Difference between revisions of "2000 AMC 8 Problems/Problem 23"
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==Solution== | ==Solution== | ||
− | Remember that if a list of <math>n</math> numbers has an average of <math>k</math>, then | + | Remember that if a list of <math>n</math> numbers has an average of <math>k</math>, then the sum <math>S</math> of all the numbers on the list is <math>S = nk</math>. |
So if the average of the first <math>4</math> numbers is <math>5</math>, then the first four numbers total <math>4 \cdot 5 = 20</math>. | So if the average of the first <math>4</math> numbers is <math>5</math>, then the first four numbers total <math>4 \cdot 5 = 20</math>. | ||
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Algebraically, if <math>a + b + c + d = 20</math>, and <math>d + e + f + g = 32</math>, you can add both equations to get <math>a + b + c + 2d + e + f + g = 52</math>. You know that <math>a + b + c + d + e + f + g = 46</math>, so you can subtract that from the last equation to get <math>d = 6</math>, and <math>d</math> is the number that appeared twice. | Algebraically, if <math>a + b + c + d = 20</math>, and <math>d + e + f + g = 32</math>, you can add both equations to get <math>a + b + c + 2d + e + f + g = 52</math>. You know that <math>a + b + c + d + e + f + g = 46</math>, so you can subtract that from the last equation to get <math>d = 6</math>, and <math>d</math> is the number that appeared twice. | ||
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+ | Yay! :) | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2000|num-b=22|num-a=24}} | {{AMC8 box|year=2000|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:28, 21 July 2017
Problem
There is a list of seven numbers. The average of the first four numbers is , and the average of the last four numbers is . If the average of all seven numbers is , then the number common to both sets of four numbers is
Solution
Remember that if a list of numbers has an average of , then the sum of all the numbers on the list is .
So if the average of the first numbers is , then the first four numbers total .
If the average of the last numbers is , then the last four numbers total .
If the average of all numbers is , then the total of all seven numbers is .
If the first four numbers are , and the last four numbers are , then all "eight" numbers are . But that's counting one number twice. Since the sum of all seven numbers is , then the number that was counted twice is , and the answer is
Algebraically, if , and , you can add both equations to get . You know that , so you can subtract that from the last equation to get , and is the number that appeared twice.
Yay! :)
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.