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

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

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: @@
 ==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>
-==Solution==
+[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex]
-We '''could''' go at it by just [[Multiplication|multiplying]] it out, dividing, etc, but there is a much more simple method.
-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>
+==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]
-Notice that each number is still there, and nothing has been changed - other than the order.
+==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>.
-Finally, since each fraction is equal to one, we have <math>1\times1\times1\times1\times1</math>, which is equal to <math>1</math>.
+~ cxsmi
+==Video Solution by BoundlessBrain!==
+https://youtu.be/eC_Vu3vogHM
-Thus, <math>\boxed{\text{A}}</math> is the answer.
 ==See Also==

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