# 2017 AMC 8 Problems/Problem 14

## Problem 14

Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers? $\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$

## Solution 1

Let the number of questions that they solved alone be $x$. Let the percentage of problems they correctly solve together be $a$%. As given, $$\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}$$.

Hence, $a = 96$.

Zoe got $\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}$ problems right out of $2x$. Therefore, Zoe got $\frac{\frac{186x}{100}}{2x} = \frac{93}{100} = \boxed{\textbf{(C) } 93}$ percent of the problems correct.

## Solution 2

Assume the total amount of problems is $100$ per half homework assignment, since we are dealing with percentages, and not values. Then, we know that Chloe got $80$ problems correct by herself, and got $176$ problems correct overall. We also know that Zoe had $90$ problems she did correct alone. We can see that the total amount of correct problems Chloe had when Zoe and she did the homework together is $176-80=96$, which is the total amount of problems she got correct, subtracted by the number of correct problems she did alone. Therefore Zoe has $96+90=186$ problems out of $200$ problems correct. This is $\boxed{\textbf{(C) } 93}$ percent.

## Solution 3

In the problem, we can see that Chloe solved 80% of the problems she solved alone, but 88% of her answers are correct. If 80 and another number's average is 88, the other number must be 96. Then Zoe solved 90% of the problems she did alone, but 96% of her answers were correct. Then the average of 90 and 96 is $\boxed{\textbf{(C) } 93}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 