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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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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Bunuel: Can you please share some thought on how to come up with such numbers for plugging in.

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.

0

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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