Difference between revisions of "2007 AMC 10B Problems/Problem 16"

(See Also)
 
(13 intermediate revisions by 6 users not shown)
Line 8: Line 8:
 
==Solution==
 
==Solution==
  
We can assume there are <math>10</math> people in the class. Then there will be <math>1</math> junior and <math>9</math> seniors. The sum of everyone's scores is <math>10 \cdot 84 = 840.</math> Since the average score of the seniors was <math>83,</math> the sum of all the senior's scores is <math>9 \cdot 83 = 747.</math> The only score that has not been added to that is the junior's score, which is <math>840 - 747 = \boxed{\mathrm{(C) \ } 93}</math>
+
We can assume there are <math>10</math> people in the class. Then there will be <math>1</math> junior and <math>9</math> seniors. The sum of everyone's scores is <math>10 \cdot 84 = 840</math>. Since the average score of the seniors was <math>83</math>, the sum of all the senior's scores is <math>9 \cdot 83 = 747</math>. The only score that has not been added to that is the junior's score, which is <math>840 - 747 = \boxed{\textbf{(C) } 93}</math>
 +
 
 +
==Solution 2==
 +
 
 +
Let the average score of the juniors be <math>j</math>. The problem states the average score of the seniors is <math>83</math>. The equation for the average score of the class (juniors and seniors combined) is <math>\frac{j}{10} + \frac{83 \cdot 9}{10} = 84</math>. Simplifying this equation yields <math>j = \boxed{\textbf{(C) } 93}</math>
 +
 
 +
~mobius247
  
 
==See Also==
 
==See Also==
  
{{AMC12 box|year=2007|ab=B|num-b=10|num-a=12}}
+
{{AMC12 box|year=2007|ab=B|num-b=11|num-a=13}}
  
 
{{AMC10 box|year=2007|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2007|ab=B|num-b=15|num-a=17}}
 +
{{MAA Notice}}

Latest revision as of 16:11, 6 July 2023

The following problem is from both the 2007 AMC 10B #16 and 2007 AMC 12B #12, so both problems redirect to this page.

Problem

A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?

$\textbf{(A) } 85 \qquad\textbf{(B) } 88 \qquad\textbf{(C) } 93 \qquad\textbf{(D) } 94 \qquad\textbf{(E) } 98$

Solution

We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \cdot 84 = 840$. Since the average score of the seniors was $83$, the sum of all the senior's scores is $9 \cdot 83 = 747$. The only score that has not been added to that is the junior's score, which is $840 - 747 = \boxed{\textbf{(C) } 93}$

Solution 2

Let the average score of the juniors be $j$. The problem states the average score of the seniors is $83$. The equation for the average score of the class (juniors and seniors combined) is $\frac{j}{10} + \frac{83 \cdot 9}{10} = 84$. Simplifying this equation yields $j = \boxed{\textbf{(C) } 93}$

~mobius247

See Also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 12 Problems and Solutions
2007 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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

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