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On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]
18 Aug 2009, 13:51
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A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
42% (03:20) correct
58% (02:03) wrong based on 312 sessions
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: Compounded interest [#permalink]
18 Aug 2009, 14:50
1
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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 _________________
Re: Compounded interest [#permalink]
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.
Re: Compounded interest [#permalink]
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?
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 \((\frac{5}{7})^n*x<\frac{x}{4}\).
Re: Compounded interest [#permalink]
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
Re: Compounded interest [#permalink]
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. _________________
On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]
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 \((\frac{5}{7})^n*x<\frac{x}{4}\).
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]
01 Sep 2014, 00:30
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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 \((\frac{5}{7})^n*x<\frac{x}{4}\).
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]
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?
Re: On January 1, 2076, Lake Loser contains x liters of water. B [#permalink]
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... _________________
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