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Between 1980 and 1985, Pierre’s investment portfolio

Expert replies
by guerrero » Tue Sep 10, 2013 8:40 pm
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

OAE
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Source: — Data Sufficiency |

by sanju09 » Wed Sep 11, 2013 1:39 am
guerrero wrote: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

OAE
It's a YES/NO DS, a perfect opportunity to keep plugging away. Let's keep giving x, y, and z some smooth looking numbers that go with statement(s) in hand.

I. x + y > z. Even 10 + 20 > 25.

A 10% rise in 100 takes it to 110, a 20% rise in 110 takes it to 132, and a 25% fall in 132 takes it directly to 99; thus answering the target question as NO.

But we also have 50 + 20 > 25.

A 50% rise in 100 takes it to 150, a 20% rise in 150 takes it to 180, and a 25% fall in 180 takes it directly to 135; thus answering the target question as YES.

Insufficient

Get rid of A,D.

II. y - x > z. Now we can't recycle, 10, 20, 25 but we can recycle 50, 20, 25 for sure, which gave us a YES in some past we don't know.

Shall we take 20 - 10 > 5?

A 10% rise in 100 takes it to 110, a 20% rise in 110 takes it to 132, and a 5% fall in 132 still keeps it well over 100. Hence, again a YES.

[spoiler]A consistent YES; hence sufficient


Pick B

[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
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by ani781 » Wed Sep 11, 2013 4:27 am
Hi Sanju,
I am very much impressed by the way you solve these problems. Can you pls tell me how to decide between plugging in numbers and what all numbers.
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by ani781 » Wed Sep 11, 2013 7:21 am
Experts , Is there any algebraic approach to this problem , or the fastest method is the method of pluggin. This would be of immense help, as I keep on approaching these problems in the algebraic way and end up messing things up, especially in these types of Percent DS questions.
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by Java_85 » Wed Sep 11, 2013 8:50 am
IMO also E. it tool me 5 mins.
BTW, Nice Question, it was really a tricky one.
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by sanju09 » Wed Sep 11, 2013 10:31 pm
ani781 wrote:Experts , Is there any algebraic approach to this problem , or the fastest method is the method of pluggin. This would be of immense help, as I keep on approaching these problems in the algebraic way and end up messing things up, especially in these types of Percent DS questions.


Hi ani,

As per my experience on GMAT so far, I found numbers more helpful than variables while attempting questions, and found variables more beautiful than numbers while explaining solutions to the person at remote. But this question is just like crying for numbers only. When to plug in? Whenever you have a recommend to set equation(s) in some sweet variable(s) with a feel that the question is very hard. Sort of things, but happy you liked that.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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by bigb62387 » Thu Sep 19, 2013 8:17 am
Has this question been officially solved? IMO, the answer is E.

Algebra
Let's assume portfolio value is P, X% increase, Y% increase, Z% decrease

From 1985 -> 1990 the value went from P ---> P*(1+(x/100))
From 1990 -> 1995 the value went from P*(1+(x/100)) ---> P*(1+(x/100))*(1+(y/100))
From 1995 -> ???? the value went from P*(1+(x/100))*(1+(y/100)) ---> P*(1+(x/100))*(1+(y/100))*(1-z/100))

Target question asks: is P*(1+(x/100))*(1+(y/100))*(1-z/100)) > P ?
Divide by P on both sides: is (1+(x/100))*(1+(y/100))*(1-z/100)) > 1 ?

You could keep working with the algebra to simplify if you feel comfortable, but the math can get tricky if you're trying to keep it under 2 minutes. I would suggest looking at the equation above and testing MAXIMUM numbers and visualize/approximate to keep it around 2 minutes.

STATEMENT 1
X + Y > Z
Case A: make these far apart: X=Y=100 and Z=1. If you start at $100, you double to $200, then double to $400, and take a small hit (1%) and up around ~$400 > $100. Target answer: YES
Case B: make these close together X=Y=100 and Z=199. If you start at $100, you double to $200, then double to $400, and then take a massive hit (199%) and end up negative ~(-$400) < $100. Target answer: NO
Two different target answers INSUFFICIENT

STATEMENT 2
Y - X > Z
Personally, I would rearrange to Y > X + Z but that's personal preference.
Case A: make these far apart: Y=100 and X=Z=1. If you start at $100, add 1% and then double it you're at $202 and take a small hit (1%) you end up ~$200 > $100. Target answer: YES
Case B: make these close together X=1, Y=100, Z=98. If you start at $100, add 1% and then double it you're at $202 and take a massive hit (98%) you end up slightly more than ~$0 < $100. Target answer: NO
Two different target answers INSUFFICIENT

STATEMENTS 1 & 2
Combining X + Y + Y - X > Z + Z reduces to 2Y > 2Z so Y > Z and X does not matter. Since X does not matter, let's just set it to 0 so that we know 1980=1985 and there is no change at first.
Case A: make these far apart: Y=100 and Z=1. If you start at $100 and then double it you're at $200 and take a small hit (1%) you end up ~$200 > $100. Target answer: YES
Case B: make these close together Y=100 and Z=99. If you start at $100 and then double it you're at $200 and take a massive hit (99%) you end up ~$0 < $100. Target answer: NO
Two different target answers INSUFFICIENT

Therefore, both statements together are INSUFFICIENT, I would choose E.

Those calculations would probably take more than 2 minutes, but you get the idea of choosing extremes (1%, 99%, 100%) and then approximating them.
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