1991 AMC 8 Problem 25

Attempt Problem 25 of the 1991 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 1991 AMC 8 solutions, or check the answer key.

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

25.

An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black?
11024\dfrac{1}{1024}
1564\dfrac{15}{64}
2431024\dfrac{243}{1024}
14\dfrac14
81256\dfrac{81}{256}
Answer: C
Concepts:geometric sequencearea ratio
Small hint:
After one change the middle fourth turns white, so 34\dfrac34 of the black area remains
Big hint:
Each change multiplies the black area by 34,\dfrac34, so after five changes multiply by (34)5\left(\dfrac34\right)^5
Solution:
Each change leaves 34\dfrac34 of the current black area black. After five changes the black fraction is (34)5=2431024.\left(\dfrac34\right)^5 = \dfrac{243}{1024}.
Thus, the correct answer is C .
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