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On January 1, 2076, Lake Loser contains x liters of water. B

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On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post 18 Aug 2009, 13:51
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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?

A. 2077
B. 2078
C. 2079
D. 2080
E. 2081
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Apr 2014, 08:13, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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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 ?
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Re: Compounded interest [#permalink] New post 19 Aug 2009, 05:40
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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?
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Re: Compounded interest [#permalink] New post 22 Apr 2014, 07:20
I did it the following way

We have that 1 - (2/7)^n >3/4

Therefore, n hast to be at least >=3, in order for the fractions of the remaining water to be higher than 3/4

Thus I got C as the correct answer choice

Unless I may be wrong in any of the arithmetic which would surprise me :)

Cheers
J :)
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Re: Compounded interest [#permalink] New post 22 Apr 2014, 08:54
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jlgdr wrote:
I did it the following way

We have that 1 - (2/7)^n >3/4

Therefore, n hast to be at least >=3, in order for the fractions of the remaining water to be higher than 3/4

Thus I got C as the correct answer choice

Unless I may be wrong in any of the arithmetic which would surprise me :)

Cheers
J :)


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?

A. 2077
B. 2078
C. 2079
D. 2080
E. 2081

Each year 2/7 of the water evaporates, thus after each year 5/7 of the water remains compared to previous year. So:

By the end of 2076: \frac{5}{7}*x liters of water will be left;
By the end of 2077: \frac{5}{7}*\frac{5}{7}*x=\frac{25}{49}*x\approx{\frac{1}{2}*x} liters of water will be left;
By the end of 2078: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{125}{343}*x\approx{\frac{1}{3}*x} liters of water will be left;
By the end of 2079: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{625}{2401}*x\approx{\frac{1}{4}*x} liters of water will be left (still a bit more than 1/4);
By the end of 2080: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x liters of water will be left (this must be less than 1/4 since the previous one was just a bit more);
....

Basically after n evaporation the amount of water left is (\frac{5}{7})^n*x. Thus the question asks to find the smallest n for which (\frac{5}{7})^n*x<\frac{x}{4}.

Answer: D.

Similar questions to practice:
if-2-3-of-the-air-in-a-tank-is-removed-with-each-stroke-of-a-128432.html
if-3-4-of-the-mineral-deposits-in-a-reservoir-of-water-are-97300.html
if-1-2-of-the-air-in-a-tank-is-removed-with-each-stroke-of-a-114943.html

Hope it helps.
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Re: Compounded interest [#permalink] New post 31 May 2014, 23:28
Bunuel wrote:
jlgdr wrote:
I did it the following way

We have that 1 - (2/7)^n >3/4

Therefore, n hast to be at least >=3, in order for the fractions of the remaining water to be higher than 3/4

Thus I got C as the correct answer choice

Unless I may be wrong in any of the arithmetic which would surprise me :)

Cheers
J :)


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?

A. 2077
B. 2078
C. 2079
D. 2080
E. 2081

Each year 2/7 of the water evaporates, thus after each year 5/7 of the water remains compared to previous year. So:

By the end of 2076: \frac{5}{7}*x liters of water will be left;
By the end of 2077: \frac{5}{7}*\frac{5}{7}*x=\frac{25}{49}*x\approx{\frac{1}{2}*x} liters of water will be left;
By the end of 2078: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{125}{343}*x\approx{\frac{1}{3}*x} liters of water will be left;
By the end of 2079: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{625}{2401}*x\approx{\frac{1}{4}*x} liters of water will be left (still a bit more than 1/4);
By the end of 2080: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x liters of water will be left (this must be less than 1/4 since the previous one was just a bit more);
....

Basically after n evaporation the amount of water left is (\frac{5}{7})^n*x. Thus the question asks to find the smallest n for which (\frac{5}{7)}^n*x<\frac{x}{4}.

Answer: D.

Hope it helps.


Hi Bunell,

If i do this by A.P. , then a(n)=(1/4)x , a1=x and d=-(2/7)x

hence an=a+(n-1)d and on solving i get n=3.6.

So shouldn't be the answer C as the 3rd year is still going on i.e. in the .6 part of the 3rd year, we will get 1/4th of x , hence in 2079 post .6, we will have the water level reduced
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Re: Compounded interest [#permalink] New post 01 Jun 2014, 01:52
Expert's post
Manik12345 wrote:
Bunuel wrote:
jlgdr wrote:
I did it the following way

We have that 1 - (2/7)^n >3/4

Therefore, n hast to be at least >=3, in order for the fractions of the remaining water to be higher than 3/4

Thus I got C as the correct answer choice

Unless I may be wrong in any of the arithmetic which would surprise me :)

Cheers
J :)


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?

A. 2077
B. 2078
C. 2079
D. 2080
E. 2081

Each year 2/7 of the water evaporates, thus after each year 5/7 of the water remains compared to previous year. So:

By the end of 2076: \frac{5}{7}*x liters of water will be left;
By the end of 2077: \frac{5}{7}*\frac{5}{7}*x=\frac{25}{49}*x\approx{\frac{1}{2}*x} liters of water will be left;
By the end of 2078: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{125}{343}*x\approx{\frac{1}{3}*x} liters of water will be left;
By the end of 2079: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{625}{2401}*x\approx{\frac{1}{4}*x} liters of water will be left (still a bit more than 1/4);
By the end of 2080: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x liters of water will be left (this must be less than 1/4 since the previous one was just a bit more);
....

Basically after n evaporation the amount of water left is (\frac{5}{7})^n*x. Thus the question asks to find the smallest n for which (\frac{5}{7)}^n*x<\frac{x}{4}.

Answer: D.

Hope it helps.


Hi Bunell,

If i do this by A.P. , then a(n)=(1/4)x , a1=x and d=-(2/7)x

hence an=a+(n-1)d and on solving i get n=3.6.

So shouldn't be the answer C as the 3rd year is still going on i.e. in the .6 part of the 3rd year, we will get 1/4th of x , hence in 2079 post .6, we will have the water level reduced


We have there a geometric progression not an arithmetic progression.
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On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post 30 Aug 2014, 07:25
Bunuel wrote:
jlgdr wrote:
I did it the following way

We have that 1 - (2/7)^n >3/4

Therefore, n hast to be at least >=3, in order for the fractions of the remaining water to be higher than 3/4

Thus I got C as the correct answer choice

Unless I may be wrong in any of the arithmetic which would surprise me :)

Cheers
J :)


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?

A. 2077
B. 2078
C. 2079
D. 2080
E. 2081

Each year 2/7 of the water evaporates, thus after each year 5/7 of the water remains compared to previous year. So:

By the end of 2076: \frac{5}{7}*x liters of water will be left;
By the end of 2077: \frac{5}{7}*\frac{5}{7}*x=\frac{25}{49}*x\approx{\frac{1}{2}*x} liters of water will be left;
By the end of 2078: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{125}{343}*x\approx{\frac{1}{3}*x} liters of water will be left;
By the end of 2079: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{625}{2401}*x\approx{\frac{1}{4}*x} liters of water will be left (still a bit more than 1/4);
By the end of 2080: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x liters of water will be left (this must be less than 1/4 since the previous one was just a bit more);
....

Basically after n evaporation the amount of water left is (\frac{5}{7})^n*x. Thus the question asks to find the smallest n for which (\frac{5}{7})^n*x<\frac{x}{4}.

Answer: D.

Similar questions to practice:
if-2-3-of-the-air-in-a-tank-is-removed-with-each-stroke-of-a-128432.html
if-3-4-of-the-mineral-deposits-in-a-reservoir-of-water-are-97300.html
if-1-2-of-the-air-in-a-tank-is-removed-with-each-stroke-of-a-114943.html

Hope it helps.


Love that approach but I don't know why can we do this. Can someone explain why this is working?
I did it tediously with the following approach: 5/7 - (5/7)*(2/7) = 25/49
25/49 - (25/49)*(2/7) = 125/343
etc.

I mean we always have to subtract 2/7 of the water remaining. If we always multipy with (5/7), why does this operation give us the same result?

Thanks for all the help.
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Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post 01 Sep 2014, 00:30
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pipe19 wrote:
Bunuel wrote:
jlgdr wrote:
I did it the following way

We have that 1 - (2/7)^n >3/4

Therefore, n hast to be at least >=3, in order for the fractions of the remaining water to be higher than 3/4

Thus I got C as the correct answer choice

Unless I may be wrong in any of the arithmetic which would surprise me :)

Cheers
J :)


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?

A. 2077
B. 2078
C. 2079
D. 2080
E. 2081

Each year 2/7 of the water evaporates, thus after each year 5/7 of the water remains compared to previous year. So:

By the end of 2076: \frac{5}{7}*x liters of water will be left;
By the end of 2077: \frac{5}{7}*\frac{5}{7}*x=\frac{25}{49}*x\approx{\frac{1}{2}*x} liters of water will be left;
By the end of 2078: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{125}{343}*x\approx{\frac{1}{3}*x} liters of water will be left;
By the end of 2079: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{625}{2401}*x\approx{\frac{1}{4}*x} liters of water will be left (still a bit more than 1/4);
By the end of 2080: \frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x liters of water will be left (this must be less than 1/4 since the previous one was just a bit more);
....

Basically after n evaporation the amount of water left is (\frac{5}{7})^n*x. Thus the question asks to find the smallest n for which (\frac{5}{7})^n*x<\frac{x}{4}.

Answer: D.

Similar questions to practice:
if-2-3-of-the-air-in-a-tank-is-removed-with-each-stroke-of-a-128432.html
if-3-4-of-the-mineral-deposits-in-a-reservoir-of-water-are-97300.html
if-1-2-of-the-air-in-a-tank-is-removed-with-each-stroke-of-a-114943.html

Hope it helps.


Love that approach but I don't know why can we do this. Can someone explain why this is working?
I did it tediously with the following approach: 5/7 - (5/7)*(2/7) = 25/49
25/49 - (25/49)*(2/7) = 125/343
etc.

I mean we always have to subtract 2/7 of the water remaining. If we always multipy with (5/7), why does this operation give us the same result?

Thanks for all the help.


x - 2/7*x = 5/7*x.

5/7*x - 2/7*(5/7*x) = 5/7*x(1 - 2/7) = 5/7*5/7*x;

5/7*5/7*x - 2/7*(5/7*5/7*x;) = 5/7*5/7*x(1 - 2/7) = 5/7*5/7*5/7*x;
...
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post 01 Sep 2014, 00:34
Thanks! I would have never thought of that.
Re: On January 1, 2076, Lake Loser contains x liters of water. B   [#permalink] 01 Sep 2014, 00:34
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On January 1, 2076, Lake Loser contains x liters of water. B

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