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by hitmewithgmat » Thu Sep 03, 2009 1:49 pm
If x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year, the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $56. If $2,000 is the total amount invested, how much is invested at 8 percent?
A) 280
B) 800
C) 892
D) 1,108
E) 1,200
Explanation is needed.
OA B

This is how I did..
1.1x - 1.08y = 56
x + y = 2,000
But no answer is found.
This question is a retired paper test (official one).
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Source: — Problem Solving |

by Gdieterling » Thu Sep 03, 2009 3:39 pm
By picking numbers :

800 at 8% is 64, thus leaving 1200 at 10%, which is 120.

120 - 64 = 56
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by hitmewithgmat » Thu Sep 03, 2009 4:15 pm
thanks Gdieterling.

is there any algebraic way to solve this problem? any takers? thanks.
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by tom4lax » Thu Sep 03, 2009 6:55 pm
10%x - 8%y = 56

x + y = 2,000

10%x - (2,000-x)%8 = 56

18%x = 216

x=1200
y=800
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