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.
$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%
To find the investments at each rate, we can set up a system of equations based on the given information.
Let's say the first amount of money invested is x. Therefore, the second amount invested would be x + P5000.
Now let's calculate the interest earned from each investment:
Interest from the first investment = x × 6% = 0.06x
Interest from the second investment = (x + P5000) × 8% = 0.08(x + P5000)
We know that the interest from the second investment exceeds the income from the first investment by P500. So we can set up the following equation:
0.08(x + P5000) - 0.06x = P500
Now let's solve this equation:
0.08x + 0.08P5000 - 0.06x = P500
0.02x + 0.08P5000 = P500
0.02x = P500 - 0.08P5000
0.02x = P500(1 - 0.08)
0.02x = P500(0.92)
x = (P500(0.92))/0.02
x = P500(46)
Therefore, the investment at the lower interest rate is P500(46).
Now we can substitute x back into one of the earlier equations to find the investment at the higher rate:
x + P5000 = P500(46) + P5000
x + P5000 = P500(46 + 10)
x + P5000 = P500(56)
The investment at the higher interest rate is P500(56).