Difference between revisions of "2020 AMC 12B Problems/Problem 2"

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== Solution ==
 
== Solution ==
 
Using difference of squares to factor the left term, we get
 
Using difference of squares to factor the left term, we get
<cmath>\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = </cmath>
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<cmath>\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = \frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}.</cmath>
<cmath>\frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}</cmath>
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Cancelling all the terms, we get <math>\boxed{\textbf{(A) } 1}</math> as the answer.
Cancelling all the terms, we get <math>\boxed{\textbf{(A) 1}}</math> as the answer.
 
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 01:18, 12 July 2021

Problem

What is the value of the following expression?

\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}\]

$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{9950} \qquad \textbf{(C) } \frac{4780}{4779} \qquad \textbf{(D) } \frac{108}{107} \qquad \textbf{(E) } \frac{81}{80}$

Solution

Using difference of squares to factor the left term, we get \[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = \frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}.\] Cancelling all the terms, we get $\boxed{\textbf{(A) } 1}$ as the answer.

Video Solution

https://youtu.be/WfTty8Fe5Fo

~IceMatrix

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 AMC 12 Problems and Solutions

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