Difference between revisions of "2020 AMC 10A Problems/Problem 2"
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| − | ==Problem | + | ==Problem== |
The numbers <math>3, 5, 7, a,</math> and <math>b</math> have an average (arithmetic mean) of <math>15</math>. What is the average of <math>a</math> and <math>b</math>? | The numbers <math>3, 5, 7, a,</math> and <math>b</math> have an average (arithmetic mean) of <math>15</math>. What is the average of <math>a</math> and <math>b</math>? | ||
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== Solution == | == Solution == | ||
| − | The arithmetic mean of the numbers <math>3, 5, 7, a,</math> and <math>b</math> is equal to <math>\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15</math>. Solving for <math>a+b</math>, we get <math>a+b=60</math>. Dividing by <math>2</math> to find the average of the two numbers <math>a</math> and <math>b</math> gives <math>\frac{60}{2}=\boxed{\ | + | The arithmetic mean of the numbers <math>3, 5, 7, a,</math> and <math>b</math> is equal to <math>\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15</math>. Solving for <math>a+b</math>, we get <math>a+b=60</math>. Dividing by <math>2</math> to find the average of the two numbers <math>a</math> and <math>b</math> gives <math>\frac{60}{2}=\boxed{\textbf{(C) }30}</math>. |
| − | ~ | + | == 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. | ||
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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. | ||
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| + | ~Dilip | ||
| + | |||
| + | ==Video Solution 1== | ||
| + | |||
| + | Education, The Study of Everything | ||
| + | |||
| + | https://youtu.be/e8Qfe5GpEUg | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | https://youtu.be/WUcbVNy2uv0 | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | ==Video Solution 3== | ||
| + | https://www.youtube.com/watch?v=7-3sl1pSojc | ||
| + | |||
| + | ~bobthefam | ||
| + | |||
| + | ==Video Solution 4== | ||
| + | https://youtu.be/zVppmKOvx_w | ||
| + | |||
| + | ~savannahsolver | ||
== See Also == | == See Also == | ||
| − | {{AMC10 box|year=2020|ab=A| | + | {{AMC10 box|year=2020|ab=A|num-b=1|num-a=3}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest 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. 
