Mr. Wilson invested money in two accounts. His total investment was $40,000. If one account pays 4% in interest and the other pays 8% in interest, how much does he have in each account if he earned a total of $2,720 in interest in 1 year?
invest x at 4%, y at 8%
x + y = 40000
.04x + .08y = 2720
.08x + .08y = 3200
.04x + .08y = 2720
now subtract:
.04x = 480
x = 12000
y = 28000
.04*12000 + .08*28000 = 480 + 2240 = 2720
x + y = 40,000
.04x + 0.08y = 2,720
x = 40,000 - y
.04(40,000 - y) + 0.08y = 2,720
1,600 - .04y + 0.08y = 2,720
1,600 + 0.04y = 2,720
0.04y = 1,120
y = 28,000
x = 40,000 - 28,000
x = 12,000
y = 28,000
I want to inverted
To solve this problem, we can use a system of equations. Let's say Mr. Wilson has x dollars in the account that pays 4% interest, and (40,000 - x) dollars in the account that pays 8% interest.
The interest earned from the first account at 4% interest for one year would be 0.04x.
Similarly, the interest earned from the second account at 8% interest for one year would be 0.08(40,000 - x).
According to the problem, the total interest earned from both accounts combined is $2,720. Therefore, we can write the equation:
0.04x + 0.08(40,000 - x) = 2,720
Now, let's solve this equation step by step:
0.04x + 0.08(40,000 - x) = 2,720
0.04x + 3,200 - 0.08x = 2,720
-0.04x + 3,200 = 2,720
-0.04x = 2,720 - 3,200
-0.04x = -480
x = (-480)/(-0.04)
x = 12,000
So, Mr. Wilson has $12,000 in the account that pays 4% interest, and $40,000 - $12,000 = $28,000 in the account that pays 8% interest.
Therefore, Mr. Wilson has $12,000 in one account and $28,000 in the other account.