November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. November 17, 2018 November 17, 2018 09:00 AM PST 11:00 AM PST Join the Quiz Saturday November 17th, 9 AM PST. The Quiz will last approximately 2 hours. Make sure you are on time or you will be at a disadvantage.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 29 May 2008
Posts: 109

On January 1, 2076, Lake Loser contains x liters of water. B
[#permalink]
Show Tags
Updated on: 22 Apr 2014, 08:13
Question Stats:
44% (02:47) correct 56% (02:52) wrong based on 529 sessions
HideShow 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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by TheRob on 18 Aug 2009, 13:51.
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.




Math Expert
Joined: 02 Sep 2009
Posts: 50619

Re: Compounded interest
[#permalink]
Show Tags
22 Apr 2014, 08:54
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: if23oftheairinatankisremovedwitheachstrokeofa128432.htmlif34ofthemineraldepositsinareservoirofwaterare97300.htmlif12oftheairinatankisremovedwitheachstrokeofa114943.htmlHope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat 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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




VP
Joined: 16 Jul 2009
Posts: 1062
Schools: CBS
WE 1: 4 years (Consulting)

Re: Compounded interest
[#permalink]
Show Tags
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
_________________
The sky is the limit 800 is the limit
GMAT Club Premium Membership  big benefits and savings



Senior Manager
Joined: 08 Jan 2009
Posts: 297

Re: Compounded interest
[#permalink]
Show Tags
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.



Director
Joined: 01 Apr 2008
Posts: 783
Name: Ronak Amin
Schools: IIM Lucknow (IPMX)  Class of 2014

Re: Compounded interest
[#permalink]
Show Tags
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.



Director
Joined: 01 Apr 2008
Posts: 783
Name: Ronak Amin
Schools: IIM Lucknow (IPMX)  Class of 2014

Re: Compounded interest
[#permalink]
Show Tags
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
Joined: 29 May 2008
Posts: 109

Re: Compounded interest
[#permalink]
Show Tags
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.



Manager
Joined: 13 Jan 2009
Posts: 162
Schools: Harvard Business School, Stanford

Re: Compounded interest
[#permalink]
Show Tags
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?



VP
Joined: 16 Jul 2009
Posts: 1062
Schools: CBS
WE 1: 4 years (Consulting)

Re: Compounded interest
[#permalink]
Show Tags
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.
_________________
The sky is the limit 800 is the limit
GMAT Club Premium Membership  big benefits and savings



Intern
Joined: 10 Jul 2009
Posts: 42

Re: Compounded interest
[#permalink]
Show Tags
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?



SVP
Joined: 06 Sep 2013
Posts: 1745
Concentration: Finance

Re: Compounded interest
[#permalink]
Show Tags
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



Intern
Joined: 25 Feb 2014
Posts: 9

Re: Compounded interest
[#permalink]
Show Tags
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+(n1)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



Math Expert
Joined: 02 Sep 2009
Posts: 50619

Re: Compounded interest
[#permalink]
Show Tags
01 Jun 2014, 01:52
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+(n1)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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat 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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 05 Feb 2014
Posts: 12

On January 1, 2076, Lake Loser contains x liters of water. B
[#permalink]
Show Tags
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: if23oftheairinatankisremovedwitheachstrokeofa128432.htmlif34ofthemineraldepositsinareservoirofwaterare97300.htmlif12oftheairinatankisremovedwitheachstrokeofa114943.htmlHope 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.



Math Expert
Joined: 02 Sep 2009
Posts: 50619

Re: On January 1, 2076, Lake Loser contains x liters of water. B
[#permalink]
Show Tags
01 Sep 2014, 00:30
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: if23oftheairinatankisremovedwitheachstrokeofa128432.htmlif34ofthemineraldepositsinareservoirofwaterare97300.htmlif12oftheairinatankisremovedwitheachstrokeofa114943.htmlHope 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; ...
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat 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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 05 Feb 2014
Posts: 12

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



Manager
Status: Please do not forget to give kudos if you like my post
Joined: 19 Sep 2008
Posts: 97
Location: United States (CA)

Re: On January 1, 2076, Lake Loser contains x liters of water. B
[#permalink]
Show Tags
16 Nov 2014, 17:01
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
_________________
Please Help with Kudos, if you like my post.



Director
Joined: 10 Mar 2013
Posts: 513
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

Re: On January 1, 2076, Lake Loser contains x liters of water. B
[#permalink]
Show Tags
09 Dec 2015, 08:32
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...
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



Manager
Joined: 09 Aug 2016
Posts: 66

Re: On January 1, 2076, Lake Loser contains x liters of water. B
[#permalink]
Show Tags
11 Jan 2017, 11:52
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



NonHuman User
Joined: 09 Sep 2013
Posts: 8791

Re: On January 1, 2076, Lake Loser contains x liters of water. B
[#permalink]
Show Tags
28 Mar 2018, 16:50
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: On January 1, 2076, Lake Loser contains x liters of water. B &nbs
[#permalink]
28 Mar 2018, 16:50






