The annual rent collected by a corporation from a certain building was

\(r<r*(1+\frac{x}{100})*(1-\frac{y}{100})\), \(r\) cancels out. Then \((1+a)(1-b)=1-b+a-ab\).

\(1<1-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}\), 1 cancels out --> \(0<-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}\), multiplu by 100 both sides --> \(0<-y+x-\frac{xy}{100}\), rearrange -->\(x-y>\frac{xy}{100}\).

Hope it's clear.

You can use the formula in the question in a very straightforward manner...
Rent charged in 1998 was x% more
Rent charged in 1999 was y % less
So overall % change in rent charged was x - y - xy/100 (since y is a decrease use (-y) in place of y)
Question: Was rent collected in 1999 > 1997 i.e. was overall percentage of change positive?
Was x - y - xy/100 > 0 ?

(1) x > y
Not enough info

(2) xy/100 *< x – y
which is basically just x - y - xy/100 > 0 when you re-arrange.
So this statement gives you 'Yes' immediately. Sufficient.

Though, make sure that you know how the formula was derived and the basic concept behind it... Its good to know short cut formulas but they are not applicable everywhere... in some questions you might need to use ingenuity...

I do not recommend using algebra to evaluate condition #2--it's too complicated for most. Here is a visual that should help, showing you that you can evaluate condition #2 by plugging in to determine the "boundary line" (the value that separates the "Yes" answers from the "No" answers).

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, that means

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, we have:

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

Video solution from Quant Reasoning starts at 23:03 here:

VeritasKarishma -- why can we cancel out the r? We dont necessarily know it is not equal to zero.

r is the rent collected in 1997. It cannot be 0. If it were, how would you collect x% more in 1998? x% of 0 is still 0 so rent collected in 1998 will again be 0 + 0 = 0 (which is not more than rent collected in 1997).
In such real world questions, the base is not 0.

Hi GMATters,

Here is my video solution to this question:



Best,

Rowan

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