Difference between revisions of "1991 AHSME Problems/Problem 8"

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== Solution ==
 
== Solution ==
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<math>\fbox{C}</math> The volume of liquid <math>X</math> is <math>6 \times 3 \times 12 = 216</math>, so if the radius is <math>r</math>, then using the formula for the volume of a cylinder, we get <math>\pi r^2 \times 0.1 = 216 \implies \pi r^2 = 2160 \implies r = \sqrt{\frac{2160}{\pi}}.</math>
  
 
== See also ==
 
== See also ==

Revision as of 17:21, 23 February 2018

Problem

Liquid $X$ does not mix with water. Unless obstructed, it spreads out on the surface of water to form a circular film $0.1$cm thick. A rectangular box measuring $6$cm by $3$cm by $12$cm is filled with liquid $X$. Its contents are poured onto a large body of water. What will be the radius, in centimeters, of the resulting circular film?

(A) $\frac{\sqrt{216}}{\pi}$ (B) $\sqrt{\frac{216}{\pi}}$ (C) $\sqrt{\frac{2160}{\pi}}$ (D) $\frac{216}{\pi}$ (E) $\frac{2160}{\pi}$

Solution

$\fbox{C}$ The volume of liquid $X$ is $6 \times 3 \times 12 = 216$, so if the radius is $r$, then using the formula for the volume of a cylinder, we get $\pi r^2 \times 0.1 = 216 \implies \pi r^2 = 2160 \implies r = \sqrt{\frac{2160}{\pi}}.$

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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