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# 80 pie square inches of material of negligible thickness are required

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Joined: 02 Sep 2009
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80 pie square inches of material of negligible thickness are required  [#permalink]

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12 Jun 2019, 23:59
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Difficulty:

95% (hard)

Question Stats:

40% (02:53) correct 60% (03:01) wrong based on 45 sessions

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80 pie square inches of material of negligible thickness are required to construct a 1/16 scale model of a cylindrical barrel. If the diameter of the base of the barrel is 160 inches, then what is the volume of the barrel, to the nearest cubic foot?

A. $$2\pi$$

B. $$11\pi$$

C. $$178\pi$$

D. $$711\pi$$

E. $$1280\pi$$

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80 pie square inches of material of negligible thickness are required  [#permalink]

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18 Jun 2019, 21:31
5
Hungluu92vn azoz7 Aim015 ayush19 stne
Here is how i solved it:
Original barrel diameter = 160 inches, hence radius = 80 inches
Now we are told scaled model is 1/16 times original. Hence the radius of scale model is 80/16 = 5 inches.
Now surface area = 2(pi)r*h + 2*pi*r^2 = 80*pi
substitute r = 5 in above formula we get h = 3 inches.
Now this is the height of the model.
So height of original barrel is 3*16 = 48 inches.

Now we know height and radius of original barrel. so we can convert into feet, since final answer is required in feet.
Height = 48 inches = 4 feet
Radius = 80 inches = 20/3 feet.

Now volume of barrel = pi*r^2*h = pi*(20/3)^2*4 = (1600/9)*pi = ~178*pi

Please give kudos if you like the explanation.
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Re: 80 pie square inches of material of negligible thickness are required  [#permalink]

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13 Jun 2019, 02:23
1
Bunuel wrote:
80 pie square inches of material of negligible thickness are required to construct a 1/16 scale model of a cylindrical barrel. If the diameter of the base of the barrel is 160 inches, then what is the volume of the barrel, to the nearest cubic foot?

A. $$2\pi$$

B. $$11\pi$$

C. $$178\pi$$

D. $$711\pi$$

E. $$1280\pi$$

2*pi*r*(h+r)= 80*pi*16*16
solve for h = 48 inches = 4 feet
and radius = 80 inches ; 20/3 feet
so vol ; pi * r^2*h = pi*20*20/9 * 4 = 178 pi
IMO C
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Re: 80 pie square inches of material of negligible thickness are required  [#permalink]

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14 Jun 2019, 04:52
Archit3110 wrote:
Bunuel wrote:
80 pie square inches of material of negligible thickness are required to construct a 1/16 scale model of a cylindrical barrel. If the diameter of the base of the barrel is 160 inches, then what is the volume of the barrel, to the nearest cubic foot?

A. $$2\pi$$

B. $$11\pi$$

C. $$178\pi$$

D. $$711\pi$$

E. $$1280\pi$$

2*pi*r*(h+r)= 80*pi*16*16
solve for h = 48 inches = 4 feet
and radius = 80 inches ; 20/3 feet
so vol ; pi * r^2*h = pi*20*20/9 * 4 = 178 pi
IMO C

Hi, How come you have multiplied 2 16s in your first line ?? Could you please explain that part?
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Joined: 27 May 2012
Posts: 814
80 pie square inches of material of negligible thickness are required  [#permalink]

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Updated on: 16 Jun 2019, 01:17
Archit3110 wrote:
Bunuel wrote:
80 pie square inches of material of negligible thickness are required to construct a 1/16 scale model of a cylindrical barrel. If the diameter of the base of the barrel is 160 inches, then what is the volume of the barrel, to the nearest cubic foot?

A. $$2\pi$$

B. $$11\pi$$

C. $$178\pi$$

D. $$711\pi$$

E. $$1280\pi$$

2*pi*r*(h+r)= 80*pi*16*16
solve for h = 48 inches = 4 feet
and radius = 80 inches ; 20/3 feet
so vol ; pi * r^2*h = pi*20*20/9 * 4 = 178 pi
IMO C

Can anyone else explain this in a better manner, sorry, this solution is not making sense to me.
Why are we multiplying by 16 twice ?
The total surface area is $$2\pi r*h +2\pi r^2$$ this is equal to $$80\pi$$, right? And this is 1/16 of the actual thing or what ????
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- Stne

Originally posted by stne on 15 Jun 2019, 00:55.
Last edited by stne on 16 Jun 2019, 01:17, edited 1 time in total.
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Re: 80 pie square inches of material of negligible thickness are required  [#permalink]

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15 Jun 2019, 08:32
Bunuel wrote:
80 pie square inches of material of negligible thickness are required to construct a 1/16 scale model of a cylindrical barrel. If the diameter of the base of the barrel is 160 inches, then what is the volume of the barrel, to the nearest cubic foot?

A. $$2\pi$$

B. $$11\pi$$

C. $$178\pi$$

D. $$711\pi$$

E. $$1280\pi$$

Hi Bunuel, Could you please explain the solution of this problem?
Intern
Joined: 11 May 2015
Posts: 2
Re: 80 pie square inches of material of negligible thickness are required  [#permalink]

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15 Jun 2019, 11:03
ayush19 wrote:
Archit3110 wrote:
Bunuel wrote:
80 pie square inches of material of negligible thickness are required to construct a 1/16 scale model of a cylindrical barrel. If the diameter of the base of the barrel is 160 inches, then what is the volume of the barrel, to the nearest cubic foot?

A. $$2\pi$$

B. $$11\pi$$

C. $$178\pi$$

D. $$711\pi$$

E. $$1280\pi$$

2*pi*r*(h+r)= 80*pi*16*16
solve for h = 48 inches = 4 feet
and radius = 80 inches ; 20/3 feet
so vol ; pi * r^2*h = pi*20*20/9 * 4 = 178 pi
IMO C

Hi, How come you have multiplied 2 16s in your first line ?? Could you please explain that part?
Manager
Joined: 07 May 2016
Posts: 60
Re: 80 pie square inches of material of negligible thickness are required  [#permalink]

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15 Jun 2019, 21:44
Interesting question. I always struggle with questions that use difficult language.
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Location: Viet Nam
GPA: 3.5
WE: General Management (Education)
Re: 80 pie square inches of material of negligible thickness are required  [#permalink]

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16 Jun 2019, 00:56
BM and waiting for the answer
Intern
Joined: 27 Nov 2017
Posts: 19
Location: India
GMAT 1: 580 Q47 V25
GPA: 4
Re: 80 pie square inches of material of negligible thickness are required  [#permalink]

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20 Jun 2019, 07:39
ruchik wrote:
Hungluu92vn azoz7 Aim015 ayush19 stne
Here is how i solved it:
Original barrel diameter = 160 inches, hence radius = 80 inches
Now we are told scaled model is 1/16 times original. Hence the radius of scale model is 80/16 = 5 inches.
Now surface area = 2(pi)r*h + 2*pi*r^2 = 80*pi
substitute r = 5 in above formula we get h = 3 inches.
Now this is the height of the model.
So height of original barrel is 3*16 = 48 inches.

Now we know height and radius of original barrel. so we can convert into feet, since final answer is required in feet.
Height = 48 inches = 4 feet
Radius = 80 inches = 20/3 feet.

Now volume of barrel = pi*r^2*h = pi*(20/3)^2*4 = (1600/9)*pi = ~178*pi

Please give kudos if you like the explanation.

Thanks ruchik for the explanation.
Re: 80 pie square inches of material of negligible thickness are required   [#permalink] 20 Jun 2019, 07:39
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