## FDP Word Bank #4 Lake Loser

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guest612

### FDP Word Bank #4 Lake Loser

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?

Isn't there an easier way to do this than the solution stated below? The numbers get messy. I can't imagine this is the best way, is it?

Answer: 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 Ã— (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) Ã— (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) Ã— (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) Ã— (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) Ã— (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.
rohit801

### alt approach

this is what i can think. this is akin to an interest rate problem where the money increases annually at a given rate [here is it a negative rate]. the regular formula is:
for simple interest...compounded per yr etc

A = P [1 + rate]^n..where A is the amount [or the accumulated result after n years]. not going too deep here as u might already know this. P= original amount

so, i looked at the rate of decrease....2/7th - i estimtes it to 30% decrease as 21 divide nice with 7 and the number is close enough to approximate.
so, here A = P/4 and we need to find N [number of years it'll take for the decrease]

so, P/4 = P [1 -30/100] ^n => .25 = .7 ^n. basically - how many times do we need to multiply 0.7 to get less than .25- it comes out to 4.

need an expoer Gmat person to comment on this

thanks
guest612

### interesting

very interesting. thanks for your input. i tried doing it out. really appreciate it. waiting and hoping gmat person will respond! :)
StaceyKoprince
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Rohit, really like your reasoning. I agree that our explanation, as written out, is way too computation-intensive.

guest612, please remember to post the entire text of the problem, including answer choices. The answer choices can make a major difference in determining the best way to do a problem! (Eg, can I estimate? etc.)
Stacey Koprince
Instructor
ManhattanPrep
guest612

### ok!

thank you both for your input! :)
guest612

the answer choices were as follows:
A. 2077
B. 2078
C. 2079
D. 2080
E. 2081
They are close together so I suppose it would be hard. But I welcome any thoughts of eliminating any of the first or last answer choices. Answer is C. 2079. (FDP Manhattan GMAT word bank #4)
guest612

Sorry! Answer is D (not C).
StaceyKoprince
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Estimation is not too bad for the first couple. If it has x liters at the beginning of 76 and 2/7 evaporates at the 2nd of 76, that's still more than half left. If another 2/7 (of what's left) evaporates by end of 77, that's till not going to be enough to get me down to 1/4 of my original. If you think about it, you can generally see that the first couple of years won't be enough time (because, each time, a smaller volume is evaporating, since it's always only 2/7 of what's left).

I'll tell you something that should you relieve your mind about the test (if not our materials): the test wouldn't give you the question in quite this way. Either the numbers would be easier to work with and the answer choices would be close together OR the numbers would be hard just like this but the answer choices would be further apart.

Let's assume we had the latter scenario: what I'd really want to do here is pick an answer choice and work backwards.

If C 2079 is the answer, and every year I have 5/7 left at the end of the year (compared to what I started with) 5/7 is approximately 70%. So I lose 30% over the year and have 70% left, each year. Let's say I have 7 (gallons?) at year end (YE) 79. Then I finished 78 / started 79 with 10 gallons. If YE78 was 10 gallons (or 70%), then I finished 77 / started 78 with about 14 gallons (just estimate here). If YE77 was 14 gallons (or 70%), then I finished 76 / started 77 with about 20 gallons. If YE76 was 20 gallons, then I finished 75 / started 76 with about 28 gallons.

(how am I doing that math by the way? Each time, divide by 7 - very roughly - then add that number three times to your starting point. eg, for 10: 10/7 is about 1.3 or 1.4. 10 + 1.3 + 1.3 + 1.3 is about 14. etc.)

So I started with 28 and ended with 7, which is about 1/4. Now, in this case, that's not close enough to tell us the answer definitively (technically, we want LESS THAN 1/4 so this isn't quite enough time - except I was estimating...)

But on the real test, when your years would be further apart with this kind of math, this would be close enough to tell.
Stacey Koprince
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sonygmat
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### Re: FDP Word Bank #4 Lake Loser

We can see that this is an exponential progression:

X, (5/7)*X, (5/7)^2*X, etc...

Therefore we know that An=X*(5/7)^n, where n>=0.

We want An=X/4, so: X/4=X*(5/7)^n which is 1/4=(5/7)^n

so we start by testing numbers and evaluating using benchmark values:

n=1: 5/7 not equal to 1/4, its obvious
n=2: 25/49 approximately 1/2
n=3 125/(49*7)=125/(50*7)=125/350 approximately 1/3
n=4 625/(49*49)=625/(50*50)=625/2500 which is approximately 1/4.

So it needs 4 years.
jnelson0612
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### Re: FDP Word Bank #4 Lake Loser

Looks good, sonygmat! To all who may be reading this, keep in mind that it needs year 2076, 2077, 2078, and 2079 to achieve the desired level, which it does in year 2079.
Jamie Nelson
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### Re: FDP Word Bank #4 Lake Loser

jamie above^^ , how can it be 2079?

year 1: 5/7 left
year 2: 25/49 left
year 3: 125/343 left
year 4: 625/2401 left (26.01%)
year 5: 3125/16807 left

in year 4 there is still more than 1/4 left even at the end of the year, so the answer should be year 5 that it breaks through to 1/4 left... which would be 2081... is that GMAT really asking us to gestimate as the answer above with 625/2500 states? 343*7 is not a difficult arithmetic calculation...
tim
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### Re: FDP Word Bank #4 Lake Loser

I think it was confirmed above that the answer is 2080. I think Jamie just missed a year.
Tim Sanders
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https://www.manhattanprep.com/gmat/forums/a-few-tips-t31405.html
Atul_manhattan
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### Re: FDP Word Bank #4 Lake Loser

If one takes approximation, one is mistaken. Answers are so close enough that the approximation will give wrong answer. So, the easy and flawless method would be to take x as 4 and solve. Answer is 2080.
Sage Pearce-Higgins
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### Re: FDP Word Bank #4 Lake Loser

Atul, I agree with you. Not only are the answer choices close together, but approximating with a repeated change is risky - the approximation error is quickly magnified and becomes more significant. Comparing the precise change (reduces by 2/7) with the approximation suggested by Rohit above (reduces by 30%) shows that they give different answers.