Between 1980 and 1985, Pierre’s investment portfolio increas

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

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.

Same approach for the second statement.

Clearly E.

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

I just want to know how you come up with these numbers,Bunuel?Is there any rule of thumb?

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.

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 !

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.

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.

Can you please elaborate on the (1)+(2) case Bunuel?

Thank you!

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.

%age as mentioned.
Pb Pa
1980 --> +x% --> 1985 --> +y% --> 1990 --> -z%
is Pa > Pb?
Assume Pb = 100
(1) Plug in test values
50 + 1 > 49 -> Pb > Pa
25 + 25 > 2 -> Pb < Pa -- (Insufficient)

(2)Same scenario as A -- (Insufficient)

Combining 1 & 2 also does give both scenarios. Ans. E

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