1994 AMC 8 Problem 2

Attempt Problem 2 of the 1994 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1994 AMC 8 solutions, or check the answer key.

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2.

What is the value of the following expression?
110+210+310+410+510+610+710+810+910+5510 \begin{aligned} &\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} \\ &\quad {}+ \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} \\ &\quad {}+ \frac{9}{10} + \frac{55}{10} \end{aligned}
4124\dfrac{1}{2}
6.46.4
99
1010
1111
Answer: D
Concepts:fraction
Small hint:
All the terms share the denominator 1010, so add the numerators first
Big hint:
1+2+⋯+9=45,1 + 2 + \cdots + 9 = 45, and then add the 5555
Solution:
The numerators sum to 1+2+⋯+9+551 + 2 + \cdots + 9 + 55 =45+55= 45 + 55 =100.= 100.
So the total is 10010=10.\dfrac{100}{10} = 10.
Thus, the correct answer is D .
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