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If x dollars is invested at 10 percent for one year and y dollars is

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If x dollars is invested at 10 percent for one year and y dollars is  [#permalink]

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New post 17 May 2011, 09:33
2
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

71% (01:58) correct 29% (01:56) wrong based on 85 sessions

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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. $1108
e. $1200
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Re: If x dollars is invested at 10 percent for one year and y dollars is  [#permalink]

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New post 17 May 2011, 09:41
3
2 equations with 2 unknowns

10x / 100 - 8y / 100 = 56

and

x + y = 2000

Solving these 2 equations, x = 1200 and y = 800

Answer B.
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Re: If x dollars is invested at 10 percent for one year and y dollars is  [#permalink]

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New post 17 May 2011, 10:29
annual income = principal + interest

for x @ 10% x + 10/100 * x = 1.1 x
for y @ 8% y + 8/100 * y = 1.08y

x+y = 2000

1.1x - 1.08y = 56
1.1 (2000-y) - 1.08y = 56
2200-1.1y-1.08y = 56
2.08y = 2144

y= 2144/2.18 = 983.46

so the question stem must say the annual interest difference rather than annual income difference.
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Re: If x dollars is invested at 10 percent for one year and y dollars is  [#permalink]

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New post 18 May 2011, 21:11
Lucky twins -the answer is B or E ;-)
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Re: If x dollars is invested at 10 percent for one year and y dollars is  [#permalink]

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New post 13 Jun 2017, 12:12
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tonebeeze 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. $1108
e. $1200


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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Re: If x dollars is invested at 10 percent for one year and y dollars is  [#permalink]

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New post 28 Jul 2018, 04:27
everyone above has mentioned the conventional method. Which almost takes 2 mins. if anyone could suggest an approach for easy POE. Bunuel chetan2u sayantanc2k
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Re: If x dollars is invested at 10 percent for one year and y dollars is  [#permalink]

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New post 28 Jul 2018, 06:29
AdityaHongunti wrote:
everyone above has mentioned the conventional method. Which almost takes 2 mins. if anyone could suggest an approach for easy POE. Bunuel chetan2u sayantanc2k



Aditya two ways ..
1) 8% would require at least two 0s to ensure that the difference is an integer
Otherwise say amount is 780...8% will result in a decimal
So answer will be 800, 1000 or 1200..
If it is 1000 each.. difference is 100 for 10% and 80 for 8%, so 100-80=20
But we want 56>20
So if you want to increase the difference, the greater should increase and smaller decrease, so only possible value 800

Otherwise
1) 56 has a units digit of 6 which is 10-4..
So 8*z should have units digit as 4.. 8*800 /100 will give 4 as units digit
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Re: If x dollars is invested at 10 percent for one year and y dollars is &nbs [#permalink] 28 Jul 2018, 06:29
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