If the maximum number of identical cylinders, standing upright and tou

Question

If the maximum number of identical cylinders, standing upright and touching the edges, that can be fit into a rectangular box with a square base is 200, what is the volume of the box?

(1) The radius of each cylinder is 5 centimeters. 
(2) The height of the box is 80 centimeters.

chetan2u · Accepted Answer

Hi  Shank18 ,

what does the main statement mean..

identical cylinders, standing upright and touching the edges  ..
 same radius , the circular portion is base

can be fit into a rectangular box with a square base  ...
the base of rectangular box is square, so if x cylinders are placed along a side of base, the length of base is  DIA*x=2rx  and area of base =  4r^2x^2 
and volume will depend the side of base and height =  4r^2x^2*h  so THREE variables

let us see the statements-

(1) The radius of each cylinder is 5 centimeters. 
Nothing about x and h
insuff

(2) The height of the box is 80 centimeters
nothing about r and h 
insuff

combined..
 total number is 200..

I.   let the square base be of 10 cylinders  .. so x=10, r=5 and h=80
so 10*10 =100 on the base and therefore another 100 on top of these 100, so volume =  4r^2x^2*h , where h=80..thus height of cylinder=40, as 2 cylinders fit in standing upright
 4r^2x^2*h=4*25*10^2*80=800,000

II.   let the square base be of 5 cylinders  .. so x=5, r=5 and h=80
so 5*5 =25 on the base and therefore another 7 sets of 25 on top of these 25, so volume =  4r^2x^2*h , where h=80..thus height of cylinder=80/8=10, as 8 cylinders fit in standing upright
 4r^2x^2*h=4*25*5^2*80=200,000 
so different volumes possible, which will depend on the height of cylinder
insuff

E

AnupamKT · Answer

From statement 1: radious of cylinder is known which can be used to find size of square base but we require height of cylinder to find the height of box.
From statement 2: Height of box is known but radious if cylinder is unknown.

Hence combining both the statements, we can find the size of rectangular box, hence volumn. So, answer is C.

Posted from my mobile device

ShankSouljaBoi · Answer

Hi  chetan2u

Please guide with this one.

Regards

taniad · Answer

Questions like these are trick. You need to know the dimensions of each item going inside the crate, box, etc, and the crate/box volume as well.

In statement 1= the cylinder radius is known, but height is not known. 
In statement 2= the height of the box is known, from the stem, we know the other 2 dimensions (length and width) will be same, as it is a sq. base. but no mention of the dimension.

Together nothing more can be found out., hence E.

PS: Has there ever been a question similar to this where the answer has been anything other than E?

ShankSouljaBoi · Answer

taniad  the stem mentions cylinders as identical .

Posted from my mobile device

ShankSouljaBoi · Answer

chetan2u Thank you ji. This was very insightful

Posted from my mobile device

taniad · Answer

Thanks- that's my downfall for careless mistakes, reading quickly and not noting properly

But this still doesnt give us the height of the cylinder. Hence both statement together does not give any other information that can help us find the volume of the container.

PinakiRDas · Answer

And this radius means the shorter edge is 10 but no mention of longer edge - doesn't that rule out the other options pretty early?

ProfDas · Answer

Hi Mentors, ( chetan2u, VeritasKarishma, mikemcgarry, Bunuel, amanvermagmat, gmatbusters )

Please share some insight on this question.

Regards,

chetan2u · Answer

Responding to the PM above here as it may help others with same doubt.

Now we are given there are 200 cylinders but we do not know the height of these cylinders but we know the height of the box
We know the radius of cylinder but not side of the base..
So we can easily vary unknowns that is height of cylinder and base of box to get different combinations..
Now when I take there are 10 cylinders along one side, the number of cylinders become 10*10=100 in one row so I can have one more row on top as total are 200 and now I have 100+100=200 thus the height of cylinder becomes 80/2=40 and side of square base becomes 10*2r=10*2*5=100..
Volume of box = 100*100*80

Now when I take there are 5 cylinders along one side, the number of cylinders become 5*5=25 in one row so I can have 200/25=8 more row on top as total are 200 and thus the height of cylinder becomes 80/8=10 and side of square base becomes 5*2r=5*2*5=50..
Volume of box = 50*50*80

As you can see the volume is different everytime, thus we cannot have a unique box
Insuff

If the maximum number of identical cylinders, standing upright and tou

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