A woman invested $15,000, part at 7% annual interest and the rest at 8%. If she earned $1,100 in income over a one-year period, how much did she invest at 7%?
Invested $P @ 7%.
Invested $(15,000-P) @ 8%.
P*r*t + (15000-P)*r*t = $1100.
P*0.07*1 + (15000-P)*0.08*1 = 1100
0.07P + 1200 -0.08P = 1100
-0.01P = 1100-1200 = -100
P = $10,000 @ 7%.
To find out how much the woman invested at 7%, we can set up a system of equations based on the given information.
Let's assume that the amount the woman invested at 7% is "x" dollars. Therefore, the amount she invested at 8% would be "(15,000 - x)" dollars, as the total investment is $15,000.
Now, we can calculate the income from each investment.
The income from the investment at 7% is given by the formula: x * 0.07.
The income from the investment at 8% is given by the formula: (15,000 - x) * 0.08.
According to the problem, the total income earned over a one-year period is $1,100. So we can write the equation:
x * 0.07 + (15,000 - x) * 0.08 = 1,100.
Now, we can solve this equation to find the value of x.
0.07x + 0.08(15,000 - x) = 1,100.
0.07x + 1,200 - 0.08x = 1,100.
Combine like terms:
-0.01x + 1,200 = 1,100.
Subtract 1,200 from both sides:
-0.01x = -100.
Divide both sides by -0.01 to isolate x:
x = -100 / -0.01.
Simplify the expression:
x = 10,000.
Therefore, the woman invested $10,000 at 7%.