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A cylindrical can placed on a table leaves a water ring that is 10% la

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A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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New post 02 Sep 2018, 23:14
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Question Stats:

83% (01:59) correct 17% (01:31) wrong based on 77 sessions

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A cylindrical can placed on a table leaves a water ring that is 10% larger than the circumference of the bottom of the can. If the area of the bottom of the can is 25π, which of the following is area encircled by the water ring?


A. \(121\pi\)
B. \(110\pi\)
C. \(55\pi\)
D. \(30.25\pi\)
E. \(11\pi\)

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Re: A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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New post 02 Sep 2018, 23:19
Shouldn't the answer be 30.25pi

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A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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New post Updated on: 03 Sep 2018, 23:26
1
25 pi = pi *r^2

r = 5

2*pi*r (circumference of can) = 10*pi

water ring that is 10% larger = 1.1(10*pi)
= 11*pi

2*pi*R = 11*pi
R= 11/2

The area is pi∗(11/2)^2 = 30.25*pi

Answer is D

Thanks PKN

Originally posted by Natty97 on 03 Sep 2018, 09:23.
Last edited by Natty97 on 03 Sep 2018, 23:26, edited 1 time in total.
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Re: A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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New post 03 Sep 2018, 19:23
1
Natty97 wrote:
25 pi = pi *r^2

r = 5

2*pi*r (circumference of can) = 10*pi

water ring that is 10% larger = 1.1(10*pi)
= 11*pi<<-------------This is the measure of circumference of water ring. We are asked to determine the area of water ring.

Answer is E


Hi Natty97,
Please refer the highlighted portion above, circumference of water ring=\(11\pi\)
Or, \(2\pi\)*radius of circular water ring=\(11\pi\)
Or, Radius of water ring=\(\frac{11}{2}\)

So, Area of water ring=\(\pi*(\frac{11}{2})^2=30.25\pi\)

Ans. (D)
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Re: A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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New post 03 Sep 2018, 19:46
Bunuel wrote:
A cylindrical can placed on a table leaves a water ring that is 10% larger than the circumference of the bottom of the can. If the area of the bottom of the can is 25π, which of the following is area encircled by the water ring?


A. \(121\pi\)
B. \(110\pi\)
C. \(55\pi\)
D. \(30.25\pi\)
E. \(11\pi\)



Circumference is 2πr..
So if we increase circumference by 10%, the effect will be on the radius only as other term 2π is constant...
So r becomes 1.1r

Now initial area =\(\pi*r^2=25\pi\)
And the new area is \(π*(1.1r)^2=π*1.21*r^2=1.21*πr^2=1.21*25=30.25\)

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Re: A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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New post 03 Sep 2018, 23:28
Natty97 wrote:
25 pi = pi *r^2

r = 5

2*pi*r (circumference of can) = 10*pi

water ring that is 10% larger = 1.1(10*pi)
= 11*pi

2*pi*R = 11*pi
R= 11/2

The area is pi∗(11/2)^2 = 30.25*pi

Answer is D

Thanks PKN


You are welcome.
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Re: A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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New post 13 Jan 2019, 10:15
Bunuel wrote:
A cylindrical can placed on a table leaves a water ring that is 10% larger than the circumference of the bottom of the can. If the area of the bottom of the can is 25π, which of the following is area encircled by the water ring?


A. \(121\pi\)
B. \(110\pi\)
C. \(55\pi\)
D. \(30.25\pi\)
E. \(11\pi\)



pi * r2 = 25 pi
r= 5

1.1* 5 = 5.5
5.5^2 * pi = 30.25 pi
IMO D
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Re: A cylindrical can placed on a table leaves a water ring that is 10% la   [#permalink] 13 Jan 2019, 10:15
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