Robert invested $10000 and earned a total of $550 interest. He invested part of the $10000 in an account paying 5% and rest in an account paying 6%. How much did he invest in each account.
(x * .05) + [(10000 - x) * .06] = 550
if x at 5% then the rest (10000-x) is at 6%. So, add up the interest:
.05x + .06(10000-x) = 550
.05x +.06(10,000-x) = 550
.05 x + 600 -.06 x = 550
50 = .01 x
x = 5000
To solve this problem, we can use a system of equations based on the information given. Let's assume Robert invested x dollars in the 5% account and ($10000 - x) dollars in the 6% account.
The interest earned from the 5% account can be calculated by multiplying the amount invested (x) by the interest rate (5%) and dividing by 100:
Interest from 5% account = (x * 5) / 100 = 0.05x
Similarly, the interest earned from the 6% account can be calculated using ($10000 - x) as the amount invested in the 6% account:
Interest from 6% account = (($10000 - x) * 6) / 100 = 0.06(10000 - x)
According to the problem, the total interest earned is $550. Therefore, we can set up the equation:
0.05x + 0.06(10000 - x) = 550
To solve this equation, we can distribute 0.06 to (10000 - x):
0.05x + 600 - 0.06x = 550
Combine like terms:
-0.01x = -50
Divide by -0.01:
x = 5000
Therefore, Robert invested $5000 in the 5% account and ($10000 - $5000) = $5000 in the 6% account.