Difference between revisions of "2007 AMC 10B Problems/Problem 16"
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==Solution 2== | ==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 $j = | + | 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 $j = \boxed{\mathrm{(C) \ } 93} |
==See Also== | ==See Also== |
Revision as of 15:34, 4 June 2021
- The following problem is from both the 2007 AMC 10B #16 and 2007 AMC 12B #12, so both problems redirect to this page.
Contents
Problem
A teacher gave a test to a class in which of the students are juniors and
are seniors. The average score on the test was
The juniors all received the same score, and the average score of the seniors was
What score did each of the juniors receive on the test?
Solution
We can assume there are people in the class. Then there will be
junior and
seniors. The sum of everyone's scores is
Since the average score of the seniors was
the sum of all the senior's scores is
The only score that has not been added to that is the junior's score, which is
Solution 2
Let the average score of the juniors be . The problem states the average score of the seniors is
. The equation for the average score of the class (juniors and seniors combined) is
. Simplifying this equation yields $j = \boxed{\mathrm{(C) \ } 93}
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
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 (Problems • Answer Key • Resources) | ||
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.