# ratio problem

• Nov 3rd 2009, 08:17 PM
ziggy777
ratio problem
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.
• Nov 3rd 2009, 08:49 PM
ukorov
let r be radius of the base of cone,
so the original volume of oil = $\displaystyle \frac{1}{3}(\pi)(r^2)(50)$
the volume of remaining oil = $\displaystyle \frac{1}{3}(\pi)(r^2)(20)$

hence the volume of oil removed:
$\displaystyle \frac{50r^2}{3}(\pi) - \frac{20r^2}{3}(\pi)$
= $\displaystyle \frac{30r^2}{3}(\pi)$

hence, the percentage concerned:
($\displaystyle \frac{30r^2}{3}\pi$) / ($\displaystyle \frac{50r^2}{3}\pi$) x 100%
$\displaystyle \frac{30}{50}$ x 100%
= 60%
• Nov 3rd 2009, 09:13 PM
pflo
Quote:

Originally Posted by ukorov
let r be radius of the base of cone

But the radius is changing as the height decreases. I agree with Ziggy on this one.

$\displaystyle \frac{V_2}{V_1}=\frac{\frac{1}{3}\pi r^2 h}{\frac{1}{3}\pi R^2 H}=\frac{r^2h}{R^2H}$

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: $\displaystyle \frac{R}{H}=\frac{r}{h}$, which means $\displaystyle R=\frac{Hr}{h}$.

Plugging this in for $\displaystyle R$ in the volume ratio equation above, I get $\displaystyle \frac{V_2}{V_1}=\frac{h^2*r^2h}{H^2r^2*H}=\frac{h^ 3}{H^3}$. So the ratio of the volumes is the cube of the ratio of the heights, which is what Ziggy has done.