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

Revision as of 01:07, 23 December 2012

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.

1994 AJHSME (Problems • Answer Key • Resources)
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

@@ Line 3: / Line 3: @@
 A gumball machine contains <math>9</math> red, <math>7</math> white, and <math>8</math> blue gumballs.  The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is
-<math>\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18</math>
+<math>\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18</math>
+==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.
+==See Also==
+{{AJHSME box|year=1994|num-b=20|num-a=22}}

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

Revision as of 01:07, 23 December 2012

Problem

Solution

See Also