Question

What is the interest rate of each investment if The annual interest on a $17,000 investment exceeds the interest earned on an $8000 investment by $276. The $17,000 is invested at a 0.3% higher rate of interest than the $8000?

Answer 1

Solution-

Let's assume, the rate of interest of $8000 is x%,

then the rate of interest of $17000 is (x+0.3x) =1.3x%

Interest earned by $8000,

$i_1=\frac{8000\times x\times 1}{100} =80x$

Interest earned by $17,000,

$i_2=\frac{17000\times 1.3x\times 1}{100} =221x$

According to the question,

$\Rightarrow i_1=i_2+276$

$\Rightarrow 221x=80x+276$

$\Rightarrow 221x-80x=276$

$\Rightarrow 141x=276$

$\Rightarrow x=1.96$

∴ Rate of interest of $8000 is 1.96% and rate of interest of $17000 is (1.3×1.96) =2.55%