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The figure represents a deflated tire 6 inches wide as shown

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The figure represents a deflated tire 6 inches wide as shown  [#permalink]

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New post Updated on: 14 Jul 2013, 02:39
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Question Stats:

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Attachment:
Tire.PNG
Tire.PNG [ 4.47 KiB | Viewed 8277 times ]
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

Ok - guys this is how I started to solve this question.

To solve this problem we need to (a) find the volume of the tire and then (b) solve a rate problem to determine how long it will take to inflate the tire.

If the radius of the hub is r, then its area equals pi \(r^2\)
The area of the entire object is then pi \((r+6) ^ 2\)
This means that the area of just the tire equals pi \((r+6)^2\)- pi \(r^2\)
The problem also tells us that the ratio of the area of the tire to the area of the entire object is 1/3. We can use this information to set up the following equation:
area of hub/area of tyre = 1/3 = pi \(r^2\)/pi \((r+6)^2\)- pi \(r^2\) => r^2 - 4r - 36 = 0. This will give r = 6 as radius can never be negative.

Now I am stuck and lost. Can someone please help?

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Originally posted by enigma123 on 13 Feb 2012, 16:42.
Last edited by Bunuel on 14 Jul 2013, 02:39, edited 2 times in total.
Edited the question
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Re: The figure represents a deflated tire 6 inches wide as shown  [#permalink]

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New post 15 Feb 2012, 01:36
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enigma123 wrote:
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
Attachment:
Tire.PNG


Ok - guys this is how I started to solve this question.

To solve this problem we need to (a) find the volume of the tire and then (b) solve a rate problem to determine how long it will take to inflate the tire.

If the radius of the hub is r, then its area equals pi \(r^2\)
The area of the entire object is then pi \((r+6) ^ 2\)
This means that the area of just the tire equals pi \((r+6)^2\)- pi \(r^2\)
The problem also tells us that the ratio of the area of the tire to the area of the entire object is 1/3. We can use this information to set up the following equation:
area of hub/area of tyre = 1/3 = pi \(r^2\)/pi \((r+6)^2\)- pi \(r^2\) => r^2 - 4r - 36 = 0. This will give r = 6 as radius can never be negative.

Now I am stuck and lost. Can someone please help?


The question itself is long and the proper solution is much longer. In such cases, try and think more creatively (practice on practice questions. In the exam do whatever comes to your mind first)
(omerrauf has given you the complete solution so I am not going to repeat it.)

I will talk of a more creative solution.

You need to find the surface area of the tire and multiply it by the thickness to get the volume of the air in the tire. Then you need to divide it by 4π (rate at which air is filled) to get the time taken.

Say the radius of big circle is R and that of the little circle (hub) is r.
Given:\({\pi}r^2 = (1/3){\pi}(R^2 - r^2)\)
or \(r^2 = R^2/4\)

Focus on what we need now: \({\pi}R^2 - {\pi}r^2\) = surface area of tire = \({\pi}R^2 - {\pi}R^2/4 = 3{\pi}R^2/4\)

Volume of air = \((3{\pi}R^2/4) * 3\)

Time taken = \((3{\pi}R^2/4) *3/4{\pi} = 9R^2/16\)

Your options give you the value of \(9R^2/16\). This means that if you multiply your options by 16 and divide by 9, you will get R^2.
16 is already a perfect square so ignore it. Divide your options by 9 and see what you get.

A. 24 (Doesn't get divided by 9. 16*24/9 cannot be R^2.)
B. 27 (You are left with 3. 16*3 cannot be R^2)
C. 48 (Doesn't get divided by 9. 48/9 cannot be R^2)
D. 81 (You are left with 9 which is a perfect square. 9*16 gives R as 12. Possible answer)
E. 108 (You are left with 12. 12 is not a perfect square.)

You might be tempted to say that 16*3 can be R^2 since \(R = 4\sqrt{3}\) is possible. Remember that we are preparing for GMAT. It tests you on concepts, not on calculations. So you will not see ugly numbers in GMAT questions. If I solve and get R as \(R = 4\sqrt{3}\), I will re-do my calculations since I would be convinced that I must have made an error somewhere! In 99.99% questions in GMAT, the numbers fall beautifully in place.
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Re: The figure represents a deflated tire 6 inches wide as shown  [#permalink]

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New post 14 Feb 2012, 00:39
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Well you did the hard part of the question. As you rightly pointed out the radius of the Hub is 6 Inches. Editing and copying some of your post to post the full solution.

If the radius of the hub is \(r\), then its area equals \({\pi}r^2\)

The area of the entire object is then \({\pi}{(r+6)}^2\)

This means that the area of just the tyre equals \({\pi}{(r+6)}^2 - {\pi}r^2\)

The problem also tells us that the ratio of the area of the tyre to the area of the entire object is \(\frac{1}{3}\). We can use this information to set up the following equation:

\(\frac{(Area of Hub)}{(Area of Tyre)}=\frac{1}{3} = \frac{({\pi}r^2)}{({\pi}{(r+6)}^2-{\pi}r^2)}\)

Which boils down to \(r^2-4r-12=0\).

So \((r-6)(r+2)=0\)

So \(r=6\) or \(r=-2\)

This will give \(r = 6\) as radius can never be negative.

So Total Radius \(=6+6=12\) inches
So Total Area \(={\pi}r^2={\pi}12*12=144{\pi}\)
So Area of Hub \(={\pi}r^2={\pi}6*6=36{\pi}\)
So Area of Just tyre \(=144{\pi}-36{\pi}=108{\pi}\)
Thickness of Tyre When inflated \(=3\) inches
So Total Volume of Trye in \({inches}^3=3*108{\pi}=324{\pi}\)

Air is filled at a rate of \(4{\pi}\)
So total time \(=\frac{324{\pi}}{4{\pi}}=81\) seconds
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The figure represents a deflated tire 6 inches wide as shown  [#permalink]

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New post 04 Jul 2015, 03:29
VeritasPrepKarishma wrote:
enigma123 wrote:
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
Attachment:
Tire.PNG


Ok - guys this is how I started to solve this question.

To solve this problem we need to (a) find the volume of the tire and then (b) solve a rate problem to determine how long it will take to inflate the tire.

If the radius of the hub is r, then its area equals pi \(r^2\)
The area of the entire object is then pi \((r+6) ^ 2\)
This means that the area of just the tire equals pi \((r+6)^2\)- pi \(r^2\)
The problem also tells us that the ratio of the area of the tire to the area of the entire object is 1/3. We can use this information to set up the following equation:
area of hub/area of tyre = 1/3 = pi \(r^2\)/pi \((r+6)^2\)- pi \(r^2\) => r^2 - 4r - 36 = 0. This will give r = 6 as radius can never be negative.

Now I am stuck and lost. Can someone please help?


The question itself is long and the proper solution is much longer. In such cases, try and think more creatively (practice on practice questions. In the exam do whatever comes to your mind first)
(omerrauf has given you the complete solution so I am not going to repeat it.)

I will talk of a more creative solution.

You need to find the surface area of the tire and multiply it by the thickness to get the volume of the air in the tire. Then you need to divide it by 4π (rate at which air is filled) to get the time taken.

Say the radius of big circle is R and that of the little circle (hub) is r.
Given:\({\pi}r^2 = (1/3){\pi}(R^2 - r^2)\)
or \(r^2 = R^2/4\)

Focus on what we need now: \({\pi}R^2 - {\pi}r^2\) = surface area of tire = \({\pi}R^2 - {\pi}R^2/4 = 3{\pi}R^2/4\)

Volume of air = \((3{\pi}R^2/4) * 3\)

Time taken = \((3{\pi}R^2/4) *3/4{\pi} = 9R^2/16\)

Your options give you the value of \(9R^2/16\). This means that if you multiply your options by 16 and divide by 9, you will get R^2.
16 is already a perfect square so ignore it. Divide your options by 9 and see what you get.

A. 24 (Doesn't get divided by 9. 16*24/9 cannot be R^2.)
B. 27 (You are left with 3. 16*3 cannot be R^2)
C. 48 (Doesn't get divided by 9. 48/9 cannot be R^2)
D. 81 (You are left with 9 which is a perfect square. 9*16 gives R as 12. Possible answer)
E. 108 (You are left with 12. 12 is not a perfect square.)

You might be tempted to say that 16*3 can be R^2 since \(R = 4\sqrt{3}\) is possible. Remember that we are preparing for GMAT. It tests you on concepts, not on calculations. So you will not see ugly numbers in GMAT questions. If I solve and get R as \(R = 4\sqrt{3}\), I will re-do my calculations since I would be convinced that I must have made an error somewhere! In 99.99% questions in GMAT, the numbers fall beautifully in place.


Hi Karishma,

So are we saying it's practically impossible for a radius to be \(R = 4\sqrt{3}\) on the GMAT ? I get that in most situations ,GMAT has whole numbers..but can we just rule it out based on that notion or are there any exceptions to the rule? Kindly advise.Thanks
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Re: The figure represents a deflated tire 6 inches wide as shown  [#permalink]

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New post 04 Jul 2015, 18:31
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VeritasPrepKarishma wrote:
The question itself is long and the proper solution is much longer. In such cases, try and think more creatively (practice on practice questions. In the exam do whatever comes to your mind first)
(omerrauf has given you the complete solution so I am not going to repeat it.)

I will talk of a more creative solution.

You need to find the surface area of the tire and multiply it by the thickness to get the volume of the air in the tire. Then you need to divide it by 4π (rate at which air is filled) to get the time taken.

Say the radius of big circle is R and that of the little circle (hub) is r.
Given:\({\pi}r^2 = (1/3){\pi}(R^2 - r^2)\)
or \(r^2 = R^2/4\)

Focus on what we need now: \({\pi}R^2 - {\pi}r^2\) = surface area of tire = \({\pi}R^2 - {\pi}R^2/4 = 3{\pi}R^2/4\)

Volume of air = \((3{\pi}R^2/4) * 3\)

Time taken = \((3{\pi}R^2/4) *3/4{\pi} = 9R^2/16\)

Your options give you the value of \(9R^2/16\). This means that if you multiply your options by 16 and divide by 9, you will get R^2.
16 is already a perfect square so ignore it. Divide your options by 9 and see what you get.

A. 24 (Doesn't get divided by 9. 16*24/9 cannot be R^2.)
B. 27 (You are left with 3. 16*3 cannot be R^2)
C. 48 (Doesn't get divided by 9. 48/9 cannot be R^2)
D. 81 (You are left with 9 which is a perfect square. 9*16 gives R as 12. Possible answer)
E. 108 (You are left with 12. 12 is not a perfect square.)

You might be tempted to say that 16*3 can be R^2 since \(R = 4\sqrt{3}\) is possible. Remember that we are preparing for GMAT. It tests you on concepts, not on calculations. So you will not see ugly numbers in GMAT questions. If I solve and get R as \(R = 4\sqrt{3}\), I will re-do my calculations since I would be convinced that I must have made an error somewhere! In 99.99% questions in GMAT, the numbers fall beautifully in place.

Great Solution by Karishma,
Such creative answers save a lot of time.

I noticed that when you reached to get the ratio of the two radii (highlighted), you could have easily calculated the actual radius of the tyre and calculated the answer instead of guessing it based on squares.

or \(r^2 = R^2/4\)
=> r = R/2
Given , R-r=6
=> r=6, R=12

Thus, from your final equation, Time taken = \(9R^2/16 = 9*12*12/16 = 81\)

I hope I am not missing anything. Is there a reason why this approach would not work everytime?
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Re: The figure represents a deflated tire 6 inches wide as shown  [#permalink]

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New post 09 Jul 2015, 23:12
1
davesinger786 wrote:
Hi Karishma,

So are we saying it's practically impossible for a radius to be \(R = 4\sqrt{3}\) on the GMAT ? I get that in most situations ,GMAT has whole numbers..but can we just rule it out based on that notion or are there any exceptions to the rule? Kindly advise.Thanks


No it is not impossible but highly unlikely. Considering that the width is 6, it is highly unlikely that the radius is not an integer. You are taking a good guess on R = 12 and then you can ensure that it is correct by putting R back. Smaller radius will be 6 which gives two areas as \(144\pi\) and \(36\pi\). The area of just the tyre is \(108\pi\) which is 3 times \(36\pi\).
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Re: The figure represents a deflated tire 6 inches wide as shown  [#permalink]

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New post 09 Jul 2015, 23:23
I see,okay thanks a lot for clearing that up.Kudos 1+
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Re: The figure represents a deflated tire 6 inches wide as shown   [#permalink] 09 Jul 2015, 23:23
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