Difference between revisions of "1985 AJHSME Problems/Problem 1"
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Notice that the <math>9 \times 11</math> in the denominator of the first fraction cancels with the same term in the second fraction, the <math>7</math>s in the numerator and denominator of the second fraction cancel, and the <math>3 \times 5</math> in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with <math>\boxed{\textbf{(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, the <math>7</math>s in the numerator and denominator of the second fraction cancel, and the <math>3 \times 5</math> in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with <math>\boxed{\textbf{(A)}~1}</math>. | ||
| − | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] |
==Video Solution by BoundlessBrain!== | ==Video Solution by BoundlessBrain!== | ||
Revision as of 11:02, 13 July 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, the 
s in the numerator and denominator of the second fraction cancel, and the 
in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with 
.
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. 
