Difference between revisions of "2020 AMC 10A Problems/Problem 2"
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Incidentally we can immediately rule out answers A and B as being too small - with these we'd never catch up to get our average of 15. An alternative solution is that to get an average of 15 we need a sum of 15*5 = 75. We have 3, 5, and 7 which adds up to 15, so a and b must make up the remaining 75-15 = 60. 60/2 = 30 so the average of a and b is 30. | Incidentally we can immediately rule out answers A and B as being too small - with these we'd never catch up to get our average of 15. An alternative solution is that to get an average of 15 we need a sum of 15*5 = 75. We have 3, 5, and 7 which adds up to 15, so a and b must make up the remaining 75-15 = 60. 60/2 = 30 so the average of a and b is 30. | ||
| + | |||
| + | ~Dilip | ||
==Video Solution 1== | ==Video Solution 1== | ||
Revision as of 20:35, 20 October 2023
Contents
Problem
The numbers 
and 
have an average (arithmetic mean) of 
. What is the average of 
and ?
Solution
The arithmetic mean of the numbers and is equal to 
. Solving for 
, we get 
. Dividing by 
to find the average of the two numbers and gives 
.
Solution (two solutions - balancing and summing)
We know the average is 15. 3, 5, and 7 are, respectively, 12, 10, and 8 below the average. So far then we are -12 - 10 - 8 or 30 below the average. We have to make this up with a and b so on average each is 30/2 = 15 above the average. The average is 15 so each is on average 15+15 = 30, or answer C.
Incidentally we can immediately rule out answers A and B as being too small - with these we'd never catch up to get our average of 15. An alternative solution is that to get an average of 15 we need a sum of 15*5 = 75. We have 3, 5, and 7 which adds up to 15, so a and b must make up the remaining 75-15 = 60. 60/2 = 30 so the average of a and b is 30.
~Dilip
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
Video Solution 4
~savannahsolver
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
