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

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Math Expert
Joined: 02 Sep 2009
Posts: 52296
A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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02 Sep 2018, 22:14
00:00

Difficulty:

25% (medium)

Question Stats:

72% (01:30) correct 28% (00:33) wrong based on 69 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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02 Sep 2018, 22:19

Posted from my mobile device
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Joined: 16 Jul 2017
Posts: 24
Location: India
Concentration: Finance, Economics
A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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Updated on: 03 Sep 2018, 22:26
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

Thanks PKN

Originally posted by Natty97 on 03 Sep 2018, 08:23.
Last edited by Natty97 on 03 Sep 2018, 22:26, 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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03 Sep 2018, 18:23
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.

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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Joined: 02 Aug 2009
Posts: 7209
Re: A cylindrical can placed on a table leaves a water ring that is 10% la  [#permalink]

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03 Sep 2018, 18:46
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 $$π*(1.1r)^2=π*1.21*r^2=1.21*πr^2=1.21*25=30.25$$

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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03 Sep 2018, 22:28
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

Thanks PKN

You are welcome.
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

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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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13 Jan 2019, 09:15
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 &nbs [#permalink] 13 Jan 2019, 09:15
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