Difference between revisions of "1985 AJHSME Problems/Problem 1"

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==Problem==
 
==Problem==
  
<math>\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=</math>
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[katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
  
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50</math>
 
  
==Soulutions==
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[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===
 
  
Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by <math>1</math>, we can rearrange the numbers in the numerator and the denominator ([[Commutative property|commutative property of multiplication]]) so that it looks like <cmath>\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}</cmath>
 
  
Notice that each number is still there, and nothing has been changed - other than the order.
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==Solution 1==
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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]
  
Finally, since all of the fractions are equal to one, we have <math>1\times1\times1\times1\times1</math>, which is equal to <math>1</math>.
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==Solution 2==
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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, 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>.
  
Thus, <math>\boxed{\text{A}}</math> is the answer.
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~ cxsmi
  
==Solution 2 (Brute force)==
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==Video Solution by BoundlessBrain!==
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https://youtu.be/eC_Vu3vogHM
  
If you want to multiply it out, then it would be <cmath>\frac{15}{99} \times \frac{693}{105}</cmath>.
 
 
That would be <cmath>\frac{10395}{10395}</cmath>, which is 1. Therefore, the answer is <math>\boxed{\text{A}}</math>
 
  
 
==See Also==
 
==See Also==

Latest revision as of 15:14, 23 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 $9 \times 11$ in the denominator of the first fraction cancels with the same term in the second fraction, and the $7$s in the numerator and denominator of the second fraction cancel. Then the expression is equal to $\frac{3 \times 5}{3 \times 5} = \boxed{\text{(A)} 1}$.

~ cxsmi

Video Solution by BoundlessBrain!

https://youtu.be/eC_Vu3vogHM


See Also

1985 AJHSME (ProblemsAnswer KeyResources)
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


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