Difference between revisions of "1993 AJHSME Problems/Problem 1"
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<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> | ||
− | ==Solution== | + | ==Solution 1== |
− | |||
− | + | Let's calculate each of the answer choices and see which one DOES NOT equal <math>36</math>. | |
− | + | <math>A</math> comes out to be <math>-4 \times -9= 36</math>, | |
− | + | <math>B</math> equals <math>-3 \times -12= 36</math>, | |
− | + | <math>C</math> is <math>\frac{1}{2} \times -72= -36</math>, | |
− | + | <math>D</math> simplifies to <math>1 \times 36= 36</math>, | |
+ | |||
+ | and <math>E</math> equals <math>\frac{3}{2} \times 24= 3 \times 12= 36</math>. | ||
+ | |||
+ | Thus, our answer is <math>\boxed{C}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | In order for a product to be positive (<math>36</math>), the numbers should either be both positive or both negative. Looking at our answer choices, the only option that does not fit this description is <math>\boxed{C}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AJHSME box|year=1993|before=First<br />Question|num-a=2}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:05, 27 June 2023
Contents
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 .
comes out to be ,
equals ,
is ,
simplifies to ,
and equals .
Thus, our answer is .
Solution 2
In order for a product to be positive (), the numbers should either be both positive or both negative. Looking at our answer choices, the only option that does not fit this description is .
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.