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Rick invested $10,000, part at 10% and part at 12%. If the total interest at the end of the year is $1,100, how much did he invest at 10%?

2007-12-02 13:21:40 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

[13]
If he invests the full amount of $10,000 at 10% ,he gets an annual interest of $1000
But it is $100 less than what he actually gets
This happens because he had invested some anount at 12%,where
He receives $2 more for every $100 invested
Therefore he has invested 100*100/2 or $5000 at 12%
If ,however,you want to solve the problem algebrically,here is the solution
Let him invest $x at 10% and the remaining 10000-x at 12%
By the problem
0.10x+0.12(10000-x)=1100
0.10x+1200-0.12x=1100
-0.02x=1100-1200= -100
x= -100/-0.02=5000
Therefore he had invested $5000 at 10% and the remaining $5000 at 12%

2007-12-02 13:30:45 · answer #1 · answered by alpha 7 · 0 0

10%x + 12%(10000 -- x) = 1100
=> 2x = 10000 => x = 5000
10000 -- 5000 = 5000
$5000 @ 10% and $5000 @ 12% gave $1100.

2007-12-02 13:30:50 · answer #2 · answered by sv 7 · 0 0

Hello,

Let t = amt. invested at 10%
Let w = amt. invested at 12%

Then t + w = 10000
and 10%t + 12%w = 1100

Now multiply the second equation by 100 to clear it of fractions.

Giving us 10t + 12w = 110000
Now divide by 2 and we have 5t + 6w = 55000
Okay multiply the first equation by -5 we have
-5t -5w = -50000

Add the two above equations then we have

w = $5000
Then t = $5000

Hope This Helps!!

2007-12-02 13:52:53 · answer #3 · answered by CipherMan 5 · 0 0

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2016-10-10 02:55:17 · answer #4 · answered by ? 4 · 0 0

amount invested:
10%: x
12%:10000-x

total interest = 1100
10%x +12% (10000-x) = 1100
10%x + 1200 - 12%x = 1100
2%x = 100
2x = 10000
x = 5000

So, amount invested at 10% is $5000!!!

2007-12-02 13:29:38 · answer #5 · answered by lazylady_stel 2 · 0 0

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