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09 Feb 2020, 07:35
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
26% (02:13) correct
74%
(03:11)
wrong
based on 50
sessions
History
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Not Attempted Yet
Cylindrical containers carrying cricket balls are to be packed in a box. Each of the containers has a radius of 7 cm and a height of 30 cm. The box has length 76 cm, breadth 46 cm, and height 45 cm. Fine the maximum number of containers that can fit inside the box.
A. 15
B. 17
C. 19.
D. 21
E. 22
A. 15
B. 17
C. 19.
D. 21
E. 22
09 Feb 2020, 11:19
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DisciplinedPrep
If we keep 2 such cylinders one on top of the other, the combined height becomes 30 + 30 = 60 cm
In this case, we must keep the height of the box as 76 cm (only then it can accommodate a combined height of 60 cm)
(Thereafter, we would still have an empty space of 76 - 60 = 16 cm)
Width of the cylinder = 14 cm
Thus, along the side of the box of 46 cm or 45 cm, we can fit only 46/14 or 45/14 = 3 cylinders (ignoring the decimal)
Thus, total cylinders in one layer = 3 x 3 = 9
There are 2 such layers (making the combined height 60 cm)
Thus, number of cylinders in vertical orientation = 2 x 9 = 18
However, now we have an empty space of 45 x 46 x 16 in the box
We can fill this region by placing the cylinders lying on their side (on top of the 2 layers of cylinders vertically placed earlier)
Since the width of a cylinder is 14 cm, it can be easily accommodated along the available height of 16 cm
Also, each cylinder is 30 cm long, hence can be accommodated along the length of 45 cm
The number of cylinders in horizontal orientation = 46/14 = 3 (ignoring the decimal)
Thus, total number of cylinders = 18 + 3 = 21
Answer D
12 Feb 2020, 00:23
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DisciplinedPrep
Solution
- • The box is in cuboid shape. So base area of the box \(= 76 * 46 \) sq. cm
• Since height of the box (i.e. 45 cm) is greater than the height of cylindrical container, so, on the base of the box, cylindrical containers can be kept in one of the following ways:
- o Way 1: If containers are kept along their height on the box’s base (as shown in the below figure):
- Then the number of containers that can be accommodated in a row along the box’s breadth \(= [\frac{46}{14}] = 3\) (Keeping only integral values)
Number of rows \(= [\frac{76}{30}] = 2\)
And, the number of columns along the box’s height\(= [\frac{45}{14}] = 3\)
So, total number of containers that can be accommodated in box \(= 3*2*3 = 18\)
o Way 2: If they are kept on their circular base (As shown in the below figure)
- So, number of containers that can be accommodated in a row along the box’s breadth \(= [\frac{46}{14}] = 3 \)
Number of rows \(= [\frac{76}{14}] = 5\)
Number of columns \(= [\frac{45}{30}] = 1\)
So, the number of containers kept in this way \(= 5*3*1 = 15\)
Now, the difference in the height of box and container \(= 45 – 30 = 15 \)cm which is more than 14 cm, so we can keep some more containers along with their height in this remaining space.
- • Then the number of containers that can be accommodated in a row along the box’s breadth \(= [\frac{46}{14}] = 3\)
• Number of rows \(= [\frac{76}{30}] = 2\)
• And, the number column along the box’s remaining height \(= [\frac{15}{14}] = 1\)
• So, number of containers accommodated in remaining space = \(3*2*1 = 6\)
Thus, the total number of containers that can be kept in the box \(= 15+6 = 21\)
Thus, the correct answer is Option D.
