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

 
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*Note that we are only accounting for odd numbers, thus the <math>/2</math>.
 
*Note that we are only accounting for odd numbers, thus the <math>/2</math>.
  
MegaBoy6679 :D 23:44, 26 December 2022 (EST)
+
~megaboy6679
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=7|num-a=9}}
 
{{AMC8 box|year=2011|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:44, 27 December 2022

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

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

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