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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]

Show Tags

18 Aug 2009, 14:51

2

This post received KUDOS

14

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

41% (02:03) correct
59% (01:54) wrong based on 513 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?

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
_________________

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.

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?

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

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

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.
_________________

On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]

Show Tags

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

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?

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

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?

Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]

Show Tags

16 Nov 2014, 18:01

1

This post received 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?

Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]

Show Tags

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...
_________________

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 !

Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]

Show Tags

10 Jan 2017, 21:51

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.
_________________

Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]

Show Tags

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