Difference between revisions of "1993 AJHSME Problems/Problem 1"
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Which pair of numbers does NOT have a product equal to <math>36</math>? | Which pair of numbers does NOT have a product equal to <math>36</math>? | ||
<math> \text{(A)}\ \{-4,-9\}\qquad\text{(B)}\ \{-3,-12\}\qquad\text{(C)}\ \left\{\frac{1}{2},-72\right\}\qquad\text{(D)}\ \{ 1,36\}\qquad\text{(E)}\ \left\{\frac{3}{2},24\right\} </math> | <math> \text{(A)}\ \{-4,-9\}\qquad\text{(B)}\ \{-3,-12\}\qquad\text{(C)}\ \left\{\frac{1}{2},-72\right\}\qquad\text{(D)}\ \{ 1,36\}\qquad\text{(E)}\ \left\{\frac{3}{2},24\right\} </math> | ||
Revision as of 18:32, 28 January 2022
Problem
Which pair of numbers does NOT have a product equal to 
?
Solution 1
Let's calculate each of the answer choices and see which one DOES NOT equal 36.
A comes out to be -4 x -9= 36
B equals -3 x -12= 36
C is 1/2 x -72= -36
D simplifies to 1 x 36= 36
and E equals 3/2 x 24= 3 x 12= 36
Thus, our answer is C. It is the only negative result.
Solution 2
In order for a product to be positive (36), the numbers should either be both positive or both negative. Looking at our answer choices, this description matches with C since one number is positive and the other is negative.
See Also
| 1993 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
