Question

A man earns 6% SI on his deposits in Bank A while he earns 8% simple interest on his deposits in Bank B. If the total interest he earns is Rs. 1800 in three years on an investment of Rs. 9000, what is the amount invested at 6%?

• A. 3000
• B. 6000
• C. 4000
• D. 4500

Hint:

Let the amount at 6% be x, write one simple interest equation for 3 years, and solve for x.

Answer 1

The Correct Option is B

Step 1: Understanding the Question.
A man splits Rs 9000 between two banks: Bank A at 6% simple interest and Bank B at 8% simple interest, for 3 years, and the combined interest from both banks is Rs 1800. We need the amount kept in Bank A.

Step 2: Key Formula or Approach.
Simple interest is given by \(SI=\dfrac{P\times R\times T}{100}\), where \(P\) is the principal, \(R\) the rate percent per year and \(T\) the time in years.
Let Rs \(x\) be invested at 6% in Bank A, so Rs \((9000-x)\) is invested at 8% in Bank B.

Step 3: Detailed Explanation.
Interest from Bank A over 3 years: \(\dfrac{x\times6\times3}{100}=\dfrac{18x}{100}\).
Interest from Bank B over 3 years: \(\dfrac{(9000-x)\times8\times3}{100}=\dfrac{24(9000-x)}{100}\).
Their sum is Rs 1800:
\[ \frac{18x}{100}+\frac{24(9000-x)}{100}=1800 \]
Multiply both sides by 100:
\[ 18x+216000-24x=180000 \]
\[ -6x = -36000 \]
\[ x = 6000 \]

Step 4: Final Answer.
The amount invested at 6% in Bank A is Rs 6000, leaving Rs 3000 in Bank B at 8%. Checking: interest from A \(=6000\times6\times3/100=1080\), interest from B \(=3000\times8\times3/100=720\), and \(1080+720=1800\), which matches the given data.
\[ \boxed{Rs.\ 6000} \]