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Difficulty:
95%
(hard)
Question Stats:
29% (03:02) correct
71%
(02:51)
wrong
based on 59
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An hourglass is formed from two identical cones, one kept on top of the other. When the upper cone is full of sand and the lower one is empty, it takes an hour for the sand to flow, at a constant rate, from the upper cone to lower cone. How long does it take for the depth of sand in the lower cone to be one third of the depth of sand in the upper cone? (The volume of a cone is \(\pi r^2 \frac{h}{3}\))
A. 160/9 min
B. 20 min
C. 555/16 min
D. 380/9 min
E. 40 min
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A. 160/9 min
B. 20 min
C. 555/16 min
D. 380/9 min
E. 40 min
Are You Up For the Challenge: 700 Level Questions
_________________
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Posts: 1,052
Explanation:
When a cone is divided into n equal parts of height,
The ratio of heights of individual cones = 1:2:3......:n
The ratio of volumes of individual cones = 1³:2³:3³.....:n³
One-third volume is 19 ( from 1,7,19).
Now only one option is divisible by 19 i.e D.
IMO-D
When a cone is divided into n equal parts of height,
The ratio of heights of individual cones = 1:2:3......:n
The ratio of volumes of individual cones = 1³:2³:3³.....:n³
One-third volume is 19 ( from 1,7,19).
Now only one option is divisible by 19 i.e D.
IMO-D
Posts: 304
When sand in the lower cone is one third of the depth of sand in the upper cone the height will be one quarter of the total height.
We calculate x (radius of the base of the upper cone) taking into account that we have two similar triangles:
\(\frac{\frac{3h}{4}}{h}=\frac{x}{r}\)
so \(x=\frac{3r}{4}\)
Volume of the sand in the lower cone will be (substract upper cone from total cone):
\(\pi r^2\frac{h}{3}-\pi (\frac{3r}{4})^2\frac{\frac{3h}{4}}{3}=\pi r^2\frac{h}{3}(1-\frac{9}{16}*\frac{3}{4})=\pi r^2\frac{h}{3}(1-\frac{27}{64})=\pi r^2\frac{h}{3}*\frac{37}{64}\)
So the volume is \(\frac{37}{64}\) of the total volume. As the total volume last 60 minutes to fall the answer is:
\(\frac{37}{64}*60=\frac{555}{16} min\)
We calculate x (radius of the base of the upper cone) taking into account that we have two similar triangles:
\(\frac{\frac{3h}{4}}{h}=\frac{x}{r}\)
so \(x=\frac{3r}{4}\)
Volume of the sand in the lower cone will be (substract upper cone from total cone):
\(\pi r^2\frac{h}{3}-\pi (\frac{3r}{4})^2\frac{\frac{3h}{4}}{3}=\pi r^2\frac{h}{3}(1-\frac{9}{16}*\frac{3}{4})=\pi r^2\frac{h}{3}(1-\frac{27}{64})=\pi r^2\frac{h}{3}*\frac{37}{64}\)
So the volume is \(\frac{37}{64}\) of the total volume. As the total volume last 60 minutes to fall the answer is:
\(\frac{37}{64}*60=\frac{555}{16} min\)
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