Difference between revisions of "2011 AMC 8 Problems/Problem 8"

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==Solution==
 
==Solution==
  
By adding a number from Bag A and a number from Bag B together, the values we can get are <math>3, 5, 7, 5, 7, 9, 7, 9, 11.</math> Therefore the number of different values is <math>\boxed{\text{(B)}\ 5}</math>. You can also approach this problem by noticing the sums will make odd multiples from <math>3</math> to <math>11</math> inclusive.  
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By adding a number from Bag A and a number from Bag B together, the values we can get are <math>3, 5, 7, 5, 7, 9, 7, 9, 11.</math> Therefore the number of different values is <math>\boxed{\textbf{(B)}\ 5}</math>. You can also approach this problem by noticing the sums will make odd multiples from <math>3</math> to <math>11</math> inclusive.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=7|num-a=9}}
 
{{AMC8 box|year=2011|num-b=7|num-a=9}}

Revision as of 19:21, 25 November 2011

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?

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

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.

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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