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Difference between revisions of "2018 AMC 8 Problems/Problem 13"

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==Problem 13==
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==Problem==
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Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
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Lailai took five math tests, each worth a maximum of 100 points. Lailai's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Lailai's score on the last test?
  
 
<math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18</math>
 
<math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18</math>
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==Solution 1==
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Say Lailai gets a value of <math>x</math> on her first 4 tests, and a value of <math>y</math> on her last test. Thus, <math>4x+y=82 \cdot 5=410</math>.
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Because <math>x</math> and <math>y</math> are different, <math>x</math> must be less than <math>82</math> and <math>y</math> must be greater than <math>82</math>. When <math>x</math> decreases by <math>1</math>, <math>y</math> must increase by <math>4</math> to keep the total constant.
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The greatest value for <math>y</math> is <math>98</math> (as <math>y=100</math> would make <math>x</math> non-integer). In the range <math>83\le y\le 98</math>, only <math>4</math> values for <math>y</math> result in integer values for <math>x</math>: 86, 90, 94 and 98.  Thus, the answer is <math>\boxed{\textbf{(A) }4}</math>.
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==Solution 2==
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The average score is <math>82</math> which leads us to suppose that Laila got all <math>82</math> points for the tests. We know that Laila got the same points in the first four tests and they are all lower than the last test. Let the first four tests be <math>81</math> points, then the last test must be <math>86</math> points to keep the average fixed. Continue to decrement the first four tests to identify other possible combinations. The possible points for the fifth test are <math>86</math>, <math>90</math>, <math>94</math>, <math>98</math>. The answer is <math>\boxed{\textbf{(A) }4}</math>.
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==Video Solution (CREATIVE ANALYSIS!!!)==
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https://youtu.be/LP7PQSvGLtA
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~Education, the Study of Everything
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==Video Solution by OmegaLearn==
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https://youtu.be/HISL2-N5NVg?t=3251
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~ pi_is_3.14
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== Video Solutions ==
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https://youtu.be/viShU5koGtk
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https://youtu.be/Olg8fUc1KQI
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~savannahsolver
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==See Also==
 
{{AMC8 box|year=2018|num-b=12|num-a=14}}
 
{{AMC8 box|year=2018|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 21:02, 13 October 2024

Problem

Lailai took five math tests, each worth a maximum of 100 points. Lailai's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Lailai's score on the last test?

$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$

Solution 1

Say Lailai gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=82 \cdot 5=410$.

Because $x$ and $y$ are different, $x$ must be less than $82$ and $y$ must be greater than $82$. When $x$ decreases by $1$, $y$ must increase by $4$ to keep the total constant.

The greatest value for $y$ is $98$ (as $y=100$ would make $x$ non-integer). In the range $83\le y\le 98$, only $4$ values for $y$ result in integer values for $x$: 86, 90, 94 and 98. Thus, the answer is $\boxed{\textbf{(A) }4}$.

Solution 2

The average score is $82$ which leads us to suppose that Laila got all $82$ points for the tests. We know that Laila got the same points in the first four tests and they are all lower than the last test. Let the first four tests be $81$ points, then the last test must be $86$ points to keep the average fixed. Continue to decrement the first four tests to identify other possible combinations. The possible points for the fifth test are $86$, $90$, $94$, $98$. The answer is $\boxed{\textbf{(A) }4}$.

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/LP7PQSvGLtA

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=3251

~ pi_is_3.14

Video Solutions

https://youtu.be/viShU5koGtk

https://youtu.be/Olg8fUc1KQI

~savannahsolver

See Also

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 AJHSME/AMC 8 Problems and Solutions

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