let r be radius of the base of cone,
so the original volume of oil =
the volume of remaining oil =
hence the volume of oil removed:
=
hence, the percentage concerned:
( ) / ( ) x 100%
x 100%
= 60%
OK here is a problem that was in my exam today. Just wanna check if I've done it correct.
An inverted cone with a height of 50 cm was originally filled with oil. Then some of the oil was removed and now the remaining oil has a height of 20 cm. find the percantage of the oil that was removed.
this is how i did it-
length ratio- 20:50 = 2:5 (oil:cone)
volume ratio- 8:125. (oil:cone)
oil removed- 125-8 =117
% oil removed- 11700/125 = 93.6%
is this method correct????
NOTE
some students did- 3:5 (removed:cone) and calculated the percentage.
But the radius is changing as the height decreases. I agree with Ziggy on this one.
Using similar triangles with the sides being the side of the cone and the heights and radii of the liquid, you can establish the following ratio: , which means .
Plugging this in for in the volume ratio equation above, I get . So the ratio of the volumes is the cube of the ratio of the heights, which is what Ziggy has done.
So Ziggy, if you miss this problem see what your teacher says about this explanation. I'd be interested to hear that answer (please post it).
well well...
I cannot say how GLAD i'm by hearing that ukorov is wrong!!! (sorry ukorov)
yes as pflo says the radius decreases as the height decreases. and we were tought the formula that if length ratio is x:y volume ratio is x^3:y^3
honestly 93.6% sounds a bit too much.
bad news - the exam i spoke of was the VCE exam (the end of year exam for year 12 in Victoria, Australia). So it would take days to get a correct answer.
can anyone else confirm my answer?