Question 852943
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Janine invested one part of her $20,000 savings at 8% per year and the other part at 9% per year. Her income from the two investments was $1,650 after one year. How much did she invest at each rate?
Ans:
Let her invest x at the rate of 8%
So amount invested at 9% is 20000 - x
Interest at 8% = 0.08*x
Interest at 9% = 0.09*(20000 - x)
Total interest = {{{0.08*x + 0.09*(20000 - x) = 0.08*x + 1800 - 0.09*x = 1650}}}
Simplifying
{{{0.01*x = 150}}} or {{{x = 15000}}}
So she invests $15000 at 8%, and $5000 at 9%.
Check: Interest = 15000*0.08 + 5000 * 0.09 = 1200 + 450 = 1650. Correct!
Hope this helps.
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