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Compounded interest

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Manager
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Compounded interest [#permalink] New post 18 Aug 2009, 13:51
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

50% (00:00) correct 50% (04:06) wrong based on 2 sessions
Would you please explain me this question? Thanks a lot

On January 1, 2076, Lake Loser contains x liters of water. By Dec 31 of that same year, 2/7 of the x liters have evaporated. This pattern continues such that by the end of each subsequent year the lake has lost 2/7 of the water that it contained at the beginning of that year. During which year will the water in the lake be reduced to less than 1/4 of the original x liters?

2077
2078
2079
2080
2081
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Re: Compounded interest [#permalink] New post 18 Aug 2009, 14:50
TheRob wrote:
Would you please explain me this question? Thanks a lot

On January 1, 2076, Lake Loser contains x liters of water. By Dec 31 of that same year, 2/7 of the x liters have evaporated. This pattern continues such that by the end of each subsequent year the lake has lost 2/7 of the water that it contained at the beginning of that year. During which year will the water in the lake be reduced to less than 1/4 of the original x liters?

2077
2078
2079
2080
2081


D:

It must be:

x*(5/7)^n<x/4;
(5/7)^n<1/4 --> n=5 --> 2076+5=2081, so it is reduced during 2080
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Re: Compounded interest [#permalink] New post 18 Aug 2009, 19:37
Can you expalin a little more briefly.

My question is Why did u take a Compound Interest and not an SI.

As i see the amount is going by a constant value each year.

decrease % = 2/7x / x *100 = 200/7

A = P + S.I

1/4x = x + S.I so S.I = -3x /4

-3x/4 = (x * 200/7 * n ) / 100


n = -21 / 8 = 2.625

So i thought C was the answer.
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Re: Compounded interest [#permalink] New post 18 Aug 2009, 20:05
The difference is not constant because it is 2/7 of whatever is left. And "whatever is left" is varying at the end of each year.
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Re: Compounded interest [#permalink] New post 18 Aug 2009, 22:35
TheRob wrote:
Would you please explain me this question? Thanks a lot
D:

It must be:

x*(5/7)^n<x/4;
(5/7)^n<1/4 --> n=5 --> 2076+5=2081, so it is reduced during 2080

How did you get n=5 ?
Manager
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Re: Compounded interest [#permalink] New post 19 Aug 2009, 05:40
The answer is D

here is the long explanation

This fraction problem contains an "unspecified" total (the x liters of water in the lake). Pick an easy "smart" number to make this problem easier. Usually, the smart number is the lowest common denominator of all the fractions in the problem. However, if you pick 28, you will quickly see that this yields some unwieldy computation.
The easiest number to work with in this problem is the number 4. Let's say there are 4 liters of water originally in the lake. The question then becomes: During which year is the lake reduced to less than 1 liter of water?
At the end of 2076, there are 4 × (5/7) or 20/7 liters of water in the lake. This is not less than 1.
At the end of 2077, there are (20/7) × (5/7) or 100/49 liters of water in the lake. This is not less than 1.
At the end of 2078, there are (100/49) × (5/7) or 500/343 liters of water in the lake. This is not less than 1.
At the end of 2079, there are (500/343) × (5/7) or 2500/2401 liters of water in the lake. This is not less than 1.
At the end of 2080, there are (2500/2401) × (5/7) or 12500/16807 liters of water in the lake. This is less than 1.
Notice that picking the number 4 is essential to minimizing the computation involved, since it is very easy to see when a fraction falls below 1 (when the numerator becomes less than the denominator.) The only moderately difficult computation involved is multiplying the denominator by 7 for each new year.
The correct answer is D.
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Re: Compounded interest [#permalink] New post 19 Aug 2009, 20:28
Economist wrote:
TheRob wrote:
Would you please explain me this question? Thanks a lot
D:

It must be:

x*(5/7)^n<x/4;
(5/7)^n<1/4 --> n=5 --> 2076+5=2081, so it is reduced during 2080

How did you get n=5 ?


Concur with Economist, how did you get that n=5?
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Re: Compounded interest [#permalink] New post 20 Aug 2009, 13:08
Ibodullo wrote:
Economist wrote:
TheRob wrote:
Would you please explain me this question? Thanks a lot
D:

It must be:

x*(5/7)^n<x/4;
(5/7)^n<1/4 --> n=5 --> 2076+5=2081, so it is reduced during 2080

How did you get n=5 ?


Concur with Economist, how did you get that n=5?


Testing numbers for n.
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Re: Compounded interest [#permalink] New post 24 Aug 2009, 02:39
this problem involves a lot of computations and it definitely takes more than 2 mins to solve. we need to find (5/7)^n<1/4, so every time we add one more year we need to compare this number with 1/4 and find a common denominator. so at the end we have to when 5^n*4<7^n.
I don't think GMAT requires to remember what 7^(3,4, or 5) is. May be there is an easier way to solve this problem? Anyone?
Re: Compounded interest   [#permalink] 24 Aug 2009, 02:39
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