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

(Problem 13)
(Problem 13)
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<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==
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Say Lalia 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=410.</math>
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The value <math>x</math> has to be less than 82, because then she would receive a lower score on her last test. Additionaly, the greatest value for y is 100, so therefore the smallest value <math>x</math> can be is 78. As a result, only <math>\boxed{4}</math> numbers work, <math>78, 79,80</math> and <math>81</math>. Thus the answer is <math>\textbf{(A) }</math>. - song2sons
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{{AMC8 box|year=2018|num-b=12|num-a=14}}
 
{{AMC8 box|year=2018|num-b=12|num-a=14}}

Revision as of 17:39, 21 November 2018

Problem 13

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?

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

Solution

Say Lalia gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=410.$


The value $x$ has to be less than 82, because then she would receive a lower score on her last test. Additionaly, the greatest value for y is 100, so therefore the smallest value $x$ can be is 78. As a result, only $\boxed{4}$ numbers work, $78, 79,80$ and $81$. Thus the answer is $\textbf{(A) }$. - song2sons


2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AJHSME/AMC 8 Problems and Solutions