At the beginning of a given year, Sarah and Olivia each invested in two different mutual funds at simple annual interest rates of x\% and y\% , respectively. Olivia invested p pounds, while Sarah invested 2,000 pounds less. By the end of the year, both earned the same total interest from their respective investments. Select for p an expression for Olivia’s investment p in terms of x and y, and select for x an expression for Sarah’s interest rate x in terms of p and y . Make exactly two selections, one from each column.

Official Solution: Sarah's investment of p-2,000 pounds in a year at at simple annual interest rates of x\% would earn (p-2,000)*\frac{x}{100} pounds Olivia's investment of p pounds in a year at at simple annual interest rates of y\% would earn p*\frac{y}{100} pounds Given that those earnings are equal, we'd have (p-2,000)*\frac{x}{100} =p*\frac{y}{100} (p-2,000)x = py From this, for the second column, we get x = \frac{py}{p-2,000} . For the first column we get (p-2,000)x = py px-2,000x = py px-py=2,000x p(x-y)=2,000x p=\frac{2,000x}{x-y} Correct answer: p " \frac{2,000x}{x-y} " x " \frac{py}{p-2,000} " GMAT-Club-Forum-4gm29iaj.png

1. Olivia invested p pounds at an interest rate of y%, while Sarah invested p - 2000 pounds at an interest rate of x%. 2. Since both of them earned the same total interest, we can write this as: p * y\% = (p - 2000) * x\% . 3. Let's derive p from this equation. p * y\% = (p - 2000) * x\% \rightarrow \frac{y\%}{x\%} = \frac{p - 2000}{p} \rightarrow \frac{y}{x} = 1 - \frac{2000}{p} \rightarrow \frac{2000}{p} = \frac{x - y}{x} \rightarrow p = \frac{2000x}{x - y} . 4. Now let's also derive x too. p * y\% = (p - 2000) * x\% \rightarrow \frac{p * y\%}{p - 2000} = x\% \rightarrow x = \frac{py}{p - 2000} . 5. Our answer will be: p - \frac{2000x}{x - y} and x - \frac{py}{p - 2000} . ------------------------------------------------- There is another solution by analyzing the answer choices. The first three options are equations that don't include p but have x - that means they can only be equal to p. The other three options are equations that don't include x but have p - that means they can only be equal to x. We can assume for each of the first three options that they equal p and try to derive x and see if the new equation is an answer choice: - \frac{2000x}{x - y} = p \rightarrow px - py = 2000x \rightarrow x = \frac{py}{p - 2000} , which is in the options. - \frac{2000y}{x - y} = p \rightarrow px - py = 2000y \rightarrow x = \frac{(p + 2000)y}{p} , which isn't in the options. - \frac{200x}{x - y} = p \rightarrow px - py = 200x \rightarrow x = \frac{py}{p - 200} , which isn't in the options. So, our answer would be: p - \frac{2000x}{x - y} and x - \frac{py}{p - 2000} .