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

A quicker route than writing a full equation is to use a weighted average rate of interest, since the time period (3 years) is the same for both banks.

Over 3 years, Rs 9000 earns Rs 1800 in total, so the average rate of interest that actually applied, spread over the whole amount, works out to:

\[ \text{average rate} = \frac{1800}{9000\times3}\times100 = \frac{1800}{270} \approx 6.67\% \text{ per year} \]

This average of about $6.67\%$ lies between the two given rates, 6% and 8%, and sits closer to 6% than to 8%. By the rule of alligation, the two amounts split in the inverse ratio of their distance from this average.

Distance of 6% from the average: $6.67-6=0.67$. Distance of 8% from the average: $8-6.67=1.33$. So amount at 6% : amount at 8% $= 1.33:0.67$, which simplifies to $2:1$.

Splitting Rs 9000 in the ratio $2:1$ gives $9000\times\frac{2}{3}=6000$ at 6% and $9000\times\frac{1}{3}=3000$ at 8%.

Let's summarize:

• The overall interest rate earned on the full Rs 9000 works out to about $6.67\%$ per year.
• By alligation, this splits the money in the ratio $2:1$ between the 6% and 8% accounts.

So the amount invested at 6% is Rs 6000, matching option (B).