let the rent = abeater wrote:The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y
(2) 100xy < x – y
Could you please help me break down (2)
My answer is A, by plugging in numbers:beater wrote:The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y
(2) 100xy < x – y
Could you please help me break down (2)
To calculate %change more than 1 time (say 2 times) over an amount, we use the following formula.beater wrote:The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y
(2) 100xy < x – y
Could you please help me break down (2)
With that correction: To increase something by x%, you can multiply by (1 + x/100). To decrease something by y%, you can multiply by (1 - y/100).beater wrote: Guys, I messed up the question. There is a typo.
(2) is xy/100 < x-y NOT 100xy < x-y.
Sorry..Sorry for the mess!
Yes, if x = 60 and y = 40, the rent in 1999 is less than in 1997. But if x = 60, and y = 10, the rent is higher in 1999.Vemuri wrote:Why is Statement 1 not sufficient? The question is asking if the rent collected in 1999 more than 1997. From the solutions discussed & from what Ian had to say, we can definitely answer the question.
If x=60 & y=40, then
Rent collected in 1998 = 1.6*Rent collected in 1997
Rent collected in 1999 = 1.6*0.6*Rent collected in 1997
Rent collected in 1999 = 0.96*Rent collected in 1997
Hence, rent collected in 1999 is less than rent collected in 1997. This is sufficient to answer the question (with a "No").
Oh boy !!! Ian please lend me your brilliant brian on the day of my exam.Ian Stewart wrote: Yes, if x = 60 and y = 40, the rent in 1999 is less than in 1997. But if x = 60, and y = 10, the rent is higher in 1999.
You will do me a great favor.Ian Stewart wrote:With that correction: To increase something by x%, you can multiply by (1 + x/100). To decrease something by y%, you can multiply by (1 - y/100).beater wrote: Guys, I messed up the question. There is a typo.
(2) is xy/100 < x-y NOT 100xy < x-y.
Sorry..Sorry for the mess!
Here, the rent (let's call it R) increases by x%, then decreases by y%. So our new rent is just:
R(1 + x/100)(1 - y/100) = R(1 + x/100 - y/100 - xy/10,000)
We want to know if
R(1 + x/100 - y/100 - xy/10,000) > R
or equivalently, dividing by R (which is fine, because R must be positive) if
1 + x/100 - y/100 - xy/10,000 > 1
Subtracting 1 from both sides, and multiplying by 100 on both sides, the question becomes whether
x - y - xy/100 > 0
or if xy/100 < x - y. Since Statement 2 tells us exactly that, the answer to the question must be yes if we use Statement 2, and it's sufficient. Since Statement 1 is insufficient (try x = 60, y = 40 to see that), the answer is B.
You need to understand perfectly just what you're doing when you plug numbers into a DS question. If you plug in numbers, of course you are going to get *some* answer to the question. If, when evaluating Statement 1, you plug in x=60 and y=40 here, you find that the rent is lower in 1999. *All* you have proven is that the rent is *sometimes* less in 1999 than in 1997, nothing more. You have not proven that the rent is *always* less in 1999 than in 1997, which is what you would need to do to prove Statement 1 is sufficient.siddhans wrote:
If we substitute X=60 and Y=40 in x - y - xy/100 > 0
then we get
20- 24 which is not > 0 but <0
so doesnt it give us a definite answer than rent collected by corporation in 1999 is not more than rent collected in 1997?? So isnt A sufficient too?
