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

How do you decide what value of x should be taken to make your calculations easier going down the line? Please advice.

Hi,

To solve this question I wouldn't really be looking for a value of x.

If you carefully analyse the question, you will notice that 2/7 of x will keep on evaporating every year; so at the end of year only 5/7 of x will remain.

By the end of 2076, the water will be 5/7 of x.
By the end of 2077, the water will be 5/7 x 5/7 of x
By the end of 2078, the water will be 5/7 x 5/7 x 5/7 of x and so on.

Now he asks you, in which year will the water be less than 1/4 of x.

So basically the question becomes that how many times will you have to multiply 5/7 to itself so that it becomes < 1/4

OR

4 x (5/7)^y < 1 where y is the power of 5/7. Solve for y and that should be the answer.

For this to happen, y will have to take the value of 5 so that 4 x (5/7)^5 = 12500 / 16807

Since the water 2/7th of x will evaporate in 2076, the answer will be 2080, when four other such incidences of evaporation will take place.

My answer: D

As far as pickign the best number with fraction problems here's is a rule of thumb that will work almost always.

Choose the lowest common multiple of denominators of the fractions involved. For example in thsi question if I had choosen 28 I would have got the answer D)

1/4 of 28 would be 7 so find the year in which water becomes < 7

2076 = 2/7*28 = 8 (28-8 = 20)
2077 = 2/7*20 = 5.7
2078 = 2/7*14.3) = 4.08 ~= 4
2079 = 2/7*10(2.8)
2080 = 2/7*7.2 = 2.02 so D)


The solution picks a different easy number so if you cant think of this magic number just go with the strategy above.

Pooja's solution is good too for this problem(another way of approaching the problem)!

yeah, this is a good approach: use estimation if the numbers get too ugly for actual computation.

it's somewhat rare for a real gmat problem to feature computations that are this ugly, but it's not completely unheard of; there have been gmat problems that have featured, for instance, conversions of ft/sec to mi/hr and vice versa. those are nasty computations, requiring the use of fractions like 3600/5280 or its reciprocal. yuk.

the most important point is this: DO NOT DELIBERATE. in other words, do not sit there staring at the problem, wondering whether you're going to be able to complete the computations in time; just start doing the computations. you'll be glad you did when you finish the problem within the allotted time.