Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
This problem will take some efficient calculation, knowledge of FDP connections & benchmarks, and smart use of “Smart Numbers.”
Let’s start by picking the number of Easter eggs in 2008 as 100. That number rose 20% over the next year, so in 2009 it was 120. Finally, that number fell 17% in the second year. You can compute this effect exactly by multiplying 120 by 0.83 to get 99.6 (not too hard), or you can recognize that 17% is very slightly more than 16.666…%, which is 1/6. For something to fall by that exact percent (16 and 2/3%) is the same as multiplying it by 5/6. Notice also that increasing something by 20% (1/5) is the same as multiplying it by 6/5. These two effects would perfectly undo each other:
100 + 20% = 120
120 – 16.66…% = 100.
Since you are taking away 17%, you must wind up under 100, but only by a very little bit – say, 0.5, so you could estimate 99.5 as your outcome (notice how close this is to the right answer). On top of that, be sure to recognize how the percent decrease (-17%) is smaller numerically, but it’s off of the larger base, so it actually has more impact than the percent increase of 20%.
Now let’s deal with the E/B ratio. Pick that as 1 to start (so that our number of bunnies starts at 100 as well). Take away 20%, so you have 0.8 as the new ratio. (No need to figure out the intermediate number of bunnies.) Finally, add back 22%. 20% of 0.8 would be 0.16, and 2% would be 0.016, so just add both of those to get 0.8 + 0.16 + 0.016 = 0.976.
Almost done! If E in 2010 is now just under 100 (= 99.6), and E/B fell from 1 to 0.976, what’s the new B? E ÷ (E/B) = B, so you have this:
99.6 / 0.976 = new B
Now you can do the long division, or you can estimate. This result will definitely be bigger than 100, so we can rule out all the decreases. (The fraction would have to be 97.6 / 0.076 or 99.6 / 0.996 to get 100.)
Now, is the increase approximately 2% or 5%? Well, if x is a very small positive number, then this relationship holds: 1/(1 – x) can be approximated by 1 + x. (It turns out that 1/(1 – x) is always bigger, but only by a slight amount when x is small.
So let’s approximate 99.6 as 100 and 0.976 as 0.98:
99.6 / 0.976 ? 100/0.98 = 100(1/0.98). Now pretend that x = 0.02. Approximate:
1/(1 – 0.02) ? 1 + 0.02 = 1.02
Now, substituting back in, the result is 102. (To more decimal places, you get 102.0492 – only just slightly bigger.) This corresponds to a 2% increase.
The correct answer is B.