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Between 1980 and 1985, Pierre’s investment portfolio increas

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Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] New post 14 Jan 2013, 05:56
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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
[Reveal] Spoiler: OA

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Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] New post 14 Jan 2013, 06:39
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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.

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Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] New post 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] New post 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] New post 02 Jun 2013, 04:14
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] New post 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] New post 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?
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas [#permalink] New post 04 Jun 2014, 18:22
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.
Re: Between 1980 and 1985, Pierre’s investment portfolio increas   [#permalink] 04 Jun 2014, 18:22
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