Difference between revisions of "2020 AMC 12B Problems/Problem 2"
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− | ==Problem== | + | == Problem == |
+ | What is the value of the following expression? | ||
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<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 == |
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{{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?
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.