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

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Difficulty:

85% (hard)

Question Stats:

29% (03:17) correct
71% (02:05) wrong based on 341 sessions

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?

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

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Expert's post

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}).

(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.

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?

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.

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

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

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

(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?