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Difference between revisions of "2001 AMC 8 Problems/Problem 21"

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==Solution==
 
==Solution==
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Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is <math> 18 </math>, and there are <math> 2 </math> numbers less than <math> 18 </math> and <math> 2 </math> numbers greater than <math> 18 </math>. The sum of these integers is <math> 5(15)=75 </math>, since the mean is <math> 15 </math>. To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than <math> 18 </math> must be positive and distinct, so the smallest possible numbers for these are <math> 1 </math> and <math> 2 </math>. One of the numbers also needs to be as small as possible, so it must be <math> 19 </math>. This means that the remaining number, the maximum possible value for a number in the set, is <math> 75-1-2-18-19=35, \boxed{\text{(D) 35}} </math>.
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The sum of the numbers from the first set is <math>5\cdot 13=65</math>. The sum of the numbers from the second set is <math>24\cdot 6 = 144</math>. The sum of all the numbers in the set is <math>144+65=209</math>, so the average of the 11 numbers in the set is <math>209/11=\boxed{19}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2001|num-b=20|num-a=22}}
 
{{AMC8 box|year=2001|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:48, 8 January 2019

Problem

The mean of one set of five numbers is 13, and the mean of a separate set of six numbers is 24. What is the mean of the set of all eleven numbers?

$\text{(A)}\ 19 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 40$

Solution

The sum of the numbers from the first set is $5\cdot 13=65$. The sum of the numbers from the second set is $24\cdot 6 = 144$. The sum of all the numbers in the set is $144+65=209$, so the average of the 11 numbers in the set is $209/11=\boxed{19}$.

See Also

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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

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