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Senior Manager  P
Joined: 13 Jan 2018
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What is the height of the conical bottle?  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 38% (02:05) correct 62% (02:19) wrong based on 81 sessions

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Attachment: image.jpg [ 18.5 KiB | Viewed 1598 times ]

In the figure above, when a conical bottle is rested on its flat base, the water in the bottle is 8 cm from its vertex. When the same conical bottle is turned upside down, the water level is 2 cm from its base. What is the approximate height of the bottle?

A. 5cm
B. 8cm
C. 10cm
D. 12cm
E. 15cm
Senior Manager  P
Joined: 13 Jan 2018
Posts: 341
Location: India
Concentration: Operations, General Management
GMAT 1: 580 Q47 V23 GMAT 2: 640 Q49 V27 GPA: 4
WE: Consulting (Consulting)
Re: What is the height of the conical bottle?  [#permalink]

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Attachment: image 1.jpg [ 32.27 KiB | Viewed 1514 times ]

Consider the image shown above.

Let the height of the cone be "h" and "R" be the radius of the cone. When the bottle is rested on its flat base, the volume of water is given by

Volume of the cone - volume of empty space above the water level

$$\frac{\pi}{3} R^2 h$$ - $$\frac{\pi}{3} r_1^2 (8)$$

Now by similar triangle concept:

$$\frac{r_1}{8} = \frac{R}{h}$$

$$r_1 = \frac{8R}{h}$$

Sub $$r_1$$ value in the above equation, we get

volume of water = $$\frac{\pi}{3} R^2 h - \frac{\pi}{3} (\frac{8R}{h})^2 (8)$$

$$\frac{\pi}{3} R^2 (h - \frac{512}{h^2})$$ ......................................... (1)

Now consider the cone is tilted upside down. Volume of water is

$$\frac{\pi}{3} (r_2)^2 (h-2)$$

Again by similar triangle concept:

$$\frac{r_2}{h-2} = \frac{R}{h}$$

$$r_2 = \frac{(h-2) R}{h}$$

Substituting $$r_2$$ in the above equation of volume of water when cone is tilted,

volume of water = $$\frac{\pi}{3} (\frac{(h-2) R}{h})^2 (h-2)$$

$$\frac{\pi}{3} R^2 \frac{(h-2)^3}{h^2}$$ ..................................... (2)

Since volume of water in both the cases is equal, (1) = (2)

$$\frac{\pi}{3} R^2 (h - \frac{512}{h^2})$$ = $$\frac{\pi}{3} R^2 \frac{(h-2)^3}{h^2}$$

$$h - \frac{512}{h^2} = \frac{(h-2)^3}{h^2}$$

$$h^3 - 512 = (h-2)^3$$

$$h^3 - 512 = h^3 - 6h^2 + 12h - 8$$

$$6h^2 - 12h - 504 = 0$$

$$h^2 - 2h - 84 = 0$$

Since $$h > 0 , h \approx {10 cm}$$

OPTION : C
Intern  B
Joined: 23 Oct 2017
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Re: What is the height of the conical bottle?  [#permalink]

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Hi,
Just thinking out loud (and this may sound like a stupid question). Would I be wrong if I just went straight ahead to add the height of the spaces 8cm + 2cm ?

cc: Bunuel, GMATNinja

PS: I've been struggling with the compulsion to rush into doing extreme math on my test prep and I am try to break away from that habit.
Intern  B
Joined: 02 Jul 2014
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Re: What is the height of the conical bottle?  [#permalink]

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1
2
The easiest way will be to equate the volume of the water in the 2 cases.

In the first case volume of water =1/3(pi)r^2h -1/3(pi)r^2.8 (r- radius of cone, h- height of cone)

In the second volume of water = 1/3(pi)r^2h - 1/3(pi)r^2(h-2)

equating the two, and cancelling 8 = h -2, or h=10
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Re: What is the height of the conical bottle?  [#permalink]

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ronnieshee
In the second volume of water = 1/3(pi)r^2h - 1/3(pi)r^2(h-2)

can you please explain the second case in details?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
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Location: Pune, India
Re: What is the height of the conical bottle?  [#permalink]

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eswarchethu135 wrote:
Attachment:
image.jpg

In the figure above, when a conical bottle is rested on its flat base, the water in the bottle is 8 cm from its vertex. When the same conical bottle is turned upside down, the water level is 2 cm from its base. What is the approximate height of the bottle?

A. 5cm
B. 8cm
C. 10cm
D. 12cm
E. 15cm

This is not a GMAT question.

Say the height of the bottle is H and radius is R.
The volume of the water in the two cases is same.

Case 1: Volume of water = Volume of cone - Volume of empty cone on top
Volume of water = $$(\frac{1}{3})*\pi*R^2*H - (\frac{1}{3})*\pi*[(8/H)*R]^2*8$$

Case 2: Volume of water = Volume of smaller cone
Volume of water = $$(\frac{1}{3})*\pi*[((H-2)/H)R]^2 *(H - 2)$$

$$\frac{1}{3}*\pi*R^2*H - \frac{1}{3}*\pi*[\frac{8}{H}*R]^2*8 = \frac{1}{3}*\pi*[\frac{H-2}{H}*R]^2 *(H - 2)$$

$$R^2*H - [\frac{8}{H}*R]^2*8 = [\frac{H-2}{H}*R]^2 *(H - 2)$$

$$H - \frac{8^3}{H^2} = \frac{(H - 2)^3}{H^2}$$

$$H^3 - 8^3 = (H - 2)^3$$

The value of H for which this would hold would be about 10.22 and hence 10 would be the best answer.
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Karishma
Veritas Prep GMAT Instructor

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Re: What is the height of the conical bottle?  [#permalink]

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In either diagram, larger triangle forms a similar triangle with the smaller triangle.

Now, assume height of cone as h. So, height of smaller triangle on left is 8 and on the right is (h-2)

As volumes of the liquid and the cone is same in either diagram, ratio of [volume of liquid : volume of cone] should also be constant in each. Hence, sides will also be in proportion i.e. h/8 = h/(h-2)
Solving, we get h= 0 or 10. As h cannot be 0, Ans is 10 / C Re: What is the height of the conical bottle?   [#permalink] 13 Nov 2018, 11:27
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