Difference between revisions of "1985 AJHSME Problems/Problem 1"
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| − | ==Solution== | + | ==Solution 1== |
By the [[associative property]], we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)} 1}[/katex] | By the [[associative property]], we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)} 1}[/katex] | ||
| + | ==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>. | ||
==Video Solution by BoundlessBrain!== | ==Video Solution by BoundlessBrain!== | ||
Revision as of 17:59, 8 January 2024
Problem
[katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex]
Solution 1
By the associative property, we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)} 1}[/katex]
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 
.
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. 
