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Investment Problem

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Investment Problem

by irfan_m1973 » Tue Dec 08, 2009 2:20 am
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
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by sadullaevd » Tue Dec 08, 2009 3:08 am
IMO B

x+y= 2000

0.1y-0.08x=56

By arranging we can find either y or x. Lets find x: y=2000-x,

0.1(2000-x) - 0.08x=56,

200-0.18x=56,

x=800

Hope its the right answer.

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by thephoenix » Tue Dec 08, 2009 3:12 am
imo b
x+y=2000

investment at 10 % =.1x
and at 8% = .08y

.1x-.08y=56

solving y=800(investment at 8%)
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by irfan_m1973 » Tue Dec 08, 2009 4:42 am
sadullaevd wrote:IMO B

x+y= 2000

0.1y-0.08x=56

By arranging we can find either y or x. Lets find x: y=2000-x,

0.1(2000-x) - 0.08x=56,

200-0.18x=56,

x=800

Hope its the right answer.

Cheers
Yes B is the right answer. Thank you.
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