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# Between 1980 and 1985, Pierre’s investment portfolio

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Between 1980 and 1985, Pierre’s investment portfolio increased in value by x%. Between 1985 and 1990, the portfolio increased in value by y%. Since 1990, the portfolio has decreased in value by z%. If x, y, and z are all positive integers, is the portfolio currently worth more than it was in 1980?

(1) x + y > z

(2) y - x > z

Answer :: (B) Only Statement 2 is alone sufficient.
commented Jan 12, 2015 by Beginner (13 points)

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If between 1980 and 1985, Piere's portfolio increased by x%.  That means that in 1985 Pierre's portfolio = P(1+x/100)

(with P being his original portfolio).  It then increases by y%, so after 1990 it is P(1+x/100)(1+y/100)

Since then it has decreased by z%, so it is now P(1+x/100)(1+y/100)(1-z/100)

and we want to know if P(1+x/100)(1+y/100)(1-z/100) > P

we can assume that P is a posivive value since it is money in a portfolio, so essentially we want to know if:

(1+x/100)(1+y/100)(1-z/100)>1

(1) tells us that x + y > z

This doesn't really tell us anything, because x=1 y=80 and z=80 and then you get

(101/100)(180/100)(20/100)=(101/100)(9/5)(1/5)=909/2500 which is less than one

Or you could have x=1, y=80 and z=1, which gives you

(101/100)(180/100)(99/100)=89991/50000 which is greater than one

(2) tells us that y - x > z

This also does not tell us enough, because we could have y=80 x=1 and z=75, then we get

(101/100)(180/100)(25/100)=(101/100)(9/5)(1/4)=909/2000 which is less than one

Or we could have y=80, x=1 and z=2, which give us

(101/100)(9/5)(49/50)=44541/25000 which is greater than 1

If we use them both together, we still can't determine if Pierre's portfolio was greater in 1980 or today

The scenario where y=80 x=1 and z=75 works for both and is less than one

The scenario where y=80, x=1 and z=2 works for both and is more than one.

Thus, the answer is E

Hope this helps!

Best,

Eliza, GMAT Tutor

answered Jan 14, 2015 by Associate (125 points)
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