A sum of ₹1500 is invested at 5% p.a. simple interest. If the interest is added to the principal after every 10 years, the amount will become ₹3000 after:
16 2/3 years
This question involves calculating the total time required for an initial investment to grow to a specific amount under simple interest, with the special condition that the earned interest is added back to the principal after a fixed period (10 years in this case).
Let's break down the problem:
We need to find the total time it takes for the initial ₹1500 to become ₹3000.
First, let's calculate the simple interest earned in the initial 10 years on the principal of ₹1500 at 5% p.a.
The formula for simple interest is:
\( \text{Simple Interest (SI)} = \frac{\text{Principal (P)} \times \text{Rate (R)} \times \text{Time (T)}}{100} \)
For the first 10 years:
Simple Interest earned in the first 10 years (\( SI_1 \)) is:
\( SI_1 = \frac{1500 \times 5 \times 10}{100} \)
\( SI_1 = \frac{1500 \times 50}{100} \)
\( SI_1 = 15 \times 50 \)
\( SI_1 = ₹750 \)
According to the problem, the interest earned after 10 years is added to the principal. So, the new principal for the subsequent period (\( P_2 \)) will be:
\( P_2 = P_1 + SI_1 \)
\( P_2 = 1500 + 750 \)
\( P_2 = ₹2250 \)
The target amount is ₹3000. After 10 years, the amount has become ₹2250. The additional amount needed to reach the target is ₹3000 - ₹2250 = ₹750.
This additional ₹750 needs to be earned as simple interest on the new principal \( P_2 = ₹2250 \) at the same rate of 5% p.a.
Let \( T_2 \) be the time required to earn this additional ₹750.
Using the simple interest formula again, with \( SI_2 = ₹750 \), \( P_2 = ₹2250 \), and \( R = 5\% \):
\( SI_2 = \frac{P_2 \times R \times T_2}{100} \)
\( 750 = \frac{2250 \times 5 \times T_2}{100} \)
\( 750 = \frac{225 \times 5 \times T_2}{10} \)
Multiply both sides by 10:
\( 7500 = 225 \times 5 \times T_2 \)
\( 7500 = 1125 \times T_2 \)
Now, solve for \( T_2 \):
\( T_2 = \frac{7500}{1125} \)
To simplify the fraction \( \frac{7500}{1125} \):
Divide numerator and denominator by 25:
\( T_2 = \frac{7500 \div 25}{1125 \div 25} = \frac{300}{45} \)
Divide numerator and denominator by 5:
\( T_2 = \frac{300 \div 5}{45 \div 5} = \frac{60}{9} \)
Divide numerator and denominator by 3:
\( T_2 = \frac{60 \div 3}{9 \div 3} = \frac{20}{3} \)
So, \( T_2 = \frac{20}{3} \) years.
Converting the fraction to a mixed number:
\( \frac{20}{3} = 6 \text{ with a remainder of } 2 \)
So, \( T_2 = 6 \frac{2}{3} \) years.
The total time taken to reach the target amount of ₹3000 is the sum of the first 10 years and the additional time \( T_2 \).
\( \text{Total Time (T)} = T_1 + T_2 \)
\( T = 10 \text{ years} + 6 \frac{2}{3} \text{ years} \)
\( T = 16 \frac{2}{3} \text{ years} \)
Thus, the amount will become ₹3000 after \( 16 \frac{2}{3} \) years.
| Period | Principal | Time | Rate | Simple Interest Earned | Amount at End of Period |
|---|---|---|---|---|---|
| First 10 years | ₹1500 | 10 years | 5% p.a. | \( \frac{1500 \times 5 \times 10}{100} = ₹750 \) | \( 1500 + 750 = ₹2250 \) (New Principal) |
| Subsequent years | ₹2250 | \( T_2 \) years | 5% p.a. | \( ₹750 \) (needed to reach ₹3000) | \( 2250 + 750 = ₹3000 \) (Target Amount) |
We found that \( T_2 = 6 \frac{2}{3} \) years were needed in the subsequent period.
Total Time = 10 years + \( 6 \frac{2}{3} \) years = \( 16 \frac{2}{3} \) years.
| Concept | Explanation | Formula (Simple Interest) |
|---|---|---|
| Principal (P) | The initial amount of money invested or borrowed. | N/A |
| Rate of Interest (R) | The percentage at which interest is calculated, usually per year. | N/A |
| Time (T) | The duration for which the money is invested or borrowed. | N/A |
| Simple Interest (SI) | Interest calculated only on the initial principal amount. | \( SI = \frac{P \times R \times T}{100} \) |
| Amount (A) | The total sum after adding the interest to the principal. | \( A = P + SI \) |
| Compounding (in this context) | Adding earned interest back to the principal to calculate interest for the next period. This problem demonstrates a specific type of periodic compounding, different from standard compound interest. | Principal for next period = Original Principal + Interest Earned |
It's important to distinguish this problem from standard compound interest:
In this problem, the interest is added back only after a fixed interval (10 years), not compounded annually or more frequently as is typical in standard compound interest scenarios. This requires a step-by-step simple interest calculation for each period with the adjusted principal.
Simple interest on a certain sum at 8% per annum for 5 years is ₹2400. What will be the compound interest on the same sum at the same rate for 2 years?
Find the compound interest on ₹64000 at 10% per annum for 9 months when interest is compounded quarterly.
A man invested on simple interest, 1/4 of his capital at 7% p.a., another 1/4 of the capital at 8% p.a. and the remaining capital at 10% p.a. He earned Rs. 700 as interest in one year. Find his total capital invested.
A sum of money doubles itself in 10 years on compound interest. In how many years will it become four times?
Find the compound interest on ₹42000 for 1½ years at 10% p.a. compounded annually.