2011 AMC 8 Problems/Problem 8

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

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 $\boxed{\textbf{(B)}\ 5}$ . You can also approach this problem by noticing the sums will make odd multiples from $3$ to $11$ inclusive.

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

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