Last visit was: 13 Dec 2024, 01:45 It is currently 13 Dec 2024, 01:45
Close
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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Hoozan
Joined: 28 Sep 2018
Last visit: 12 Dec 2024
Posts: 703
Own Kudos:
633
 []
Given Kudos: 248
GMAT 1: 660 Q48 V33 (Online)
GMAT 2: 700 Q49 V37
Products:
GMAT 2: 700 Q49 V37
Posts: 703
Kudos: 633
 []
Kudos
Add Kudos
11
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Fdambro294
Joined: 10 Jul 2019
Last visit: 19 Oct 2024
Posts: 1,369
Own Kudos:
637
 []
Given Kudos: 1,658
Posts: 1,369
Kudos: 637
 []
4
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
General Discussion
User avatar
IanStewart
User avatar
GMAT Tutor
Joined: 24 Jun 2008
Last visit: 09 Dec 2024
Posts: 4,126
Own Kudos:
9,915
 []
Given Kudos: 97
 Q51  V47
Expert reply
Posts: 4,126
Kudos: 9,915
 []
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Rogermoris
Joined: 13 Aug 2020
Last visit: 24 Sep 2020
Posts: 25
Own Kudos:
50
 []
Given Kudos: 12
Posts: 25
Kudos: 50
 []
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
We need to look at how many cylinders are to be stacked together. Both vertically and in a row.

So, starting with the base, we have the Length where we can put = 120/ 4 = 30 cylinders (4 is nothing but the diameter of the cylinder)
Similarly, across the Width we can put = 40/4 = 10 cylinders
In total, we have completely occupied the base with 30 cylinders along length and 10 along width, which equals to 300 on the base.

Now we stack them vertically, so we know height of cylinder as 6 and the container is 30, so 30/6 = 5 levels of 300 cylinders = 1500 cylinders can be stacked in the Box

Volume of 1 cylinder = \((pi) * r^2 * h = 24*(pi)\)

Volume of 1500 cylinders = \(1500*24*(pi) = 36000 (pi) = 113000(approx) \)

Hence OPTION [C] is right.
User avatar
Hoozan
Joined: 28 Sep 2018
Last visit: 12 Dec 2024
Posts: 703
Own Kudos:
Given Kudos: 248
GMAT 1: 660 Q48 V33 (Online)
GMAT 2: 700 Q49 V37
Products:
GMAT 2: 700 Q49 V37
Posts: 703
Kudos: 633
Kudos
Add Kudos
Bookmarks
Bookmark this Post
IanStewart
If you put each cylinder perfectly in a rectangular box with height 6, and length and width 4 (the diameter of the cylinder), those boxes would fit perfectly into the cargo container with no unused space (because the height is divisible by 6, and the other dimensions by 4). A cylinder only takes up less space than a box that perfectly contains it because of the area outside of the circular circumference of the cylinder, but inside the square confines of the box. Since a circle inscribed in a square takes up π/4 of the area of that square, all of our cylinders will occupy π/4 of the total volume of the cargo container. If we estimate π/4 to be 3/4, we'll get an answer slightly too small, so the answer here will be slightly bigger than (3/4)(40)(30)(120) = 108,000, and C must be right.


This is a really cool approach <3

IanStewart
Could you please show the calculation that lands up with "a circle inscribed in a square takes up π/4 of the area of that square"
avatar
Sagart21
Joined: 25 Oct 2020
Last visit: 13 Jan 2024
Posts: 49
Own Kudos:
Given Kudos: 38
Location: India
Concentration: Strategy, General Management
Posts: 49
Kudos: 15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Volume of cylinder = pi*r^2*h = 24pi

Now we need to find how many cylinders can be kept.
Length = 120
So 120/4 = 30 cylinders across it
Width = 40
So 40/4 = 10 cylinders
Height = 30
So 30/6 = 5 cylinders

Hence total = 30*10*5 = 1500 cylinders

Total capacity = 1500*24*3.14 = 113040

Answer - Option C

Posted from my mobile device
User avatar
IanStewart
User avatar
GMAT Tutor
Joined: 24 Jun 2008
Last visit: 09 Dec 2024
Posts: 4,126
Own Kudos:
9,915
 []
Given Kudos: 97
 Q51  V47
Expert reply
Posts: 4,126
Kudos: 9,915
 []
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hoozan
This is a really cool approach <3

IanStewart
Could you please show the calculation that lands up with "a circle inscribed in a square takes up π/4 of the area of that square"

If you draw a circle of radius 1, its area is πr^2 = π(1^2) = π. If you inscribe that circle in a square, the diameter of the circle with be the length of a side of the square. The diameter of the circle is 2, so the area of the square is 2^2 = 4, and the ratio of the area of the circle to that of the square is π/4.

I've chosen a number for the radius, because the answer will be the same no matter what the radius is, but anyone unsure of that could instead do the above using a radius of 'r', and will get the same answer; the 'r' will cancel out when you find the ratio of the two areas.
User avatar
CEdward
Joined: 11 Aug 2020
Last visit: 14 Apr 2022
Posts: 1,227
Own Kudos:
Given Kudos: 332
Posts: 1,227
Kudos: 222
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Wondering here if my approach is correct.

r = 2, h = 6 --> V = πr^2 = 24π ---> diameter = 4

Given the dimensions of the container, then we can fit this many boxes
h ---> 5
w ---> 10
l ---> 30

So 30 x 10 x 5 = 1500 boxes

1500 x 24π = ~ 1500 x 25 x 3 = 1500 x 75

15 x 75 = 1125

So C.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 35,805
Own Kudos:
Posts: 35,805
Kudos: 929
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderator:
Math Expert
97852 posts