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.