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:
https://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