Difference between revisions of "1985 AJHSME Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
Notice that the <math>9 \times 11</math> in the denominator of the first fraction cancels with the same term in the second fraction, and the <math>7</math>s in the numerator and denominator of the second fraction cancel. Then the expression is equal to <math>\frac{3 \times 5}{3 \times 5} = \boxed{\text{(A)} 1}</math>. | Notice that the <math>9 \times 11</math> in the denominator of the first fraction cancels with the same term in the second fraction, and the <math>7</math>s in the numerator and denominator of the second fraction cancel. Then the expression is equal to <math>\frac{3 \times 5}{3 \times 5} = \boxed{\text{(A)} 1}</math>. | ||
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+ | ~ cxsmi | ||
==Video Solution by BoundlessBrain!== | ==Video Solution by BoundlessBrain!== |
Revision as of 17:59, 8 January 2024
Problem
Solution 1
By the associative property, we can rearrange the numbers in the numerator and the denominator.
Solution 2
Notice that the in the denominator of the first fraction cancels with the same term in the second fraction, and the s in the numerator and denominator of the second fraction cancel. Then the expression is equal to .
~ cxsmi
Video Solution by BoundlessBrain!
See Also
1985 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.