The annual interest on a 17000 investment exceeds the intere

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. What is the interest rate of each investment? $17,000 is invested at %

Solution

let interest earned on $8000 be x%

and interest earned on $17000 be = (x+0.3)%

And

amount earned on 8000$ be = y

and interest amount earned on $17000 be = y+276

so now

(8000*x%) = y

and (17000 * (x+0.3) % ) = y+276

so solving we get

80x = y ...(1)

170x+51 = y+276

170x-y = 225 ..........(2)

170x-80x = 225

90x = 225

x = 2.5

y = 80(2.5) = 200

so interest rate on 8000$ is x = 2.5%

and interest rate on 17000$ is (x+0.3) = 2.8%

interest amount earned on $8000 = (8000*2.5)/100 = 200

interest amount earned on $17000 = (17000*2.8) /100= 476