Difference between revisions of "1985 AJHSME Problems/Problem 17"
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If we increase the seventh test score by <math>7</math>, the average will increase by <math>\frac{7}{7}=1</math>. We need the average to increase by <math>85-72=13</math>, so the seventh test score is <math>7\times 13=91</math> more than <math>0</math>, which is clearly <math>91</math>. This is choice <math>\boxed{\text{D}}</math> | If we increase the seventh test score by <math>7</math>, the average will increase by <math>\frac{7}{7}=1</math>. We need the average to increase by <math>85-72=13</math>, so the seventh test score is <math>7\times 13=91</math> more than <math>0</math>, which is clearly <math>91</math>. This is choice <math>\boxed{\text{D}}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/PMlL7M0HzTY | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Latest revision as of 08:14, 13 January 2023
Problem
If your average score on your first six mathematics tests was 
and your average score on your first seven mathematics tests was 
, then your score on the seventh test was
Solution 1
If the average score of the first six is , then the sum of those six scores is 
.
The average score of the first seven is , so the sum of the seven is 
Taking the difference leaves us with just the seventh score, which is 
, so the answer is 
Solution 2
Let's remove the condition that the average of the first seven tests is , and say the 7th test score was a 
. Then, the average of the first seven tests would be ![\[\frac{6\times 84}{7}=72\]](http://latex.artofproblemsolving.com/0/9/d/09d3847e2f9297bc19ff2efb059af08ae39e1817.png)
If we increase the seventh test score by 
, the average will increase by 
. We need the average to increase by 
, so the seventh test score is 
more than , which is clearly 
. This is choice
Video Solution
~savannahsolver
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
