2024 AMC 10A Problems/Problem 5

Problem

Andrea is taking a series of several exams. If Andrea earns $61$ points on her next exam, her average score will decrease by $3$ points. If she instead earns $93$ points on her next exam, her average score will increase by $1$ point. How many points should Andrea earn on her next exam to keep her average score constant?

$\textbf{(A)}~80 \qquad\textbf{(B)}~82 \qquad\textbf{(C)}~83 \qquad\textbf{(D)}~85 \qquad\textbf{(E)}~86$

Solution 1

Let Andrea's current average be $a$ , and assume there have been $n$ tests already. The $(n + 1)$ th test must have a score of $a + k(n + 1)$ in order to increase her average by $k$ and $a - k(n + 1)$ in order to decrease her average by $k$ . Hence, we see that the average must be $\tfrac{3}{4}$ of the way from $61$ to $93$ , which is $\boxed{\textbf{(D)}~85}$ .

~joshualiu315

Solution 2

Let $s$ be the sum of Andrea's scores currently and let $n$ be the number of tests. We have $\begin{array}{r} \frac{s + 61}{n + 1} = \frac{s}{n} - 3 \\ \frac{s + 93}{n + 1} = \frac{s}{n} + 1 \end{array}$ Equate $s$ to get a quadratic in $n$ to find $n = 7$ , $s = 595$ so $\frac{595}{7} = \boxed{\textbf{(D)}~85}$ .

~eg4334

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2024 AMC 10A Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

See also