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AnkitK
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The figure represents a deflated tire (6 inches wide as show 02 Jun 2011, 03:09
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D
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The figure represents a deflated tire (6 inches wide as shown) with a hub (the center circle). The area of the hub surface shown in the picture is 1/3 the area of the tire surface shown in the picture. The thickness of the tire, when fully inflated is 3 inches. (Assume the tire material itself has negligible thickness.) Air is filled into the deflated tire at a rate of 4π inches3 / second. How long (in seconds) will it take to inflate the tire?

A. 24
B. 27
C. 48
D. 81
E. 108

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D
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jamifahad
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The figure represents a deflated tire (6 inches wide as show 02 Jun 2011, 04:53
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Is it B? 27.

Please let me know OA otherwise no point detailing calculation here if my answer is not correct.
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bellcurve
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The figure represents a deflated tire (6 inches wide as show 02 Jun 2011, 06:35
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Rt = radius of the tire including hub & Rh = radius of the hub
eq 1 => Rt-6=Rh
eq 2 => 4 pi Rh^2 = pi Rt^2
Solving we get Rt = 12 or 4
4 can't be right so the Rt is 12
so the area of the tire is 108
the volume of the air in the tire = 108*3
time it takes to fill up = (108*3)/4 = 81

Shorter way would be:

pi Rt^2:pi Rh^2 = 4:1 pi cancels. take square root of both sides
Rt:Rh = 2:1
since Rt-Rh=6, Rt = 12
after this same as above.
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fluke
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The figure represents a deflated tire (6 inches wide as show 02 Jun 2011, 06:46
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Sol:

Volume of tire = (Volume of tire+hub)-Volume of hub

Let the radius of the hub\(=r\)

Area of hub\(=\pi*r^2\)

Area of tire+hub \(= \pi*(r+6)^2\)

Area of just tire\(=\pi*(r+6)^2-\pi*r^2=\pi(r^2+12r+36-r^2)=\pi(12r+36)\)

\(\frac{Area of tire}{Area of hub}=3\)

\(\frac{\pi(12r+36)}{\pi*r^2}=3\)

\(12r+36=3r^2\)

\(3r^2-12r-36=0\)

\(r^2-4r-12=0\)

Upon solving;
r=6 or r=-2

Radius can't be negative; r=6

Now,
Volume of tire=Total Volume-Hub Volume
Total Volume\(=\pi*r^2*h=\pi*(r+6)^2*h=\pi*12^2*h=144 \pi*h=144*\pi*3=432*\pi\) (Note: h=3(given))

Hub Volume\(=\pi*r^2*h=\pi*(r)^2*h=\pi*6^2*h=36 \pi*h=36*\pi*3=108*\pi\)

Volume of tire\(=432 \pi-108 \pi=324*\pi\)

This is what we wanted.
\(4\pi -> 1 sec\)

\(1 -> \frac{1}{4\pi} sec\)

\(324\pi -> \frac{324\pi}{4\pi} sec=81 sec\)

Ans: "D"
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The figure represents a deflated tire (6 inches wide as show 02 Jun 2011, 07:50
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fluke
Volume of tire=Total Volume-Hub Volume
Show more

Ah! this is what i wasn't doing.

Thanks.
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toughmat
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The figure represents a deflated tire (6 inches wide as show 06 Jun 2011, 23:02
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Good question I got D too
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puneetj
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The figure represents a deflated tire (6 inches wide as show 11 Jun 2011, 11:20
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D for me...did the same way as fluke

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