Difference between revisions of "1999 CEMC Gauss (Grade 7) Problems/Problem 8"
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The average of <math>10</math>, <math>4</math>, <math>8</math>, <math>7</math>, and <math>6</math> is | The average of <math>10</math>, <math>4</math>, <math>8</math>, <math>7</math>, and <math>6</math> is | ||
| − | <math>\text{(A)}\ 33 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ </math> | + | <math>\text{(A)}\ 33 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 7</math> |
==Solution 1== | ==Solution 1== | ||
The sum of all the listed numbers is | The sum of all the listed numbers is | ||
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The only answer choice that is less than <math>10</math> is <math>\boxed {\textbf {(E)} 7}</math>. | The only answer choice that is less than <math>10</math> is <math>\boxed {\textbf {(E)} 7}</math>. | ||
==Solution 3== | ==Solution 3== | ||
| − | From the list of numbers, we can | + | From the list of numbers, we can observe that the numbers are all centered around 7. <math>8</math> and <math>6</math> are both <math>1</math> away from <math>7</math>, and <math>4</math> and <math>10</math> are both <math>3</math> away from <math>7</math>. <math>7</math> is equal to itself. |
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| + | Thus, the answer is <math>\boxed {\textbf {(E)} 7}</math>. | ||
Latest revision as of 10:49, 15 April 2025
Problem
The average of 
, 
, 
, 
, and 
is
Solution 1
The sum of all the listed numbers is
Since there are five numbers, we can divide the sum by 
to get the average:
Solution 2 (answer choices)
is too large of an answer because the largest number listed is but there are numbers that are less than , so the average will be lower than .
The only answer choice that is less than is 
.
Solution 3
From the list of numbers, we can observe that the numbers are all centered around 7. and are both 
away from , and and are both 
away from . is equal to itself.
Thus, the answer is .
