Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included!
10 Jun 2020, 03:30
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
55% (03:02) correct
45%
(03:23)
wrong
based on 105
sessions
History
Date
Time
Result
Not Attempted Yet
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%
Are You Up For the Challenge: 700 Level Questions
Attachment:
1.png [ 30.46 KiB | Viewed 3748 times ]
Updated on: 10 Jun 2020, 04:04
Originally posted by RaghavKhanna on 10 Jun 2020, 03:44.
Last edited by RaghavKhanna on 10 Jun 2020, 04:04, edited 2 times in total.
Last edited by RaghavKhanna on 10 Jun 2020, 04:04, edited 2 times in total.
Kudos
Bookmarks
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) - (160/3)*π*125] / 20*20*100
=> Empty space % = 47% = 50% approx
Answer: C
Posted from my mobile device
=> 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) - (160/3)*π*125] / 20*20*100
=> Empty space % = 47% = 50% approx
Answer: C
Posted from my mobile device
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 10 Aug 2026
Posts: 3,094
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
10 Jun 2020, 03:58
Kudos
Bookmarks
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
Posted from my mobile device
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
Posted from my mobile device
