Difference between revisions of "2024 AMC 10A Problems/Problem 5"
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| − | + | ==Problem== | |
| − | == Problem == | + | Andrea is taking a series of several exams. If Andrea earns <math>61</math> points on her next exam, her average score will decrease by <math>3</math> points. If she instead earns <math>93</math> points on her next exam, her average score will increase by <math>1</math> point. How many points should Andrea earn on her next exam to keep her average score constant? |
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| − | <math>\textbf{(A) } | + | <math>\textbf{(A)}~80 \qquad\textbf{(B)}~82 \qquad\textbf{(C)}~83 \qquad\textbf{(D)}~85 \qquad\textbf{(E)}~86</math> |
| − | == Solution== | + | ==Solution 1== |
| − | + | Let Andrea's current average be <math>a</math>, and assume there have been <math>n</math> tests already. The <math>(n + 1)</math>th test must have a score of <math>a + k(n + 1)</math> in order to increase her average by <math>k</math> and <math>a - k(n + 1)</math> in order to decrease her average by <math>k</math>. Hence, we see that the average must be <math>\tfrac{3}{4}</math> of the way from <math>61</math> to <math>93</math>, which is <math>\boxed{\textbf{(D)}~85}</math>. | |
| − | + | ~joshualiu315 | |
| − | + | ==Solution 2== | |
| + | Let <math>s</math> be the sum of Andrea's scores currently and let <math>n</math> be the number of tests. We have \begin{align*} \frac{s + 61}{n + 1} = \frac{s}{n} - 3 \\ \frac{s+93}{n+1} = \frac{s}{n} + 1\end{align*} Equate <math>s</math> to get a quadratic in <math>n</math> to find <math>n = 7</math>, <math>s = 595</math> so <math>\frac{595}{7} = \boxed{\textbf{(D)}~85}</math>. | ||
| − | ~ | + | ~eg4334 |
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==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2024|ab=A|num-b=4|num-a=6}} | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:27, 20 March 2025
Contents
Problem
Andrea is taking a series of several exams. If Andrea earns 
points on her next exam, her average score will decrease by 
points. If she instead earns 
points on her next exam, her average score will increase by 
point. How many points should Andrea earn on her next exam to keep her average score constant?
Solution 1
Let Andrea's current average be 
, and assume there have been 
tests already. The 
th test must have a score of 
in order to increase her average by 
and 
in order to decrease her average by . Hence, we see that the average must be 
of the way from to , which is 
.
~joshualiu315
Solution 2
Let 
be the sum of Andrea's scores currently and let be the number of tests. We have \begin{align*} \frac{s + 61}{n + 1} = \frac{s}{n} - 3 \\ \frac{s+93}{n+1} = \frac{s}{n} + 1\end{align*} Equate to get a quadratic in to find 
, 
so 
.
~eg4334
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
