Answer:
D. $150 in 4%, $850 in 6%
Step-by-step explanation:
1. The amount available to invest in the accounts, A = $1,000
The amount of average interest rate needed = 5.7 percent = 0.057
The interest paid by one of the accounts, account 1, I₁ = 4 percent = 0.04
The interest paid by the other account, account 2, I₂ = 6 percent = 0.06
Let 'x' represent the amount invested in account 1 that pays 4% and let 'y' represent the amount invested in account 2, that pays 6% we get;
x + y = 1,000...(1)
0.04·x + 0.06·y = 0.057 × 1,000 = 57
∴ 0.04·x + 0.06·y = 57...(2)
Making 'y' the subject of equation (1) and (2), we have;
For equation (1), y = 1,000 - x...(3)
For equation (2), y = 950 - (2/3)·x...(4)
Subtract equation (4) from equation (3) gives;
y - y = 1,000 - x - (950 - (2/3)·x)
0 = 1,000 - 950 - x + (2/3)·x
0 = 50 - (1/3)·x
(1/3)·x = 50
x = 3 × 50 = 150
x = 150
The amount invested in the 4% account, x = $150
From equation (3), we have;
y = 1,000 - x
∴ y = 1,000 - 150 = 850
y = 850
The amount invested in the 6% account, y = $850.
Therefore;
$150 should be invested in the account that pays 4% while $850 should be invested in the account that pays 6%.
