Difference between revisions of "2020 AMC 12B Problems/Problem 2"
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| − | ==Problem== | + | == Problem == |
| + | What is the value of the following expression? | ||
| − | |||
<cmath>\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}</cmath><math>\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} </math> | <cmath>\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}</cmath><math>\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} </math> | ||
| − | ==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> | <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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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 == |
https://youtu.be/WfTty8Fe5Fo | https://youtu.be/WfTty8Fe5Fo | ||
~IceMatrix | ~IceMatrix | ||
| − | ==See Also== | + | == See Also == |
| − | |||
{{AMC12 box|year=2020|ab=B|num-b=1|num-a=3}} | {{AMC12 box|year=2020|ab=B|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:29, 19 October 2020
Contents
Problem
What is the value of the following expression?
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Solution
Using difference of squares to factor the left term, we get
Cancelling all the terms, we get
as the answer.
Video Solution
~IceMatrix
See Also
| 2020 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.