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• ### $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### FREE Quant Workshop by e-GMAT! December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. # Between 1980 and 1985, Pierre’s investment portfolio increas  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Status: Never ever give up on yourself.Period. Joined: 23 Aug 2012 Posts: 137 Location: India Concentration: Finance, Human Resources GMAT 1: 570 Q47 V21 GMAT 2: 690 Q50 V33 GPA: 3.5 WE: Information Technology (Investment Banking) Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] ### Show Tags 14 Jan 2013, 05:56 3 17 00:00 Difficulty: 95% (hard) Question Stats: 31% (02:58) correct 69% (02:40) wrong based on 769 sessions ### HideShow timer Statistics 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 _________________ Don't give up on yourself ever. Period. Beat it, no one wants to be defeated (My journey from 570 to 690) : http://gmatclub.com/forum/beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] ### Show Tags 14 Jan 2013, 06:39 3 6 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? Say the value of the portfolio in 1980 was$1. then:

Price in 1980 = 1;
Price in 1985 = $$(1+\frac{x}{100})$$;
Price in 1990 = $$(1+\frac{x}{100})(1+\frac{y}{100})$$;
Price in now = $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})$$.

Question asks whether $$1<(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})$$.

(1) x + y > z. If $$x=1$$, $$y=100$$, and $$z=1$$, then $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})=1.01*2*0.99>1$$ BUT if $$x=1$$, $$y=100$$, and $$z=90$$, then $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})=1.01*2*0.1<1$$. Not sufficient.

(2) y − x > z. Consider the same cases. Not sufficient.

(1)+(2) Consider the same cases. Not sufficient.

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Intern
Joined: 27 Mar 2013
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GMAT 1: 480 Q32 V23
Re: Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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01 Jun 2013, 08:33
can someone explain me the difference in reasoning between the question above and this question?:

gmatclub. com/forum/the-annual-rent-collected-by-a-corporation-from-a-certain-89184.html

the look similar, but the questions meant to be different:

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) xy/100 < x-y

cant we make the reasoning there the same as here, r=1 and solve similiar as above?
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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01 Jun 2013, 10:20
Kyuss wrote:
can someone explain me the difference in reasoning between the question above and this question?:

gmatclub. com/forum/the-annual-rent-collected-by-a-corporation-from-a-certain-89184.html

the look similar, but the questions meant to be different:

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) xy/100 < x-y

cant we make the reasoning there the same as here, r=1 and solve similiar as above?

Yes you can. But whatever value of r you pick will eventually not matter.

I will go directly to the solution here, so we can write the question as:
$$R(1+\frac{x}{100})(1-\frac{y}{100})>R$$, as you see now we can safely divide by R (which is positive) and obtain
$$(1+\frac{x}{100})(1-\frac{y}{100})>1$$. So you can assume $$R=1$$ at the beginning if this makes your calculus easier.

Hope it's clear
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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02 Jun 2013, 04:14
1
My approach:

Simple pick-numbers.

Origin value of the investment portfolio: 100. Increase by 10% and then also by 10% = 100*1,1 = 110
110 * 1,1 = 121. Decrease by 19% ~ 20% = 1/5 = ~ 24 so the total value is below 100.

Same approach for the second statement.

Clearly E.
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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18 Mar 2014, 02:43
Bunuel wrote:
(1) x + y > z. If $$x=1$$, $$y=100$$, and $$z=1$$, then $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})=1.01*2*0.99>1$$ BUT if $$x=1$$, $$y=100$$, and $$z=90$$, then $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})=1.01*2*0.1<1$$. Not sufficient.

(2) y − x > z. Consider the same cases. Not sufficient.

(1)+(2) Consider the same cases. Not sufficient.

Bunuel: Can you please share some thought on how to come up with such numbers for plugging in.
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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30 Mar 2014, 00:00
Bunuel 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?

Say the value of the portfolio in 1980 was $1. then: Price in 1980 = 1; Price in 1985 = $$(1+\frac{x}{100})$$; Price in 1990 = $$(1+\frac{x}{100})(1+\frac{y}{100})$$; Price in now = $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})$$. Question asks whether $$1<(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})$$. (1) x + y > z. If $$x=1$$, $$y=100$$, and $$z=1$$, then $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})=1.01*2*0.99>1$$ BUT if $$x=1$$, $$y=100$$, and $$z=90$$, then $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})=1.01*2*0.1<1$$. Not sufficient. (2) y − x > z. Consider the same cases. Not sufficient. (1)+(2) Consider the same cases. Not sufficient. Answer: E. I just want to know how you come up with these numbers,Bunuel?Is there any rule of thumb? Intern Joined: 01 Jun 2014 Posts: 5 Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] ### Show Tags 04 Jun 2014, 18:22 1 E. Pick the most extreme values of z: z = 1 and z = 100. If z = 1, pick pretty much any large values of x and y to satisfy both statements 1 and 2, and you'll see that the portfolio has net grown. If z = 100, then pick any large values of x and y to satisfy both statements 1 and 2, and the portfolio has become 0. Both cases satisfy statements 1 and 2, but differ in the overall result. Therefore E. Manager Joined: 13 Apr 2014 Posts: 61 Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] ### Show Tags 25 Oct 2014, 03:37 Bunuel 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? Say the value of the portfolio in 1980 was$1. then:

Price in 1980 = 1;
Price in 1985 = $$(1+\frac{x}{100})$$;
Price in 1990 = $$(1+\frac{x}{100})(1+\frac{y}{100})$$;
Price in now = $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})$$.

Question asks whether $$1<(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})$$.

(1) x + y > z. If $$x=1$$, $$y=100$$, and $$z=1$$, then $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})=1.01*2*0.99>1$$ BUT if $$x=1$$, $$y=100$$, and $$z=90$$, then $$(1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})=1.01*2*0.1<1$$. Not sufficient.

(2) y − x > z. Consider the same cases. Not sufficient.

(1)+(2) Consider the same cases. Not sufficient.

Can you please tell how did you choose these numbers ? I mean within two minutes finding these number may be little difficult. So is there are some number which I have to always take care of. Please help !
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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25 Oct 2014, 04:09
0
daviesj 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

E.

The quickest way to do this ques is by plugging in values for x,y,and z

1) x+y > z
Now looking at FS1 we can easily say that the portfolio value would easily be greater than that it was in 1980 (assuming x and y to be very large numbers and z to be really small)
To make the value smaller, we need negative growth...for that x,y,and z should be as close as possible.
let x=y=z=1
compound % growth in first 2 periods = 1+1+(1*1/100) = 2.01
compound % growth in 3rd period = 2.01 - 1 - .201 = .809 which is negative growth.
so insufficient.

2) y-x > z
Again positive growth can be show by assuming x and y to be large and z to be small
for negative growth: x=1, y=100, and z=98

(1)+(2) --> same can be done here.
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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