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

(Video Solution by BoundlessBrain!)
(Jost playing, don't use.)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
<math>\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=</math>
+
[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>
+
 
 +
[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==
 
==Solution 1==

Revision as of 19:19, 14 December 2023

Problem

3×59×11×7×9×113×5×7=\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=


(A) 1(B) 0(C) 49(D) 149(E) 50\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50

Solution 1

By the associative property, we can rearrange the numbers in the numerator and the denominator. \[\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)} 1}\]

Solution 2

Multlipication gives:\[\frac{15}{99}\cdot\frac{693}{105}=\frac{10395}{10395}=\boxed{\text{(A)}\ 1}.\]

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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png