Difference between revisions of "1993 AJHSME Problems/Problem 12"

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==Solution==
 
==Solution==
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There are a reasonable number of ways to place the operation signs, so guess and check to find that <math>5-4+6 \times 3 = \boxed{\text{(E)}\ 19}</math>.
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==See Also==
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{{AJHSME box|year=1993|num-b=11|num-a=13}}
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{{MAA Notice}}

Revision as of 00:11, 5 July 2013

Problem

If each of the three operation signs, $+$, $-$, $\times$, is used exactly ONCE in one of the blanks in the expression

\[5\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}4\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}6\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}3\]

then the value of the result could equal

$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 19$

Solution

There are a reasonable number of ways to place the operation signs, so guess and check to find that $5-4+6 \times 3 = \boxed{\text{(E)}\ 19}$.

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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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