Simple Interest

gmattester · Posted: Thu Aug 21, 2008 6:23 pm Post subject: Simple Interest

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

Here is my strategy but I am not getting answer
P1-P2=56------------1
P1=x+x/10=11x/10
P2=y+8y/100=108y/100

x+ y=2000---------------2
From 1 and 2 find value of x

Suyog · Posted: Thu Aug 21, 2008 6:48 pm Post subject:

I would solve using the given option...
1. 280 seems very less.

2. 800

8% of 800 = 64

remaining amount 2000-800 = 1200

10% of 1200 = 120

120 - 64 = 56

Choose(b)

gmattester · Posted: Thu Aug 21, 2008 6:54 pm Post subject:

Suyog , your answer is right .
Is someone having some other method to solve this problem

dnairo1981 · Posted: Thu Aug 21, 2008 6:55 pm Post subject:

Is it B?

.10X - .08(2000-X) = 56......Get rid of decimals(multiply each side by 100)

10X - 16000 + 8X = 5600

18X = 21600

x = 1200

2000-x = 2000 - 1200 = 800

Sunny22uk · Posted: Thu Aug 21, 2008 6:55 pm Post subject: Re: Simple Interest

gmattester · Posted: Thu Aug 21, 2008 7:08 pm Post subject: Re: Simple Interest