Problem: What is 10⋅(12+15+110)−110 \cdot\left(\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{10}\right)^{-1}10⋅(21+51+101)−1 ?
Answer Choices:
A. 333 B. 888 C. 252\dfrac{25}{2}225 D. 1703\dfrac{170}{3}3170 E. 170170170 Solution:
Note that
10⋅(12+15+110)−1=10⋅(810)−1=25210 \cdot\left(\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{10}\right)^{-1}=10 \cdot\left(\dfrac{8}{10}\right)^{-1}= \boxed{\dfrac{25}{2}} 10⋅(21+51+101)−1=10⋅(108)−1=225
The problems on this page are the property of the MAA's American Mathematics Competitions