approximate sum.
Rs. 24000
This problem asks us to find the approximate original sum of money (the principal) given the difference between the Simple Interest (SI) earned over a period and the Compound Interest (CI) earned over a different period on the same sum, but possibly at different rates.
Simple Interest is calculated only on the principal amount. It does not compound, meaning interest earned in previous periods is not added to the principal for calculating interest in the next period.
The formula for Simple Interest is:
SI = $\frac{P \times R \times T}{100}$
Compound Interest is calculated on the principal amount and also on the accumulated interest from previous periods. This means the interest "compounds" over time.
The formula for the total amount (Principal + CI) after T years compounded annually is:
Amount = $P \left(1 + \frac{R}{100}\right)^T$
The Compound Interest is then:
CI = Amount - P = $P \left(1 + \frac{R}{100}\right)^T - P$
Let the principal sum be P.
The SI is for 4 years at 4 percent per annum.
SI = $\frac{P \times 4 \times 4}{100} = \frac{16P}{100} = 0.16P$
The CI is for 3 years at 5 percent per annum.
CI = $P \left(1 + \frac{5}{100}\right)^3 - P$
CI = $P \left(1 + 0.05\right)^3 - P$
CI = $P (1.05)^3 - P$
Calculate $(1.05)^3$:
So, CI = $P (1.157625) - P = P (1.157625 - 1) = 0.157625P$
The problem states that the SI exceeds the CI by Rs. 57.
SI - CI = 57
Substitute the calculated values of SI and CI:
$0.16P - 0.157625P = 57$
Combine the terms involving P:
$(0.16 - 0.157625)P = 57$
$0.002375P = 57$
Now, solve for P:
$P = \frac{57}{0.002375}$
Let's perform the division:
$P \approx 24000$
The approximate sum is Rs. 24000.
Let's quickly verify the options:
| Option | Value (Rs.) | SI (4 yrs at 4%) | CI (3 yrs at 5%) | Difference (SI - CI) |
|---|---|---|---|---|
| 1 | 30000 | $0.16 \times 30000 = 4800$ | $0.157625 \times 30000 = 4728.75$ | $4800 - 4728.75 = 71.25$ |
| 2 | 16000 | $0.16 \times 16000 = 2560$ | $0.157625 \times 16000 = 2522$ | $2560 - 2522 = 38$ |
| 3 | 20000 | $0.16 \times 20000 = 3200$ | $0.157625 \times 20000 = 3152.5$ | $3200 - 3152.5 = 47.5$ |
| 4 | 24000 | $0.16 \times 24000 = 3840$ | $0.157625 \times 24000 = 3783$ | $3840 - 3783 = 57$ |
The calculation with P = 24000 yields a difference of 57, matching the problem statement exactly. Therefore, the approximate sum is Rs. 24000.
| Concept | Formula | Description |
|---|---|---|
| Simple Interest (SI) | $SI = \frac{P \times R \times T}{100}$ | Interest calculated only on the principal. |
| Compound Interest (CI) | $CI = P \left(1 + \frac{R}{100}\right)^T - P$ | Interest calculated on principal plus accumulated interest. |
| Principal (P) | - | The initial sum of money. |
| Rate (R) | - | The annual percentage rate. |
| Time (T) | - | The duration in years. |
Some problems in mathematics, especially those involving compound interest over several periods, can result in complex calculations. When asked for an "approximate sum," this indicates that the final answer might be rounded or that intermediate calculations could involve some level of rounding.
In this specific problem, calculating $(1.05)^3$ yields a precise value of 1.157625. The difference $0.16 - 0.157625 = 0.002375$ is also precise. Dividing 57 by 0.002375 gives exactly 24000. So, in this case, the "approximate" sum is actually the exact value among the given options. However, in other problems, you might need to round your final calculated value to match the closest option provided.
Always read the question carefully to understand if an exact value or an approximation is required. If options are given, they can help guide your level of precision in calculations.
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