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

Latest revision as of 13:41, 4 April 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, the $7$ s in the numerator and denominator of the second fraction cancel, and the $3 \times 5$ 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 $\boxed{\textbf{(A)}~1}$ .

~ cxsmi

Video Solution by BoundlessBrain!

https://youtu.be/eC_Vu3vogHM

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.

@@ Line 1: / Line 1: @@
-==Question==
+==Problem==
-<math>\frac{3\times5}{9\times11} \times \frac{7\times9\times11}{3\times5\times7} </math><br><br>
-<math>(A) 1 (B) 0 (C) 49 (D) \frac{1}{49} (E) 50</math>
-==Solution==
+[katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
-We '''could''' go at it by just multiplying it out, dividing, etc, but there is a much more obvious, simpler method.<br>Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by 1, we can rearrange the numbers in the numerator and the denominator (commutative property of multiplication) so that it looks like...<br><br><math>\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}</math><br><br>Notice that each number is still there, and nothing has been changed - other than the order.<br>Finally, since each fraction is equal to one, we have <math>1\times1\times1\times1\times1</math>, which is equal to 1.
-Thus, <math>A</math> is the answer.
+[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 <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]
+==Video Solution by BoundlessBrain!==
+https://youtu.be/eC_Vu3vogHM
+==See Also==
+{{AJHSME box|year=1985|before=First <br> Question|num-a=2}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

Page

Toolbox

Search