Four spheres and three cubes are arranged in a line according to incre : GMAT Data Sufficiency (DS)
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# Four spheres and three cubes are arranged in a line according to incre

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Four spheres and three cubes are arranged in a line according to incre [#permalink]

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19 Nov 2012, 13:33
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Four spheres and three cubes are arranged in a line according to increasing volume, with no two solids of the same type adjacent to each other. The ratio of the volume of one solid to that of the next largest is constant. If the radius of the smallest sphere is ¼ that of the largest sphere, what is the radius of the smallest sphere?

(1) The volume of the smallest cube is 72pi.
(2) The volume of the second largest sphere is 576pi.
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Oct 2014, 07:10, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Four spheres and three cubes are arranged in a line according to incre [#permalink]

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19 Nov 2012, 16:01
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himanshuhpr wrote:
Four spheres and three cubes are arranged in a line acc. to increasing vol. , with no two solids of the same type adjacent to each other . The ratio of the volume of one solid to that of the next largest is constant . if the radius of the smallest sphere is 1/4 of the largest sphere, what is the radius of smallest sphere?
1. Vol. of smallest cube is 72 pi .
2. Vol. of second largest sphere is 576 pi.

I'm happy to help with this.

From the prompt, we know the shapes are arranged in order of increasing volume, and the shapes are in the order:
1) sphere
2) cube
3) sphere
4) cube
5) sphere
6) cube
7) sphere

The ratios of the volumes are constant, so what we have is a geometric sequence. Call that ratio r. Let's say the volume of #1 is V. Then, the volumes are
1) sphere = V
2) cube = r*V
3) sphere = (r^2)*V
4) cube = (r^3)*V
5) sphere = (r^4)*V
6) cube = (r^5)*V
7) sphere = (r^6)*V

Then, the prompt tells us that radius of #1 is 1/4 the radius of #7. If the lengths from #1 to #7 increase by 4 times (scale factor = 4), then the volume increases by 4^3 = 64. For more on scale factors, see:
http://magoosh.com/gmat/2012/scale-fact ... decreases/

In other words, (r^6)*V = 64V ----> r^6 = 64 -----> r = 2

As it happens, we were easily able to solve for the numerical value of r here. Even if that were not the case, even if r were some ugly decimal, it would be the same. At this point, we know all the ratios, and all we need is a single value, any value on the list, and that would allow us to calculate every number on the list. In other words, any value would allow us to solve for V and, since we already know r, knowing V would allow us to calculate every term.

Now, that statements:
1. Vol. of smallest cube is 72 pi .
2. Vol. of second largest sphere is 576 pi.
Each statement gives us the numerical value of the volume of a specific shape in the arrangement, so each statement, all alone and by itself, will be sufficient for determining the answer to the prompt question.

That's why the OA is D

Does all this make sense?

Mike
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Re: Four spheres and three cubes are arranged in a line according to incre [#permalink]

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19 Nov 2012, 17:31
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himanshuhpr wrote:
Four spheres and three cubes are arranged in a line acc. to increasing vol. , with no two solids of the same type adjacent to each other . The ratio of the volume of one solid to that of the next largest is constant . if the radius of the smallest sphere is 1/4 of the largest sphere, what is the radius of smallest sphere?

1. Vol. of smallest cube is 72 pi .
2. Vol. of second largest sphere is 576 pi.

Since there are 4 spheres and 3 cubes. Only possible way to arrange them in alternate position is possible when we have S C S C S C S
Thus we know smallest in volume is sphere and largest one is also sphere and the order of placement.

Also, ratio of the volume of one solid to that of the next largest is constant. So its a GP. and from "if the radius of the smallest sphere is 1/4 of the largest sphere" we would know the ratio between volume of largest and smallest sphere and from it can deduce the ratio between any two solids.

Statement 1: Vol of smallest cube is 72pi.
We know ratio between any two solids. We can find out volume of smallest sphere and thus radius. Sufficient
Statement 2:Vol. of second largest sphere is 576 pi.
We know ratio between any two solids. We can find out volume of smallest sphere and thus radius. Sufficient

Ans D it is.
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Re: Four spheres and three cubes are arranged in a line according to incre [#permalink]

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03 Nov 2014, 08:56
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18 Jun 2016, 03:17
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Re: Four spheres and three cubes are arranged in a line according to incre [#permalink]

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20 Jul 2016, 09:04
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himanshuhpr wrote:
Four spheres and three cubes are arranged in a line according to increasing volume, with no two solids of the same type adjacent to each other. The ratio of the volume of one solid to that of the next largest is constant. If the radius of the smallest sphere is ¼ that of the largest sphere, what is the radius of the smallest sphere?

(1) The volume of the smallest cube is 72pi.
(2) The volume of the second largest sphere is 576pi.

Please check the solution as attached
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Re: Four spheres and three cubes are arranged in a line according to incre [#permalink]

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26 Aug 2016, 23:14
mikemcgarry wrote:
himanshuhpr wrote:
Four spheres and three cubes are arranged in a line acc. to increasing vol. , with no two solids of the same type adjacent to each other . The ratio of the volume of one solid to that of the next largest is constant . if the radius of the smallest sphere is 1/4 of the largest sphere, what is the radius of smallest sphere?
1. Vol. of smallest cube is 72 pi .
2. Vol. of second largest sphere is 576 pi.

I'm happy to help with this.

From the prompt, we know the shapes are arranged in order of increasing volume, and the shapes are in the order:
1) sphere
2) cube
3) sphere
4) cube
5) sphere
6) cube
7) sphere

The ratios of the volumes are constant, so what we have is a geometric sequence. Call that ratio r. Let's say the volume of #1 is V. Then, the volumes are
1) sphere = V
2) cube = r*V
3) sphere = (r^2)*V
4) cube = (r^3)*V
5) sphere = (r^4)*V
6) cube = (r^5)*V
7) sphere = (r^6)*V

Then, the prompt tells us that radius of #1 is 1/4 the radius of #7. If the lengths from #1 to #7 increase by 4 times (scale factor = 4), then the volume increases by 4^3 = 64. For more on scale factors, see:
http://magoosh.com/gmat/2012/scale-fact ... decreases/

In other words, (r^6)*V = 64V ----> r^6 = 64 -----> r = 2

As it happens, we were easily able to solve for the numerical value of r here. Even if that were not the case, even if r were some ugly decimal, it would be the same. At this point, we know all the ratios, and all we need is a single value, any value on the list, and that would allow us to calculate every number on the list. In other words, any value would allow us to solve for V and, since we already know r, knowing V would allow us to calculate every term.

Now, that statements:
1. Vol. of smallest cube is 72 pi .
2. Vol. of second largest sphere is 576 pi.
Each statement gives us the numerical value of the volume of a specific shape in the arrangement, so each statement, all alone and by itself, will be sufficient for determining the answer to the prompt question.

That's why the OA is D

Does all this make sense?

Mike

hi mike,

how do you know sphere is the first shape in the order. Even cube might be first in the order. In the question they have just mentioned that smallest sphere radius is 1/4 the radius of largest sphere
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Re: Four spheres and three cubes are arranged in a line according to incre [#permalink]

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26 Aug 2016, 23:22
SunthoshiTejaswi wrote:

hi mike,

how do you know sphere is the first shape in the order. Even cube might be first in the order. In the question they have just mentioned that smallest sphere radius is 1/4 the radius of largest sphere

Since there are no two solid of same type adjacent to each other For 4 spheres and 3 cubes be arranged so that no two spheres to be adjacent, any two consecutive spheres must be separated by a cube between them hence the arrangement will be alternate starting from a sphere.

Sphere-------Sphere-------Sphere-------Sphere

Sphere----Cube----Sphere----Cube----Sphere----Cube----Sphere

I hope this helps!!!
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Re: Four spheres and three cubes are arranged in a line according to incre   [#permalink] 26 Aug 2016, 23:22
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