Difference between revisions of "2002 AMC 10B Problems/Problem 25"
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<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math> | <math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math> | ||
− | == Solution == | + | == Solution 1== |
Let <math>x</math> be the sum of the integers and <math>y</math> be the number of elements in the list. Then we get the equations <math>\dfrac{x+15}{y+1}=\dfrac{x}{y}+2</math> and <math>\dfrac{x+15+1}{y+1+1}=\dfrac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1</math>. With a lot of algebra, the solution is found to be <math>y= \boxed{\textbf{(A)}\ 4} </math>. | Let <math>x</math> be the sum of the integers and <math>y</math> be the number of elements in the list. Then we get the equations <math>\dfrac{x+15}{y+1}=\dfrac{x}{y}+2</math> and <math>\dfrac{x+15+1}{y+1+1}=\dfrac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1</math>. With a lot of algebra, the solution is found to be <math>y= \boxed{\textbf{(A)}\ 4} </math>. | ||
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==Solution 2== | ==Solution 2== | ||
We let <math>m</math> be the original number of elements in the set and we let <math>n</math> be the original average of the terms of the original list. Then we have <math>mn</math> is the sum of all the elements of the list. So we have two equations: <cmath>mn+15=(m+2)(n+1)=mn+m+2n+2</cmath> and <cmath>mn+16=(m+1)(n+2)=mn+2m+n+2.</cmath>Simplifying both equations and we get, | We let <math>m</math> be the original number of elements in the set and we let <math>n</math> be the original average of the terms of the original list. Then we have <math>mn</math> is the sum of all the elements of the list. So we have two equations: <cmath>mn+15=(m+2)(n+1)=mn+m+2n+2</cmath> and <cmath>mn+16=(m+1)(n+2)=mn+2m+n+2.</cmath>Simplifying both equations and we get, |
Revision as of 19:50, 22 October 2018
Contents
[hide]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 .
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 |
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