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Math Expert V
Joined: 02 Sep 2009
Posts: 60515
A clown blows up a spherical balloon such that its volume increases at  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 57% (02:25) correct 43% (02:40) wrong based on 86 sessions

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A clown blows up a spherical balloon such that its volume increases at a constant rate. It takes 3 seconds for the radius of the balloon to increase from 1 inch to 2 inches. How many seconds does it take for the radius of the balloon to increase from 3 inches to 5 inches?

NOTE: The volume of a sphere is 4/3*πr^3.

A. 6
B. 9
C. 24
D. 30
E. 42

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Senior PS Moderator V
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Posts: 3294
Location: India
GPA: 3.12
Re: A clown blows up a spherical balloon such that its volume increases at  [#permalink]

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1
Volume of sphere = 4/3*pi*R^3
Change in volume(from 1 inch to 2 inch)
= 4/3*pi*(8*R^3 - R^3) = 4/3*pi*(7*R^3)
This takes 3 seconds.

Similarly we need how much time it would take for the sphere to increase from 3 to 5inch

Change in volume(from 3 inch to 5 inch)
= 4/3*pi*(125*R^3 - 27*R^3) = 4/3*pi*(98*R^3)

Time taken = (98*3)/7 = 42seconds(Option E)
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Re: A clown blows up a spherical balloon such that its volume increases at  [#permalink]

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Hi All,

We're told that a clown blows up a spherical balloon such that its volume increases at a constant rate; it takes 3 seconds for the radius of the balloon to increase from 1 inch to 2 inches. We're asked how many seconds it takes for the radius of the balloon to increase from 3 inches to 5 inches. This is a variation on a 'rate' question, but it involves VOLUME, so you have to think about the rate in a slightly different way. There's also a math 'shortcut' that you can take advantage of at the end of your calculations...

To start, we need to calculate the Volumes that occur when the radius of the balloon is 1 inches and when it is 2 inches.

Volume of a 1 inch balloon = (4/3)(pi)(1^3) = 4pi/3
Volume of a 2 inch balloon = (4/3)(pi)(2^3) = 32pi/3

Since that increase occurred in 3 seconds, we now know the rate that the volume increases: (32pi/3) - (4pi/3) = 28pi/3 in 3 seconds = 28pi/9 each second.

Now we have to calculate the volumes of a 3-inch and 5-inch balloon:

Volume of a 3 inch balloon = (4/3)(pi)(3^3) = 108pi/3
Volume of a 5 inch balloon = (4/3)(pi)(5^3) = 500pi/3

We need the volume to increase by 392pi/3 at a rate of 28pi/9 per second. At this point, the calculation 'looks' a bit ugly, but the answer choices are sufficiently 'spread out' that we can use estimation...

130pi/3pi is GREATER than 40, so there's only one answer that makes sense...

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Re: A clown blows up a spherical balloon such that its volume increases at  [#permalink]

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Bunuel wrote:
A clown blows up a spherical balloon such that its volume increases at a constant rate. It takes 3 seconds for the radius of the balloon to increase from 1 inch to 2 inches. How many seconds does it take for the radius of the balloon to increase from 3 inches to 5 inches?

NOTE: The volume of a sphere is 4/3*πr^3.

A. 6
B. 9
C. 24
D. 30
E. 42

If the radius of the balloon is 1 inch, the volume of the balloon is 4/3*π(1)^3 = 4π/3. If the radius is 2 inches, the volume is 4/3*π(2)^3 = 32π/3. Since it takes 3 seconds for the radius to increase from 1 inch to 2 inches, the rate at which the volume is increasing is (32π/3 - 4π/3)/3 = 28π/9 cubic inches per second.

Now, if the radius of the balloon is 3 inches, the volume of the balloon is 4/3*π(3)^3 = 108π/3. If the radius is 5 inches, the volume is 4/3*π(5)^3 = 500π/3. Since the rate at which the volume is increasing is 28π/9 cubic inches per second, then it takes (500π/3 - 108π/3)/(28π/9) = 392π/3 * 9/(28π) = 14 * 3 = 42 seconds to increase the radius of the balloon from 3 inches to 5 inches.

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