1989 AMC 8 Problem 1 — Solution | LIVE by Po-Shen Loh

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1989 AMC 8 Problem 1

Attempt Problem 1 of the 1989 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 1989 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

What is the value of

$\begin{aligned} &(1 + 11 + 21 + 31 + 41) \\ &\quad {}+ (9 + 19 + 29 + 39 + 49)? \end{aligned}$

$150$

$199$

$200$

$249$

$250$

Answer: E

Concepts:pairing and grouping

Difficulty rating: 560

Solution:

Pair the terms so each pair sums to $50$ : $1 + 49,$ $11 + 39,$ $21 + 29,$ $31 + 19,$ and $41 + 9.$

There are $5$ such pairs, so the total is $5 \times 50 = 250.$

Thus, the correct answer is E .

Full Exam Problem 2#2 →

Problem 1 in Other Years

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