Question

Hope invests $\$16,000$ and $\$32,000$ at $6\%$ and $4\%$ rates of interest p.a., respectively. Both the investments are done for the same time period fetching a net interest of $\$13,440.$ What is the period of investment?

• A. $6\text{years}$
• B. $5\text{years}$
• C. $4\text{years}$
• D. $8\text{years}$

Answer 1

The correct option is A $6\text{years}$

Let $‘‘t^{\prime \prime }$ be the time period for which the amount is invested.
We know,
$SI=\frac{Prt}{100}$

Simple interest from $\$16,000\text{in}“t”$ year at the rate of $6\%p.a.:$

$SI=\frac{16000\times 6\times t}{100}$

$SI=160\times 6\times t=\$960t$

Simple interest from $\$32,000\text{in}“t”$ years at the rate of $4\%p.a.:$
$SI=\frac{32,000\times 4\times t}{100}$

$SI=320\times 4\times t=\$1280t$

Net interest earned in ‘‘t’’ years =$\$960t+\$1280t=\$2240t$

The net interest earned is given as $\$13,440.$

Hence , $2240t=13,440$

Time period of investment, $t=\frac{13440}{2240}=\frac{1344}{224}=6\text{years}$