Difference between revisions of "1985 AJHSME Problems/Problem 17"
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<math>\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92</math> | <math>\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92</math> | ||
| − | + | ==Solution 1== | |
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If the average score of the first six is <math>84</math>, then the [[sum]] of those six scores is <math>6\times 84=504</math>. | If the average score of the first six is <math>84</math>, then the [[sum]] of those six scores is <math>6\times 84=504</math>. | ||
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Taking the difference leaves us with just the seventh score, which is <math>595-504=91</math>, so the answer is <math>\boxed{\text{D}}</math> | Taking the difference leaves us with just the seventh score, which is <math>595-504=91</math>, so the answer is <math>\boxed{\text{D}}</math> | ||
| − | + | ==Solution 2== | |
Let's remove the condition that the average of the first seven tests is <math>85</math>, and say the 7th test score was a <math>0</math>. Then, the average of the first seven tests would be <cmath>\frac{6\times 84}{7}=72</cmath> | Let's remove the condition that the average of the first seven tests is <math>85</math>, and say the 7th test score was a <math>0</math>. Then, the average of the first seven tests would be <cmath>\frac{6\times 84}{7}=72</cmath> | ||
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> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/PMlL7M0HzTY | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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{{AJHSME box|year=1985|num-b=16|num-a=18}} | {{AJHSME box|year=1985|num-b=16|num-a=18}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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| + | {{MAA Notice}} | ||
Latest revision as of 07:14, 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. 
