olymmath

AMC 8 1994 Problem 2

  • Paper

110+210+310+410+510+610+710+810+910+5510=\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{55}{10}= (A) 412(B) 6.4(C) 9(D) 10(E) 11\text{(A)}\ 4\dfrac{1}{2} \text{(B)}\ 6.4 \text{(C)}\ 9 \text{(D)}\ 10 \text{(E)}\ 11

Dataset text NuminaMath-1.5. Quoted with source.

Show the answer

10

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Similar problems

These problems have the most words in common with this statement.

Locator amc8/1994/p2 · Tag F69T7AAYMX · Edit this page on GitHub · Keys: j next, k previous