Question

Two investments totaling $ 18,000 produce an annual income of $1370. One investment yields 7% per year, while the other yields 8% per year. How much is invested at each rate?

Answer 1

To determine the amounts invested at 7% and 8% rates, we set up two equations based on the total investment and the income from the investments. By solving these equations, we find that $7,000 is invested at the 7% rate and $11,000 is invested at the 8% rate.

Step-by-step explanation:

To solve this problem, we can set up a system of equations. Let x be the amount invested at 7% and y be the amount invested at 8%. We are given that:

• x + y = $18,000 (The total investment)
• 0.07x + 0.08y = $1,370 (The total income from investments)

We can solve the system of equations by substitution or elimination. We'll use the substitution method here:

1. Solve the first equation for y: y = $18,000 - x.
2. Substitute for y in the second equation: 0.07x + 0.08($18,000 - x) = $1,370.
3. Simplify and solve for x:

0.07x + $1,440 - 0.08x = $1,370

-0.01x = -$70

x = $7,000 (Amount invested at 7%)

Substitute x back into the first equation to find y:

y = $18,000 - $7,000 = $11,000 (Amount invested at 8%)

Therefore, $7,000 is invested at 7% and $11,000 is invested at 8%.

Answer 2

Let amount invested at 6% be S
Then amount invested at 8% = 16,000 - S
We therefore have: .06(S) + .08(16,000 - S) = 1,140 .06S + 1,280 - .08S = 1,140 .06S - .08S = 1,140 - 1,280 - .02S = - 140 S, or amount invested at 6% = $ 7,000Amount invested at 8% = $16,000 - 7,000, or $ 9,000