2011 AMC 8 Problems/Problem 8

Problem

Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?

$\textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9$

Solution 1

By adding a number from Bag A and a number from Bag B together, the values we can get are $3, 5, 7, 5, 7, 9, 7, 9, 11.$ Therefore the number of different values is $\boxed{\textbf{(B)}\ 5}$ .

Solution 2

The sum of an even number added to an odd number is always odd. The smallest possible sum is $3$ , and the largest possible sum is $11$ . The odd numbers in between can be achieved by replacing chips with ${\displaystyle \pm }2$ within the same bag. Therefore, we can conclude that there are $(11-3)/2(*)+1=\boxed{\textbf{(B)}\ 5}$ possible sums.

Note that we are only accounting for odd numbers, thus the $/2$ .

~megaboy6679

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AJHSME/AMC 8 Problems and Solutions

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2011 AMC 8 Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

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