NEW HERE? LEARN MORE
closeclose
Last visit was: 25 Apr 2024, 05:58 It is currently 25 Apr 2024, 05:58
Decision Tracker
My Rewards
New posts
New comers' posts

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

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

SORT BY:
Date
Kudos
 1   2   
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
TheRob
Manager
Manager
Joined: 29 May 2008
Posts: 88
Own Kudos [?]: 718 [60]
Given Kudos: 0
Send PM
On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post   Updated on: 22 Apr 2014, 09:13
6
Kudos
54
Bookmarks
00:00
A
B
C
D
E

Difficulty:

95% (hard)

Question Stats:

45% (02:51) correct 55% (02:50) wrong based on 885 sessions
Hide Show timer Statistics
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
ShowHide Answer
Official Answer
D

Originally posted by TheRob on 18 Aug 2009, 14:51.
Last edited by Bunuel on 22 Apr 2014, 09:13, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Most Helpful Reply
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618930 [21]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: Compounded interest [#permalink] New post   22 Apr 2014, 09:54
8
Kudos
13
Bookmarks
Expert Reply
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.
User avatar
TheRob
Manager
Manager
Joined: 29 May 2008
Posts: 88
Own Kudos [?]: 718 [9]
Given Kudos: 0
Send PM
Re: Compounded interest [#permalink] New post   19 Aug 2009, 06:40
7
Kudos
2
Bookmarks
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.
General Discussion
User avatar
noboru
Director
Director
Joined: 16 Jul 2009
Posts: 543
Own Kudos [?]: 8532 [2]
Given Kudos: 2
Schools:CBS
 Q50  V37
WE 1: 4 years (Consulting)
Send PM
Re: Compounded interest [#permalink] New post   18 Aug 2009, 15:50
2
Kudos
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
User avatar
tkarthi4u
Manager
Manager
Joined: 08 Jan 2009
Posts: 144
Own Kudos [?]: 305 [0]
Given Kudos: 5
Concentration: International business
Send PM
Re: Compounded interest [#permalink] New post   18 Aug 2009, 20:37
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.
User avatar
Economist
Senior Manager
Senior Manager
Joined: 01 Apr 2008
Posts: 392
Own Kudos [?]: 4074 [0]
Given Kudos: 18
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Send PM
Re: Compounded interest [#permalink] New post   18 Aug 2009, 21:05
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.
User avatar
Economist
Senior Manager
Senior Manager
Joined: 01 Apr 2008
Posts: 392
Own Kudos [?]: 4074 [1]
Given Kudos: 18
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Send PM
Re: Compounded interest [#permalink] New post   18 Aug 2009, 23:35
1
Kudos
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 ?
User avatar
Ibodullo
Manager
Manager
Joined: 13 Jan 2009
Posts: 120
Own Kudos [?]: 32 [0]
Given Kudos: 9
Concentration: Finance
Schools:Harvard Business School, Stanford
Send PM
Re: Compounded interest [#permalink] New post   19 Aug 2009, 21:28
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?
User avatar
noboru
Director
Director
Joined: 16 Jul 2009
Posts: 543
Own Kudos [?]: 8532 [0]
Given Kudos: 2
Schools:CBS
 Q50  V37
WE 1: 4 years (Consulting)
Send PM
Re: Compounded interest [#permalink] New post   20 Aug 2009, 14:08
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.
User avatar
tomirisk
Intern
Intern
Joined: 10 Jul 2009
Posts: 17
Own Kudos [?]: 36 [0]
Given Kudos: 4
 Q45  V38
Send PM
Re: Compounded interest [#permalink] New post   24 Aug 2009, 03:39
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?
User avatar
jlgdr
VP
VP
Joined: 06 Sep 2013
Posts: 1345
Own Kudos [?]: 2391 [0]
Given Kudos: 355
Concentration: Finance
Schools: Wharton '17 (A) HBS '17 (WA$)
Send PM
GMAT ToolKit User
Re: Compounded interest [#permalink] New post   22 Apr 2014, 08:20
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 :)
User avatar
Manik12345
Intern
Intern
Joined: 25 Feb 2014
Posts: 8
Own Kudos [?]: 16 [0]
Given Kudos: 2
Send PM
Re: Compounded interest [#permalink] New post   01 Jun 2014, 00:28
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
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618930 [1]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: Compounded interest [#permalink] New post   01 Jun 2014, 02:52
1
Bookmarks
Expert Reply
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.
avatar
pipe19
Intern
Intern
Joined: 05 Feb 2014
Posts: 11
Own Kudos [?]: 11 [0]
Given Kudos: 22
Send PM
GMAT ToolKit User
On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post   30 Aug 2014, 08:25
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.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618930 [2]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post   01 Sep 2014, 01:30
1
Kudos
1
Bookmarks
Expert Reply
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;
...
avatar
pipe19
Intern
Intern
Joined: 05 Feb 2014
Posts: 11
Own Kudos [?]: 11 [0]
Given Kudos: 22
Send PM
GMAT ToolKit User
Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post   01 Sep 2014, 01:34
Thanks! I would have never thought of that.
User avatar
pradeepss
Manager
Manager
Joined: 19 Sep 2008
Status:Please do not forget to give kudos if you like my post
Posts: 69
Own Kudos [?]: 206 [2]
Given Kudos: 257
Location: United States (CA)
Send PM
GMAT ToolKit User
Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post   16 Nov 2014, 18:01
2
Kudos
Its easier to choose a smart number. since we have 4 and 7 in denominator so lets choose 28 as a value for x.

so lets quarter of that is 7.
2076 -> water evaporates from 28 ---> 5/7*28 => 20
2078 -> 20*5/7 => 100/7 => ~14....
2079 -> 14*5/7 ==> 10...
2080 -> 10*5/7 => <7 so the correct answer is D.

Answer: D

TheRob wrote:
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
User avatar
B
BrainLab
Current Student
Joined: 10 Mar 2013
Posts: 360
Own Kudos [?]: 2696 [0]
Given Kudos: 200
Location: Germany
Concentration: Finance, Entrepreneurship
Schools: WHU MBA"20 (A$)
GMAT 1: 580 Q46 V24
GPA: 3.7
WE:Marketing (Telecommunications)
Send PM
GMAT ToolKit User
Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post   09 Dec 2015, 09:32
I've solved this one, actually it's not a complex one, it just requires a great amount od calculations.... --> It's not GMAT like for me, each GMAT problem can be solved faster if you identify a right path...
avatar
B
Ndkms
Intern
Intern
Joined: 09 Aug 2016
Posts: 42
Own Kudos [?]: 65 [0]
Given Kudos: 8
Send PM
Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post   11 Jan 2017, 12:52
Bunuel wrote:

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.


Very good and thorough explanation.

Realistically speaking this is a v. tough question and very few would solve it under real testing conditions... (who knows you might be one of them)...

Key things that you need to understand and derive [

1) \((\frac{5}{7})^n*x<\frac{x}{4}\)

2) All the approximations for the fractions + powers of 5

If you cannot see relatively easily 1) and 2) you wasted your time unfortunately
User avatar
L
ScottTargetTestPrep
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18756
Own Kudos [?]: 22050 [0]
Given Kudos: 283
Location: United States (CA)
Send PM
Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink] New post   23 Apr 2019, 19:40
Expert Reply
TheRob wrote:
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


Since the lake loses 2/7 of its water, we see that it retains 5/7 of its water. We can keep track of the fraction of x liters of water left in the lake as follows:

End of 2076 (or Beginning of 2077): 5/7 ≈ 71.4%

End of 2077 (or Beginning of 2078): 5/7 x 5/7 = 25/49 ≈ 51.0%

End of 2078 (or Beginning of 2079): 25/49 x 5/7 = 125/343 ≈ 36.4%

End of 2079 (or Beginning of 2080): 125/343 x 5/7 = 625/2401 ≈ 26.0%

End of 2080 (or Beginning of 2081): 625/2401 x 5/7 = 3125/16,807 ≈ 18.6%

We see that at the beginning of 2080, the lake has approximately 26.0% of the original x liters, but at the end of the same year, it has approx 18.6%; therefore, the lake must have been reduced to less than ¼ of the original x liters at some point in 2080.

Answer: D
GMAT Club Bot
gmatclubot
Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]
23 Apr 2019, 19:40
 1   2   
Moderators:
Bunuel
Math Expert
92912 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions