In a rectangular box of dimensions 20 centimetres by 20 centimetres by

Question

https://gmatclub.com/forum/download/file.php?id=55136 In a rectangular box of dimensions 20 centimetres by 20 centimetres by 100 centimetres, identical spherical balls are arranged in layers such that each layer has exactly 4 balls. The top view of this arrangement is shown in the figure above. What is the approximate percentage of space left empty inside the box, if the box contains the maximum possible number of layers of such spherical balls? (The volume of a sphere is \frac{4}{3}\pi r^3 , where r is the radius of the sphere) A. 33% B. 40% C. 50% D. 60% E. 67% Project PS Butler Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS Are You Up For the Challenge: 700 Level Questions 1.png

RaghavKhanna · Answer

Let r be the radius of a sphere

=> 4r = 20
=> r = 5cm

Let the number of layers be : N

=> Height of the box = N*2r

=> N = 100/10 = 10

Volume of 1 layer = Volume of 4 spheres = (16/3)*π*(5)^3

=> Volume of 10 layers = (160/3)* π*(5)^3

Volume of box = 20*20*100

Fraction of Empty space =   / 20*20*100

=> Empty space % = 47% = 50% approx

Answer: C
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yashikaaggarwal · Answer

Radius = 5 cm {20/4 = r}
A level consisting 4 identical spheres whose diameter is 10 = 10
Height of box = 100cm
Total no. of level = 100/10 = 10 levels
1 level = 4 balls.
10 level = 40 balls.
Volume of cuboid = l*b*h =20*20*100 = 40,000 
Volume of 1 ball(sphere) = 4/3πr^2 = 4/3*22/7*5*5*5
Volume of 40 spheres = 40*4/3*22/7*5*5*5 = 20952.38 or 20952.4 
Space left inside box = 40,000-20952.4 = 19047.6
Approximate percentage left = 19047.6*100/40,000
=> 47.619 or 50% approximate.
IMO C

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yashikaaggarwal · Answer

You might have to correct this 4*40 spheres = 160 not 1600

vipulshahi · Answer

Volume of cuboid = 20*20*100
Volume of sphere = (4/3)*3.14*r^3

Radius of sphere, r = 5cm
In one level , 4 identical spheres 
1 level = 4 spherical balls
Total level = 100/10 ; height is 100cm & diameter is 10cm
Total level = 10 it means we have 40 balls are in cuboid 
 
% space left =   / ( total area)
=   / 20*20*100
=   / 40,0000
= 19000(approx) / 40,0000
= 48%(approx) = 50%

Answer (C)

HoneyLemon · Answer

Radius of the sphere = 5 cm 
number of sphere (n)= n * 10 = 100 => n = 10

Radius of sphere, r = 5cm
In one level , 4 identical spheres

% space left =  \frac{(Volume of Cuboid - total Volume of sphere) * 100 }{ (Volume of Cuboid)} 
= {  \frac{(20*20*100 - 40*4*22*5^3) * 100 }{ (3*7) }  } / {20*20*100}
=   \frac{(40,000 - 20952) * 100 }{ 40,0000}

= 47% = 50% (approx )

Ans : = C

ScottTargetTestPrep · Answer

Solution:

We see that a box of volume 20 by 20 by 10 will contain 4 balls. Therefore, a box of volume 20 by 20 by 100 will contain 10 times as many balls, i.e., 40 balls. Since the radius of each ball is 20/4 = 5, the volume of each ball is 4/3 * π * 5^3. Using the number 3 for π, we have the volume of each ball approximately equal to 4/3 * 3 * 5^3 = 4 x 125 = 500. Therefore, the volume of 40 balls is approximately 500 x 40 = 20,000. The volume of the box is 20 x 20 x 100 = 40,000. Therefore, the empty space inside the box is approximately 40,000 - 20,000 = 20,000 or 50% of the volume of the box.

Answer: C

sambitspm · Answer

Kindly see the attachment
C ~50%

bumpbot · Answer

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