Difference between revisions of "2002 AMC 10B Problems/Problem 25"
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<cmath>14=2m+n</cmath> | <cmath>14=2m+n</cmath> | ||
Solving for <math>m</math> and <math>n</math>, we get <math>m=5</math> and <math>n=\boxed{\textbf{(A)}4}</math>. | Solving for <math>m</math> and <math>n</math>, we get <math>m=5</math> and <math>n=\boxed{\textbf{(A)}4}</math>. | ||
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+ | ==Solution 3== | ||
+ | Warning:This solution will rarley ever work however seeing that you can so easily plug and chug in probem 25 it is funny to see this | ||
+ | |||
+ | Plug and chug random numbers with the answer choices we can start with <math>4</math> numbers. You see that if you have 4 5s and you add 15 the resulting mean will be 7 we can verify this with math | ||
+ | <cmath>\frac{5+5+5+5+15}{5}=7</cmath> | ||
+ | adding in 1 to the set you result in the mean to be 6. | ||
+ | <cmath>\frac{5+5+5+5+15+1}{6}=6</cmath> | ||
+ | Thus we conclude that 4 is the correct choice or <math>\boxed{\textbf{(A)}}</math> | ||
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=B|num-b=24|after=Last problem}} | {{AMC10 box|year=2002|ab=B|num-b=24|after=Last problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:50, 8 July 2019
Problem
When is appended to a list of integers, the mean is increased by . When is appended to the enlarged list, the mean of the enlarged list is decreased by . How many integers were in the original list?
Solution 1
Let be the sum of the integers and be the number of elements in the list. Then we get the equations and . With a lot of algebra, the solution is found to be .
Solution 2
We let be the original number of elements in the set and we let be the original average of the terms of the original list. Then we have is the sum of all the elements of the list. So we have two equations: and Simplifying both equations and we get, Solving for and , we get and .
Solution 3
Warning:This solution will rarley ever work however seeing that you can so easily plug and chug in probem 25 it is funny to see this
Plug and chug random numbers with the answer choices we can start with numbers. You see that if you have 4 5s and you add 15 the resulting mean will be 7 we can verify this with math adding in 1 to the set you result in the mean to be 6. Thus we conclude that 4 is the correct choice or
See also
2002 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.