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

m (Solution)
m (Solution)
Line 8: Line 8:
 
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>  
 
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>  
  
The value <math>y</math> has to be greater than 82, because otherwise she would receive a higher score on her last test. Additionally, the greatest value for y is 98 (as y=100 would give x as a decimal), so therefore the greatest value <math>x</math> can be is 98. As a result, only <math>4</math> numbers work, <math>86, 90, 94</math> and <math>98</math>. Thus the answer is <math>\boxed{\textbf{(A) }4}</math>.
+
The value <math>y</math> has to be greater than 82, because otherwise she would receive the same score on her last test. Additionally, the greatest value for y is 98 (as y=100 would give x as a decimal), so therefore the greatest value <math>x</math> can be is 98. As a result, only <math>4</math> numbers work, <math>86, 90, 94</math> and <math>98</math>. Thus the answer is <math>\boxed{\textbf{(A) }4}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 18:33, 11 November 2019

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 $y$ has to be greater than 82, because otherwise she would receive the same score on her last test. Additionally, the greatest value for y is 98 (as y=100 would give x as a decimal), so therefore the greatest value $x$ can be is 98. As a result, only $4$ numbers work, $86, 90, 94$ and $98$. Thus the answer is $\boxed{\textbf{(A) }4}$.

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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

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