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.
See previous post.
To solve this problem, we can set up a system of equations. Let's say the first amount of money invested at 6% per year is x.
So, the second investment, which is P5000 larger than the first, would be x + P5000.
The interest earned from the first investment at 6% per year would be (0.06)(x), and the interest earned from the second investment at 8% per year would be (0.08)(x + P5000).
According to the problem, the interest from the second investment exceeds the income from the first investment by P500. This can be written as an equation:
(0.08)(x + P5000) - (0.06)(x) = P500
Now we can solve this equation to find the values of x and P5000.
Let's simplify the equation:
0.08x + 0.08(P5000) - 0.06x = P500
0.08x - 0.06x + 0.08(P5000) = P500
0.02x + 0.08(P5000) = P500
Now let's isolate x by subtracting 0.08(P5000) from both sides:
0.02x = P500 - 0.08(P5000)
0.02x = P500 - 0.08 * 5000
0.02x = P500 - 400
0.02x = P100
Now, divide both sides by 0.02 to solve for x:
x = P100 / 0.02
x = P5000
So, the first investment is P5000 at 6% per year.
To find the second investment, substitute the value of x back into one of the original equations:
Second investment = x + P5000
Second investment = P5000 + P5000
Second investment = P10000
Therefore, the second investment is P10000 at 8% per year.
To summarize:
The first investment is P5000 at 6% per year.
The second investment is P10000 at 8% per year.