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Between 1980 and 1985, Pierre’s investment portfolio increas
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14 Jan 2013, 06:56
Question Stats:
30% (02:58) correct 70% (02:41) wrong based on 564 sessions
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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
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14 Jan 2013, 07:39
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
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01 Jun 2013, 09:33
can someone explain me the difference in reasoning between the question above and this question?:
gmatclub. com/forum/theannualrentcollectedbyacorporationfromacertain89184.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 < xy
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
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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/theannualrentcollectedbyacorporationfromacertain89184.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 < xy
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
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02 Jun 2013, 05:14
My approach:
Simple picknumbers.
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
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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.



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Re: Between 1980 and 1985, Pierre’s investment portfolio increas
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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
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04 Jun 2014, 19:22
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.



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Between 1980 and 1985, Pierre’s investment portfolio increas
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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
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25 Oct 2014, 05:09
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) yx > 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
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11 Feb 2019, 10:19
Evaluate Statement (1), which states that x + y > z. Plug In values that satisfy this statement and determine the answer to the question. Plug In a starting portfolio value of $200 and values of x = 50, y = 50, and z = 10. These values satisfy Statement (1), because 50 + 50 > 10. Using these numbers the value of the portfolio would first increase 50% to $300, then increase another 50% to $450, then decrease by 10% to $405. Since $405 is greater than $200 the portfolio would be more than it was in 1980. The answer to the question is "Yes". Now Plug In different values to try to get an answer of "No." Plug In a starting portfolio value of $200 and values of x = 100, y = 100, and z = 100. These values satisfy Statement (1), because 100 + 100 > 100. The portfolio would first increase 100% to $400, then increase another 100% to $800, then decrease by 100% to have a final value of $0. Since $0 is less than $200 the portfolio would be worth less than it was in 1980. Now, the answer to the question is "No". When different numbers that satisfy a statement yield different answers to the question, the statement is insufficient. Write down BCE. Now, evaluate Statement (2), which states that y − x > z. Plug In values that satisfy this statement and determine the answer to the question. Plug In a starting portfolio value of $200 and values of x = 10, y = 50, and z = 10. These values satisfy Statement (2), because 50 – 10 > 10. Using these numbers the value of the portfolio would first increase 10% to $220, then increase another 50% to $330, then decrease by 10% to $297. Since $297 is greater than $200 the portfolio would be worth more than it was in 1980. The answer to the question is "Yes". Now Plug In different values to try to get an answer of "No." Plug In a starting portfolio value of $200 and values of x = 10, y = 200, and z = 100. These values satisfy Statement (2), because 200 – 10 > 100. Using these values the portfolio would first increase 10% to $220, then increase another 200% to $660, then decrease by 100% to have a final value of $0. Since $0 is less than $200 the value of the portfolio is worth less than it was in 1980. The answer to the question is "No". When different numbers that satisfy a statement yield different answers to the question, the statement is insufficient. Eliminate B. Now, evaluate Statement (1) and Statement (2) together. Plug In values that satisfy both of the statements at once and determine the answer to the question. The values used to evaluate Statement (2) also satisfy Statement (1), so both answers "Yes" and "No" are possible. Therefore, Statements (1) and (2) together are insufficient. Eliminate C. The correct answer is choice E.
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas
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15 Feb 2019, 12:04
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 elaborate on the (1)+(2) case Bunuel? Thank you!



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Re: Between 1980 and 1985, Pierre’s investment portfolio increas
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16 Feb 2019, 05:34
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. Hi Even for C (1 & 2 combined), We need to consider values of X, Y & Z that need to satisfy both the cases. In case of bunuel's example, he has considered \(x=1\), \(y=100\), and \(z=1\). which satisfy both 1 & 2. And as we have seen in case 1 that eqn. \((1+\frac{x}{100})(1+\frac{y}{100})(1\frac{z}{100})\)can be both less than or greater than 1. 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 \((1+\frac{x}{100})(1+\frac{y}{100})(1\frac{z}{100})=1.01*2*0.1<1\). So in case of C too the answer is Not sufficient.
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Re: Between 1980 and 1985, Pierre’s investment portfolio increas
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