Solving Simple Interest Problem: Splitting an Amount

This problem involves calculating how an initial amount of money was split and lent at two different simple interest rates, given the total interest earned over a specific period. We need to find the portion of the amount that was lent at the higher interest rate.

Understanding the Problem

We have a total amount of Rs. 7,000. This amount is divided into two parts. Let's call the amount lent at 6% per annum as 'x' rupees. Then, the remaining amount lent at 4% per annum will be (7000 - x) rupees.

Both parts are lent for the same time period, which is 5 years. The total simple interest received from both parts combined is Rs. 1,600.

We will use the simple interest formula:

$SI = \frac{P \times R \times T}{100}$

Where:

• $P$ = Principal amount
• $R$ = Rate of interest per annum
• $T$ = Time in years
• $SI$ = Simple Interest

Setting up the Equations

Let's calculate the simple interest earned from each part:

Part 1: Amount lent at 6% p.a.

• Principal ($P_1$) = x
• Rate ($R_1$) = 6%
• Time ($T_1$) = 5 years
• Simple Interest ($SI_1$) = $\frac{x \times 6 \times 5}{100} = \frac{30x}{100}$

Part 2: Amount lent at 4% p.a.

• Principal ($P_2$) = 7000 - x
• Rate ($R_2$) = 4%
• Time ($T_2$) = 5 years
• Simple Interest ($SI_2$) = $\frac{(7000 - x) \times 4 \times 5}{100} = \frac{(7000 - x) \times 20}{100}$

The total simple interest received is the sum of $SI_1$ and $SI_2$. We are given that the total simple interest is Rs. 1,600.

$SI_1 + SI_2 = 1600$

$\frac{30x}{100} + \frac{20(7000 - x)}{100} = 1600$

Solving the Equation

Now, let's solve the equation for x:

Multiply the entire equation by 100 to remove the denominators:

$30x + 20(7000 - x) = 1600 \times 100$

$30x + 140000 - 20x = 160000$

Combine the 'x' terms:

$(30x - 20x) + 140000 = 160000$

$10x + 140000 = 160000$

Subtract 140000 from both sides:

$10x = 160000 - 140000$

$10x = 20000$

Divide by 10:

$x = \frac{20000}{10}$

$x = 2000$

So, the amount lent at 6% p.a. is Rs. 2,000.

We can also find the amount lent at 4% p.a.: 7000 - x = 7000 - 2000 = Rs. 5,000.

Let's verify the total simple interest:

$SI_1$ (from Rs. 2000 at 6% for 5 years) = $\frac{2000 \times 6 \times 5}{100} = \frac{60000}{100} = 600$

$SI_2$ (from Rs. 5000 at 4% for 5 years) = $\frac{5000 \times 4 \times 5}{100} = \frac{100000}{100} = 1000$

Total Simple Interest = $SI_1 + SI_2 = 600 + 1000 = 1600$. This matches the given total interest.

Summary of Amounts Lent

The question asks for the amount lent at 6% p.a., which is 'x'. We found x = 2000.

Thus, the amount lent at 6% p.a. was Rs. 2,000.

Revision Table: Simple Interest Concepts

Additional Information: Weighted Average Rate Approach

This type of problem can also be solved using the concept of a weighted average interest rate. If the total amount (Rs. 7000) earned Rs. 1600 in simple interest over 5 years, we can find the overall average simple interest rate for the total amount.

Total Simple Interest ($SI_{total}$) = 1600

Total Principal ($P_{total}$) = 7000

Time ($T_{total}$) = 5 years

Using the formula $SI = \frac{P \times R \times T}{100}$, we can find the average rate ($R_{avg}$):

$1600 = \frac{7000 \times R_{avg} \times 5}{100}$

$1600 = \frac{35000 \times R_{avg}}{100}$

$1600 = 350 \times R_{avg}$

$R_{avg} = \frac{1600}{350} = \frac{160}{35} = \frac{32}{7} \%$ p.a.

Now, let 'x' be the fraction of the amount lent at 6% and (1-x) be the fraction lent at 4%. The weighted average rate is given by:

$x \times R_1 + (1-x) \times R_2 = R_{avg}$

$x \times 6 + (1-x) \times 4 = \frac{32}{7}$

$6x + 4 - 4x = \frac{32}{7}$

$2x + 4 = \frac{32}{7}$

$2x = \frac{32}{7} - 4 = \frac{32 - 28}{7} = \frac{4}{7}$

$x = \frac{4}{7} \times \frac{1}{2} = \frac{2}{7}$

This means $\frac{2}{7}$ of the total amount was lent at 6%. The amount is:

Amount at 6% = $\frac{2}{7} \times 7000 = 2 \times 1000 = 2000$

This confirms the previous result using algebraic equations.