SOLUTION: 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

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Question 852943:
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?
Found 2 solutions by ramkikk66, checkley79:
Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
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%2Ax+%2B+0.09%2A%2820000+-+x%29+=+0.08%2Ax+%2B+1800+-+0.09%2Ax+=+1650
Simplifying
0.01%2Ax+=+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.


Answer by checkley79(3341) About Me  (Show Source):
You can put this solution on YOUR website!
.08X+.09(20,000-X)=1,650
.08X+1,800-.09X=1,650
-.01X=1,650-1,800
-.01X=-150
X=-150/.01=15,000 INVESTED @8%.
20,000-15,000=5,00 INVESTED @9%.
PROOF:
.08*15,000+.09*5,000=1,650
1,200+450=1,650
1,650=1,650