Cylindrical tennis-ball cans are being packed into cartons.

C.

((14-2)/12)*(16/4)*(20/4)=20

Cylindrical tennis-ball can has 4*4*12 size

for 14 size of carton: 14=12+2(empty), 4*3+2(empty)
for 16 size of carton: 16=12+4(empty), 4*4+0(empty)
for 20 size of carton: 20=12+8(empty), 4*5+0(empty)

Never try doing these by solving for volume, it doesnt really work that way. Heres why.

We want the most cans so want to fit as many cans in the length and width as possible.

So we have 16x20 which are both divisible by 4. (diameter of the can is 4).

so we can have 4 one way and 5 another so 20 total.

And 12 inches is the height of each can so they all fit.

C.

Hi all. Was just going through the question. The answer would change if some cans are placed vertically and some horizontally.!!

i drew a table: hope its correct:
14 16 20
1 4 5
3 4 1
3 1 5

the height will fit only 1 time in every dimension (12 apears only once in 14/16/20) and the 4=d will apear 3 times in 14 and 4 times in 16 and 5 times in 20

hence 4*5

Experts, one clarification - If I modify the question little bit and let the new dimensions be 16* 20 * 16
then we can obviously place 20 cylinders as above on the 16*20 base.
Now, there will be 4 inches left in the height of the cuboid and we can place more cylinders in horizontal position over the vertical ones.

Please clarify that shall we consider these extra ones - y/n

I agree with you. There will be room for 6 cans I think.
However, you have to consider that there are some industrial and safety restrictions

Is it true always that selecting the face with biggest area will result in max cans fit inside the carton?

Can you please explain the above calculation?

Answer to your question is that it depends on the situation represented in the question. In case of cuboid, we have to keep all its dimensions(length, breadth and height) in mind. It is suggested that we visualize the situation given, when the question is about solid geometry.

E.g. In present question since it is give that height of the tennis-ball cans is 12 inches. Even if we insert the 4*5 = 20 cans inside the carton there would be some space that will remain empty. Because the height of can is only 12 inches and 2 inches of empty space will be remaining out of 14 inches height of carton. Since a can of 4 inches diameter can not be inserted in 2 inches space.

Now Visualize that you are trying to stand some cans on 20x16 carpet area and the condition is that diameter of each can is 4 inches. So for 20 inches length we can consider standing only 20/4 = 5 cans(rows). And for 16 inches breadth we can stand only 16/4 = 4 cans(columns). So total no of cans that can stand in 20x16 area is 5x4 = 20 cans.
You can see the diagram below for the calculation of 20.
0 represents tennis - ball cans - 5 rows x 4 columns

0000
0000
0000
0000
0000

Dimensions of can = 4 inches * 12 inches.
Dimensions of carton = 14 * 16 * 20 inches.

Since 16*20 will be the plane where we can get the maximum occupancy, lets fit in 4*5 cans.The height of each can is 12 inches leaving 2 inches on the top as clearance , so C

((14-2)/12)*(16/4)*(20/4)=20

where exactly did you get 14-2 from? and why did you divide it by 12? why did you divide 16 by 4 and 20 by 4?

and I don't get the empty, why do you say that?

if the cylinders are fit along 16 axis, 20 will also be placed in the canton if the cans are oriented on vertical directions. So 15 cylinders along 16 and then on top of that 5 cans can be placed horizontally. same thing if we put them along 20 axis, the canton can also hold 20 cylinders.

The equation\(= \frac{14 * 16 * 20}{\pi * 4 * 12} \approx \frac{12 * 16 * 20}{4 * 4 * 12} = 20\)

Answer = C

Height of Cylindrical can = 12. This gets aligned with dimension "14" of the carton (refer it as height of the box. This will give minimum wastage)

\(\pi = 3.14 \approx {4}\)

\(\frac{16 * 20}{4*4} = 20\)

Answer = C

No, not always true.



This is NOT always true.

In this case, no matter how you orient the box, you will be able to fit 20 in the box.

If you put them vertical on the 14x16 plane, then you can fit 12, but then you can fit 8 on top = 20.
If you put them vertical on the 14x20 plane, then you can fit 15, but then you can fit 5 on top = 20.
If you put them vertical on the 16x20 plane, then you can fit 20, and none on top = 20.

In other scenarios it's not always equal, but it's erroneous to state that one way will give minimum wastage all the time, simply not true.

for maximum number of tennis - ball to fit in each carton, say length = 20 inches

now, as the diameter of the can is 4 inches, maximum number of cans that will fit in the carton is (20/4) = 5 cans
and if width is 16 inches, (16/4) = 4 cans cans can be placed in the carton
finally let, height = 14 inches, so 1 can can be kept

so, things are done
( 5 * 4 * 1 ) = 20 cans will fit in each carton

= C

thanks

In this case, no matter how you orient the box, you will be able to fit 20 in the box.


If you put them vertical on the 14x16 plane, then you can fit 12, but then you can fit 8 on top = 20.
If you put them vertical on the 14x20 plane, then you can fit 15, but then you can fit 5 on top = 20.
If you put them vertical on the 16x20 plane, then you can fit 20, and none on top = 20.


This is correct approach to solve ques.