1987 AMC 8 Problem 11

Attempt Problem 11 of the 1987 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 1987 AMC 8 solutions, or check the answer key.

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11.

The sum 217+312+51192\dfrac17 + 3\dfrac12 + 5\dfrac{1}{19} is between which two values?

1010 and 101210\dfrac12

101210\dfrac12 and 1111

1111 and 111211\dfrac12

111211\dfrac12 and 1212

1212 and 121212\dfrac12

Answer: B
Concepts:fractionestimation

Difficulty rating: 860

Solution:

The whole-number parts sum to 10.10. The fractional parts are 12+17+119.\dfrac12 + \dfrac17 + \dfrac{1}{19}.

Since 17+119\dfrac17 + \dfrac{1}{19} is a small positive amount, the fractional total is more than 12\dfrac12 but well under 1.1. So the sum lies between 101210\dfrac12 and 11.11.

Thus, the correct answer is B .

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