Bob invested $30,000, part at 10% and part at 1%. If the total interest at the end of the year is $1,560, how much did he invest at 10%?
Let x be the amount invested at 10%. Then, 30,000-x is invested at 1%
0.1 x + 0.01 (30,000 - x) = 1560
Solve that algebraic equation for x.
First step:
0.09 x = 1560 - 300 = 1260
x = ?
14000
To solve the equation, we can start by subtracting 300 from both sides:
0.09x = 1560 - 300
0.09x = 1260
Next, we can divide both sides of the equation by 0.09 to isolate x:
x = 1260 / 0.09
Evaluating this expression gives:
x ≈ 14,000
Therefore, Bob invested approximately $14,000 at 10%.