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

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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?

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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 = \((1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})\).

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

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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]

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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.

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 = \((1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})\).

(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?

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

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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 = \((1+\frac{x}{100})(1+\frac{y}{100})(1-\frac{z}{100})\).

(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 !

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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04 Nov 2015, 08:51

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

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03 Dec 2016, 02:46

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