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

Revision as of 21:38, 22 December 2012

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 $5-4+6 \times 3 = \boxed{\text{(E)}\ 19}$ .

1993 AJHSME (Problems • Answer Key • Resources)
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

@@ Line 10: / Line 10: @@
 ==Solution==
-The Answer is (E), because if you put 5-4+6*3, it equals 19 <3 guess and check
+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>.
+==See Also==
+{{AJHSME box|year=1993|num-b=11|num-a=13}}

Page

Toolbox

Search

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

Revision as of 21:38, 22 December 2012

Problem

Solution

See Also