Last Updated on May 9, 2023
GMAT OFFICIAL GUIDE DS
Solution:
We are given that the rent collected in a building was x percent more in 1998 than it was in 1997 and y percent less in 1999 than it was in was in 1998. Let’s start by defining some variables.
a = the annual rent collected in 1997
b = the annual rent collected in 1998
c = the annual rent collected in 1999
We can now create the following equations, using the “percent greater than” and “percent less than” formulas:
b = [(100+x)/100]a
c = [(100-y)/100]b
We need to determine whether the annual rent collected by the corporation was more in 1999 than in 1997. Thus, we need to determine: Is c > a?
Since b = [(100+x)/100]a and c = [(100-y)/100]b, so c = [(100-y)/100][(100+x)/100]a.
Now we can rephrase the question as:
Is [(100-y)/100][(100+x)/100]a > a?
Notice if we divide the entire inequality by a, the a’s cancel out on both sides.
Is [(100-y)/100][(100+x)/100] > 1?
Is (100-y)(100+x)/10,000 > 1?
Is (100+x)(100-y) > 10,000 ?
Is 10,000 – 100y + 100x – xy > 10,000 ?
Is -100y + 100x – xy > 0 ?
Is 100 x – 100y > xy ?
Is 100(x – y) > xy ?
Statement One Alone:
x > y
Knowing only that x is greater than y is not enough to determine whether 100(x – y) > xy. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.
Statement Two Alone:
(xy/100) < (x-y)
Multiplying both sides of the inequality by 100, we have:
xy <100(x – y)
xy < 100(x – y) is exactly the same as saying 100(x – y) > xy. Statement two alone is sufficient to answer the question.
Answer: B
