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Between 1980 and 1985, Pierre’s investment portfolio increas

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Manager
Status: Never ever give up on yourself.Period.
Joined: 23 Aug 2012
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Location: India
Concentration: Finance, Human Resources
GMAT 1: 570 Q47 V21
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WE: Information Technology (Investment Banking)
Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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14 Jan 2013, 06:56
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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

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

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14 Jan 2013, 07:39
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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?

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. _________________ General Discussion Intern Joined: 27 Mar 2013 Posts: 4 GMAT 1: 480 Q32 V23 Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] Show Tags 01 Jun 2013, 09:33 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? VP Status: Far, far away! Joined: 02 Sep 2012 Posts: 1088 Location: Italy Concentration: Finance, Entrepreneurship GPA: 3.8 Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] Show Tags 01 Jun 2013, 11:20 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 _________________ It is beyond a doubt that all our knowledge that begins with experience. Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b] Intern Joined: 19 Apr 2012 Posts: 25 Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] Show Tags 02 Jun 2013, 05:14 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. Intern Joined: 21 Mar 2013 Posts: 39 GMAT Date: 03-20-2014 Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] Show Tags 18 Mar 2014, 03:43 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. Manager Joined: 20 Dec 2013 Posts: 245 Location: India Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] Show Tags 30 Mar 2014, 01:00 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?
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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04 Jun 2014, 19:22
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.
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Between 1980 and 1985, Pierre’s investment portfolio increas  [#permalink]

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25 Oct 2014, 04:37
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.

Answer: E.

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, 05:09
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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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