Difference between revisions of "2018 AMC 10A Problems/Problem 1"
m (→Solution) |
m (minor edit) |
||
(10 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
What is the value of | What is the value of | ||
− | <cmath>\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?</cmath><math>\textbf{(A) } \frac58 \qquad \textbf{(B) }\frac{11}7 \qquad \textbf{(C) } \frac85 \qquad \textbf{(D) } \frac{18}{11} \qquad \textbf{(E) } \frac{15}8 </math> | + | <cmath>\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?</cmath> |
+ | |||
+ | <math>\textbf{(A) } \frac58 \qquad \textbf{(B) }\frac{11}7 \qquad \textbf{(C) } \frac85 \qquad \textbf{(D) } \frac{18}{11} \qquad \textbf{(E) } \frac{15}8 </math> | ||
== Solution == | == Solution == | ||
− | < | + | <cmath> \left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 </cmath> |
+ | <cmath> =\left(\left(3)^{-1}+1\right)^{-1}+1\right)^{-1}+1 </cmath> | ||
+ | <cmath> =\left(\left(\frac{1}{3}+1\right)^{-1}+1\right)^{-1}+1 </cmath> | ||
+ | <cmath> =\left(\left(\frac{4}{3}\right)^{-1}+1\right)^{-1}+1 </cmath> | ||
+ | <cmath> =\left(\frac{3}{4}+1\right)^{-1}+1 </cmath> | ||
+ | <cmath> =\left(\frac{7}{4}\right)^{-1}+1 </cmath> | ||
+ | <cmath> =\frac{4}{7}+1 </cmath> | ||
+ | <cmath> =\frac{11}{7} </cmath> | ||
+ | Therefore, the answer is <math>\boxed{\textbf{(B) } \frac{11}{7} }</math>. | ||
+ | |||
+ | == Video Solutions == | ||
+ | https://youtu.be/vO-ELYmgRI8 | ||
+ | |||
+ | https://youtu.be/cat3yTIpX4k | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == | ||
− | |||
{{AMC10 box|year=2018|ab=A|before=First Problem|num-a=2}} | {{AMC10 box|year=2018|ab=A|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:45, 19 October 2020
Contents
Problem
What is the value of
Solution
Therefore, the answer is .
Video Solutions
~savannahsolver
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.