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Simple Interest

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by gmattester » Thu Aug 21, 2008 6:23 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

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
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by Suyog » Thu Aug 21, 2008 6:48 pm
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)
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by gmattester » Thu Aug 21, 2008 6:54 pm
Suyog , your answer is right .
Is someone having some other method to solve this problem
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by dnairo1981 » Thu Aug 21, 2008 6:55 pm
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
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Re: Simple Interest

by Sunny22uk » Thu Aug 21, 2008 6:55 pm
gmattester wrote: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
Equation 1 is wrong, as per the question 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 which means 10%x-8%y=56
Hope this helps
You cannot discover new oceans unless you have the courage to loose sight of the shore.
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Re: Simple Interest

by gmattester » Thu Aug 21, 2008 7:08 pm
Sunny22uk wrote:
gmattester wrote: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
Equation 1 is wrong, as per the question 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 which means 10%x-8%y=56
Hope this helps
Thanks for correction. I was taking annual income=Principal + Interest
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by Scott@TargetTestPrep » Fri Dec 15, 2017 7:16 am
gmattester wrote: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
We are given that x = the number of dollars invested at 10% and y = the number of dollars invested at 8%. We are also given that the total amount invested is $2000, and thus we have x + y = 2000 or x = 2000 - y. Finally, we are given that the annual income from the 10% investment exceeds the annual income from the 8% investment by $56, so we can create the following equation to determine the value of y:

0.10x = 0.08y + 56

0.10(2000 - y) = 0.08y + 56

200 - 0.10y = 0.08y + 56

144 = 0.18y

14400 = 18y

y = 14400/18 = 800

Answer: B

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