10
algebra

Jessica splits $15,000 between two investments. One pays 5% annually, the other 8% annually. After one year, total interest is $990. How much was invested at 5%?

A

\6,000$

B

\7,000$

C

\8,000$

D

\9,000$

Correct Answer: B

Choice B is the correct answer. Let xx = amount at 5% and yy = amount at 8%.

System: {x+y=150000.05x+0.08y=990\begin{cases} x + y = 15000 \\ 0.05x + 0.08y = 990 \end{cases}

Step 1: From first: y=15000−xy = 15000 - x

Step 2: Substitute: 0.05x+0.08(15000−x)=9900.05x + 0.08(15000 - x) = 9900.05x+1200−0.08x=9900.05x + 1200 - 0.08x = 990−0.03x=−210-0.03x = -210x=7000x = 7000

Solution: $7,000 at 5% (and $8,000 at 8%)

Verification: 0.05(7000)+0.08(8000)=350+640=9900.05(7000) + 0.08(8000) = 350 + 640 = 990 ✓

💡 Strategic Tip: Investment problems: Principal × Rate = Interest.

**Other choices fail verification.

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Linear Functions Set 60