Difference between revisions of "1994 AJHSME Problems/Problem 21"

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==Solution==
 
==Solution==
 
If a person gets three gumballs of each of the three colors, that is, <math>9</math> gumballs, then the <math>10^{th}</math> gumball must be the fourth one for one of the colors. Therefore, the person must buy <math>\boxed{\text{(C)}\ 10}</math> gumballs.
 
If a person gets three gumballs of each of the three colors, that is, <math>9</math> gumballs, then the <math>10^{th}</math> gumball must be the fourth one for one of the colors. Therefore, the person must buy <math>\boxed{\text{(C)}\ 10}</math> gumballs.
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Note that the answer key says B, but the answer is actually C, please notice this.
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1994|num-b=20|num-a=22}}
 
{{AJHSME box|year=1994|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:42, 31 October 2015

Problem

A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is

$\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$

Solution

If a person gets three gumballs of each of the three colors, that is, $9$ gumballs, then the $10^{th}$ gumball must be the fourth one for one of the colors. Therefore, the person must buy $\boxed{\text{(C)}\ 10}$ gumballs.

Note that the answer key says B, but the answer is actually C, please notice this.

See Also

1994 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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