A certain amount of money is invested at 6% per year. A second amount is P5000 larger than the first is invested at 8% per year. The interest from the investment at the higher rate exceeds the income from the lower investment by P500. Find the investment at each rate.

Answer 1

$X @ 6%.
$(X+5000) @ 8%.

(x+5000)0.08 = 0.06x + 500.
0.08x + 400 = 0.06x+500
0.02x = 500-400 = 100
X = $5000 @ 6%.
X+5000 = 5000+5000 = $10,000 @ 8%