Jack is storing a rectangular box inside a cylindrical container. The

Question

Jack is storing a rectangular box inside a cylindrical container. The container has a volume of  980\pi  cubic inches and a height of 20 inches. Which of the following dimensions could the box have in order to fit inside the cylinder?

I. 6 inches by 9 inches by 15 inches 
II. 11 inches by 15 inches by 18 inches 
III. 9 inches by 9 inches by 20 inches

A. I only
B. III only
C. I and II only
D. I and III only
E. I, II and III

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CrackverbalGMAT · Answer

Volume of the cylinder =  \pi r^2 h

\pi r^2 * 20 = 980 \pi

r^2 = 49 . Therefore r = 7 and the diameter = 14

For the rectangle the diameter becomes the diagonal and each side can have a length lesser than or equal to Diagonal /  \sqrt{2}  = 14 / 1.4 = 10 inches or a little less than 10 inches.

The height can be any length up to a maximum of 20 inches.

Statements I and III are possible.

Option D
 
Arun Kumar

dushyanta · Answer

Jack is storing a rectangular box inside a cylindrical container. The container has a volume of  980\pi  cubic inches and a height of 20 inches. Which of the following dimensions could the box have in order to fit inside the cylinder?

I. 6 inches by 9 inches by 15 inches 
II. 11 inches by 15 inches by 18 inches 
III. 9 inches by 9 inches by 20 inches

A. I only
B. III only
C. I and II only
D. I and III only
E. I, II and III

vol of cylinder = (pi)*r^2*h =980(pi)
or r^2 =980/20=> r = 7
so h= 20 and r = 7 or d = 14

to put the rectangle inside the cylinder, we should always keep one side less than equal to height and other two side less than diameter (diagonal of the rectangle or d/\sqrt{2} or ~14/1.4 or ~ 10)

I. 6 inches by 9 inches by 15 inches 
d, d, h
II. 11 inches by 15 inches by 18 inches 
d,  d , h
III. 9 inches by 9 inches by 20 inches
d, d, h

IMO (D)

abhinavsodha800 · Answer

Hi Dushyanta,
How did you got diagonal/ sqrt{2}

sanch93 · Answer

The way I solved it was:

Based on pythagoras theorem, calculate the Diagonal of the rectagle and see if its less than the diameter of the circular part of cylinder
In this case, r=7 inches, so D= 14 inches.

Stat 1 & 3 gives the length of the Hypotenuse less than root 196

dushyanta · Answer

for any rectangle diagonal = \sqrt{l^2+b^2}

to maximize the value of diagonal => l = b
or d = \sqrt{2l^2}
or d = l\sqrt{2}

baraa900 · Answer

you can consider any number here to be as length and height of the rectangle in order to calculate the hypotenuse?

dushyanta · Answer

Yes.. try personalising the question
U really want to fit that rectangular box of L*B*H area, and maximum surface area coverage would be by diagonal D in that cylinder of height h and diameter d..

As one side of that box needs to be facing cylinder bottom, and other side height of the cylinder.. u take ur pick

Posted from my mobile device

ScottTargetTestPrep · Answer

Solution:

Let the radius of the base of the cylinder be r. Since the volume of the cylinder is 980π and the height is 20, we can create the following equation:

Volume = πr^2 * h

980π = 20πr^2

r^2 = 49

r = 7

We see that the radius of the base is 7. In order for a rectangle to fit inside a circle, the diagonal of the rectangle must be less than the diameter of the circle. Examining the Roman numerals, we see that each of them has a coordinate greater than the diameter of the base of the cylinder (which is 2 x 7 = 14), so if the boxes of given dimensions can fit in the cylinder, there is only one way they can fit. In other words, the last coordinate in each of the Roman numerals must be the height and we need to check whether rectangles defined by the first two coordinates have diagonals less than 14. Notice that the square of 14 is 14^2 = 196.

I. 6 x 9 x 15

Since 6^2 + 9^2 = 36 + 81 = 117 is less than 196, the diagonal of the 6 x 9 base of this box is less than 14. This box will fit inside the cylinder.

II. 11 x 15 x 18

Since a side of the 11 x 15 base of this box is greater than 14, it is clear that the diagonal is greater than 14 as well. We don’t even need to make any calculations to see this (because the diagonal of a rectangle is always greater than the lengths of either side of the rectangle). This box will not fit inside the cylinder.

III. 9 x 9 x 20

Since 9^2 + 9^2 = 81 + 81 = 162 is less than 196, the diagonal of the 9 x 9 base of this box is less than 14. This box will fit inside the cylinder (the top base of the box will be flush with the top base of the cylinder).

Answer: D

Nandini2242 · Answer

So I thought of finding the volume of the box as if the volume of the box is more than the volume of the cylinder, it should not fit in the cylinder. With this approach, the 3rd box (9*9*20) won't fit.

Can anyone explain what I did wrong?

Can anyone explain what I did wrong?

griffith7 · Answer

The third box has a lesser volume than the cylinder in the first place. 980(pi) > 9 * 9 * 20

But the approach is flawed because all 3 of the boxes have lesser volumes than the cylinder, but the 2nd box won't fit because it does not have a base that will fit within the base of the cylinder. The 2nd smallest side is 15 while the diameter of the cylinder base is 14.

Jack is storing a rectangular box inside a cylindrical container. The

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