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Sarah invests $4500 part at 7% the balance at 8.5%. After one year, the interest earned on the 7% investment was $150 less than the interest earned on the 8.5% investment. How much was invested at each rate?
make 2 equations to answer question

2007-09-09 04:59:16 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

Let x be the amount invested at 7% and y the amount invested at 8.5%

First equation is

x + y = 4500

or x = 4500 - y

Interest earned after 1 year = 7x / 100 and 8.5y / 100 respectively.

We are given that 8.5y / 100 = 150 + 7x / 100

0.085y = 150 + 0.07x

0.085y = 150 + 0.07 (4500 - y)

0.085y = 150 + 315 - 0.07y

0.085y + 0.07y = 465

0.155y = 465

y = 465 / 0.155 = 3000

And x = 4500 - 3000 = 1500

So, Sarah invested 1500 dollars at 7% and 3000 dollars at 8.5%

2007-09-09 05:56:03 · answer #1 · answered by Swamy 7 · 0 0

x: the amount of 7% investment
y: the amount of 8% inverstment
Total investment balance:
x+y = 4500......(1)
Interest balance:
.07x = .085y-150......(2)
Plug in (1) in (2),
0.07x = 0.085(4500-x) - 150
Solve for x,
x = $1,500
y = $3,000

2007-09-09 05:14:14 · answer #2 · answered by sahsjing 7 · 0 0

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2016-12-31 17:40:04 · answer #3 · answered by ? 4 · 0 0

x: 7% investment
y: 8% inverstment
x+y = 4500......(1)
.07x = .085y-150......(2)
Plug in (1) in (2),
0.07x = 0.085(4500-x) - 150
Solve for x,
x = $1,500
y = $3,000

2007-09-09 05:15:19 · answer #4 · answered by orefilwe p 1 · 0 0

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