23202 is invested part at 14% and the rest at 6% if the interest earned from the amount invested at 14% exceeds the interest earned from the amount invested at 6% by 1479.28 how much is invested at each rate
Let's assume that x is the amount invested at 14% interest rate and 23202 - x is the amount invested at 6% interest rate.
The interest earned from the amount invested at 14% is given by x * 0.14.
The interest earned from the amount invested at 6% is given by (23202 - x) * 0.06.
According to the problem, the interest earned from the amount invested at 14% exceeds the interest earned from the amount invested at 6% by 1479.28:
x * 0.14 - (23202 - x) * 0.06 = 1479.28.
Let's simplify the equation:
0.14x - 0.06(23202 - x) = 1479.28.
0.14x - 1392.12 + 0.06x = 1479.28.
0.2x - 1392.12 = 1479.28.
0.2x = 2871.4.
x = 2871.4 / 0.2.
x = 14357.
So, $14357 is invested at 14% and $23202 - $14357 = $8845 is invested at 6%.