A person divided a sum of Rs. 17, 200 into three parts and invested at 5%, 6% and 9% per annum simple interest. At the end of two years, he got the same interest on each part of money. What is the money invested at 9%?
Rs. 4,000
Solving the Simple Interest Problem
The question asks us to find the amount of money invested at a specific interest rate (9%) when a total sum is divided into three parts, invested at different simple interest rates for the same time period, and each part earns the same amount of simple interest.
Understanding the Problem Setup
We have a total sum of Rs. 17,200 divided into three parts. Let's call these parts \(P_1\), \(P_2\), and \(P_3\). These parts are invested at simple interest rates of 5%, 6%, and 9% per annum, respectively. The investment period for all three parts is 2 years. The key condition is that the simple interest earned from each part is equal.
The total sum is:
\( P_1 + P_2 + P_3 = 17200 \)
Simple Interest Formula
The formula for simple interest (SI) is given by:
\( SI = \frac{\text{Principal} \times \text{Rate} \times \text{Time}}{100} \)
Where:
- Principal (P) is the amount invested.
- Rate (R) is the annual interest rate (in percent).
- Time (T) is the time period in years.
Setting up Equations based on Equal Interest
Let \(SI_1\), \(SI_2\), and \(SI_3\) be the simple interests earned on \(P_1\), \(P_2\), and \(P_3\) respectively. The rates are \(R_1 = 5\%\), \(R_2 = 6\%\), and \(R_3 = 9\%\). The time is \(T = 2\) years for all parts.
The condition is \(SI_1 = SI_2 = SI_3\).
Using the simple interest formula:
- \( SI_1 = \frac{P_1 \times 5 \times 2}{100} = \frac{10 P_1}{100} \)
- \( SI_2 = \frac{P_2 \times 6 \times 2}{100} = \frac{12 P_2}{100} \)
- \( SI_3 = \frac{P_3 \times 9 \times 2}{100} = \frac{18 P_3}{100} \)
Since \(SI_1 = SI_2 = SI_3\), we can write the equations:
\( \frac{10 P_1}{100} = \frac{12 P_2}{100} = \frac{18 P_3}{100} \)
We can multiply the entire equation by 100 to simplify:
\( 10 P_1 = 12 P_2 = 18 P_3 \)
Finding Relationships between the Parts
From \(10 P_1 = 12 P_2\), we get:
\( \frac{P_1}{P_2} = \frac{12}{10} = \frac{6}{5} \implies P_1 = \frac{6}{5} P_2 \)
From \(12 P_2 = 18 P_3\), we get:
\( \frac{P_2}{P_3} = \frac{18}{12} = \frac{3}{2} \implies P_2 = \frac{3}{2} P_3 \)
Now, we can express \(P_1\) in terms of \(P_3\) by substituting the expression for \(P_2\):
\( P_1 = \frac{6}{5} P_2 = \frac{6}{5} \times \left(\frac{3}{2} P_3\right) = \frac{18}{10} P_3 = \frac{9}{5} P_3 \)
Solving for the Amount Invested at 9%
We know that the total sum is \(P_1 + P_2 + P_3 = 17200\). Substitute the expressions for \(P_1\) and \(P_2\) in terms of \(P_3\):
\( \frac{9}{5} P_3 + \frac{3}{2} P_3 + P_3 = 17200 \)
To add these fractions, find a common denominator, which is 10:
\( \frac{9 \times 2}{5 \times 2} P_3 + \frac{3 \times 5}{2 \times 5} P_3 + \frac{10}{10} P_3 = 17200 \)
\( \frac{18}{10} P_3 + \frac{15}{10} P_3 + \frac{10}{10} P_3 = 17200 \)
\( \frac{(18 + 15 + 10) P_3}{10} = 17200 \)
\( \frac{43 P_3}{10} = 17200 \)
Now, solve for \(P_3\):
\( 43 P_3 = 17200 \times 10 \)
\( 43 P_3 = 172000 \)
\( P_3 = \frac{172000}{43} \)
Performing the division:
\( P_3 = 4000 \)
So, the money invested at 9% (which is \(P_3\)) is Rs. 4,000.
Verification (Optional)
We found \(P_3 = 4000\).
\( P_2 = \frac{3}{2} P_3 = \frac{3}{2} \times 4000 = 3 \times 2000 = 6000 \)
\( P_1 = \frac{9}{5} P_3 = \frac{9}{5} \times 4000 = 9 \times 800 = 7200 \)
Total sum = \(P_1 + P_2 + P_3 = 7200 + 6000 + 4000 = 17200\). This matches the total sum given in the problem.
Calculate simple interest for each part for 2 years:
- \( SI_1 = \frac{7200 \times 5 \times 2}{100} = \frac{7200 \times 10}{100} = 72 \times 10 = 720 \)
- \( SI_2 = \frac{6000 \times 6 \times 2}{100} = \frac{6000 \times 12}{100} = 60 \times 12 = 720 \)
- \( SI_3 = \frac{4000 \times 9 \times 2}{100} = \frac{4000 \times 18}{100} = 40 \times 18 = 720 \)
The simple interest earned on each part is indeed the same (Rs. 720). The calculations are correct.
Revision Table: Simple Interest Concepts
| Concept | Description | Formula |
|---|---|---|
| Principal (P) | The initial amount of money invested or borrowed. | - |
| Rate (R) | The percentage at which interest is calculated annually. | - |
| Time (T) | The duration for which the money is invested or borrowed (usually in years). | - |
| Simple Interest (SI) | Interest calculated only on the principal amount. | \( SI = \frac{P \times R \times T}{100} \) |
| Amount (A) | The total sum after adding interest to the principal. | \( A = P + SI \) |
Additional Information: Simple Interest Applications
Simple interest is a fundamental concept in finance used for calculating interest on various types of transactions. While compound interest is more common for long-term investments and loans, simple interest is often used in specific scenarios like:
- Short-term loans.
- Certain types of bonds or securities.
- Calculations for parts of a year.
Understanding how to calculate and compare simple interest across different principals, rates, and times is crucial for various financial problems. Problems like this one, where a sum is divided and conditions related to interest are given, test the understanding of the simple interest formula and algebraic manipulation.
Similar Questions
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Important Questions from Simple Interest
How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest ?
Nirav and Mehul borrowed Rs.4000 and Rs.5000 respectively for 2.5 years at the rate of x% per annum. Mehul paid Rs 125 more interest than Nirav. Find x.
If the interest on a sum of Rs.1200 is more than the interest on Rs.1000 by Rs.120 in three years, then what is the rate of interest per annum?.
The difference between the simple interest received from two banks on Rs. 500 for two years is Rs. 2.50. What is the difference between their rates?
A sum of Rs.1200 becomes Rs.1560 at a rate of simple interest in 3 years. In how many years will the sum of Rs.800 amount to Rs.1120 at the same rate of simple interest?
