Difference between revisions of "2017 AMC 10B Problems/Problem 25"
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==Solution 2== | ==Solution 2== | ||
First, we find the largest sum of scores which is <math>100+99+98+97+96+95+94</math> which equals <math>7(97)</math>. Then we find the smallest sum of scores which is <math>91+92+93+94+95+96+97</math> which is <math>7(94)</math>. So the possible sums for the 7 test scores so that they provide an integer average are <math>7(97), 7(96), 7(95)</math> and <math>7(94)</math> which are <math>679, 672, 665,</math> and <math>658</math> respectively. Now in order to get the sum of the first 6 tests, we negate <math>95</math> from each sum producing <math>584, 577, 570,</math> and <math>563</math>. Notice only <math>570</math> is divisible by <math>6</math> so, therefore, the sum of the first <math>6</math> tests is <math>570</math>. We need to find her score on the <math>6th</math> test so we have to find which number will give us a number divisible by <math>5</math> when subtracted from <math>570.</math> Since <math>95</math> is the <math>7th</math> test score and all test scores are distinct that only leaves <math>\boxed{\textbf{(E) } 100}</math>. | First, we find the largest sum of scores which is <math>100+99+98+97+96+95+94</math> which equals <math>7(97)</math>. Then we find the smallest sum of scores which is <math>91+92+93+94+95+96+97</math> which is <math>7(94)</math>. So the possible sums for the 7 test scores so that they provide an integer average are <math>7(97), 7(96), 7(95)</math> and <math>7(94)</math> which are <math>679, 672, 665,</math> and <math>658</math> respectively. Now in order to get the sum of the first 6 tests, we negate <math>95</math> from each sum producing <math>584, 577, 570,</math> and <math>563</math>. Notice only <math>570</math> is divisible by <math>6</math> so, therefore, the sum of the first <math>6</math> tests is <math>570</math>. We need to find her score on the <math>6th</math> test so we have to find which number will give us a number divisible by <math>5</math> when subtracted from <math>570.</math> Since <math>95</math> is the <math>7th</math> test score and all test scores are distinct that only leaves <math>\boxed{\textbf{(E) } 100}</math>. | ||
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==Solution 4== | ==Solution 4== |
Revision as of 14:41, 13 January 2020
Contents
[hide]Problem
Last year Isabella took math tests and received different scores, each an integer between and , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was . What was her score on the sixth test?
Solution 1
Let the sum of the scores of Isabella's first tests be . Since the mean of her first scores is an integer, then , or . Also, , so by CRT, . We also know that , so by inspection, . However, we also have that the mean of the first integers must be an integer, so the sum of the first test scores must be an multiple of , which implies that the th test score is .
Solution 2
First, we find the largest sum of scores which is which equals . Then we find the smallest sum of scores which is which is . So the possible sums for the 7 test scores so that they provide an integer average are and which are and respectively. Now in order to get the sum of the first 6 tests, we negate from each sum producing and . Notice only is divisible by so, therefore, the sum of the first tests is . We need to find her score on the test so we have to find which number will give us a number divisible by when subtracted from Since is the test score and all test scores are distinct that only leaves .
Solution 4
Since all of the scores are from , we can 'subtract' 90 off from all of the scores. Basically, we're looking at the units digits except for 100; we're looking at 10 in this case. Since the last score was a 95, the sum of the scores from the first six tests must be and . Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be because . The only possible test scores are and , so the answer is .
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
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