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

Revision as of 12:51, 22 December 2020

Problem

$\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=$

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

Solutions

Solution 1

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 $\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}.$

Notice that each number is still there, and nothing has been changed - other than the order.

Finally, since all of the fractions are equal to one, we have $1\times1\times1\times1\times1$ , which is equal to $1$ .

Thus, $\boxed{\mathbf{(A )}1}$ is the answer.

Solution 2 (Brute force)

If you want to multiply it out, then it would be $\frac{15}{99} \times \frac{693}{105}.$

That would be $\frac{10395}{10395},$ which is 1. Therefore, the answer is $\boxed{\mathbf{(A )}1}$

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 14: / Line 14: @@
 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>.
-Thus, <math>\boxed{\text{A}}</math> is the answer.
+Thus, <math>\boxed{\mathbf{(A )}1}</math> is the answer.
 ===Solution 2 (Brute force)===

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