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Math Expert V
Joined: 02 Sep 2009
Posts: 58445
A cylinder with a volume of 80π is altered so that its radius becomes  [#permalink]

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Difficulty:   15% (low)

Question Stats: 80% (01:25) correct 20% (00:47) wrong based on 47 sessions

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A cylinder with a volume of 80π is altered so that its radius becomes half of the original radius and its height becomes double the original height. What is the volume of the altered cylinder?

A. $$160\pi$$

B. $$80\pi$$

C. $$40\pi$$

D. $$20\pi$$

E. $$10\pi$$

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GMAT Club Legend  D
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Posts: 5021
Location: India
Concentration: Sustainability, Marketing
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Re: A cylinder with a volume of 80π is altered so that its radius becomes  [#permalink]

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A cylinder with a volume of 80π is altered so that its radius becomes half of the original radius and its height becomes double the original height. What is the volume of the altered cylinder?

A. 160π160π

B. 80π80π

C. 40π40π

D. 20π20π

E. 10π10π

original area : 3.14*r^2*h: 80*3.14
We can deduce
=> r^2*h: 80-----(1)

Area to be found: 3.14*(r/2)^2*2h -----(2)

substitute value from (1) in (2)
=> 3.14*(80*2)/4
=> answer 40 *3.14 option C
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A cylinder with a volume of 80π is altered so that its radius becomes  [#permalink]

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Bunuel wrote:
A cylinder with a volume of 80π is altered so that its radius becomes half of the original radius and its height becomes double the original height. What is the volume of the altered cylinder?

A. $$160\pi$$

B. $$80\pi$$

C. $$40\pi$$

D. $$20\pi$$

E. $$10\pi$$

Given, $$r^'$$=$$\frac{r}{2}$$ and $$h^'=2h$$
Volume of altered cylinder= $$\pi*r^'^2*h^'$$=$$\pi*(\frac{r}{2})^2*2h$$=$$\frac{1}{2}*\pi*r^2*h)$$=$$\frac{1}{2}*80\pi$$= $$40\pi$$

Ans. (C)
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Target Test Prep Representative G
Status: Head GMAT Instructor
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Posts: 2815
Re: A cylinder with a volume of 80π is altered so that its radius becomes  [#permalink]

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Bunuel wrote:
A cylinder with a volume of 80π is altered so that its radius becomes half of the original radius and its height becomes double the original height. What is the volume of the altered cylinder?

A. $$160\pi$$

B. $$80\pi$$

C. $$40\pi$$

D. $$20\pi$$

E. $$10\pi$$

We can let r = the original radius and h = the original height of the cylinder. So we have:

πr^2h = 80π

r^2h = 80

Now, we can make up a convenient number for r and calculate h. For example, we can let r = 4, so we have:

4^2h = 80

16h = 80

h = 5

Now, let’s halve the radius and double the height, so the new radius is 2 and the new height is 10. Therefore, the new volume is:

π x 2^2 x 10 = 40π

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: A cylinder with a volume of 80π is altered so that its radius becomes   [#permalink] 03 Sep 2018, 19:05
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