Difference between revisions of "1985 AJHSME Problem 1"

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The answer is <math>\text{(A) 1}.</math>
 
The answer is <math>\text{(A) 1}.</math>
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==Solution==
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the numeretor is 3*5*7*9*11, so is the denominator so (3*5*7*9*11)/(3*5*7*9*11)
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:57, 19 February 2023

Problem

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

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

Solution

$\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}= \frac{3 \times 5 \times 7 \times 9 \times 11}{9 \times 11 \times 3 \times 5 \times 7} = \boxed{1}$

The answer is $\text{(A) 1}.$


Solution

the numeretor is 3*5*7*9*11, so is the denominator so (3*5*7*9*11)/(3*5*7*9*11)

Video Solution

https://youtu.be/dszCk0HVWH8

~savannahsolver

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 0
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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