Last visit was: 24 Apr 2026, 05:10 It is currently 24 Apr 2026, 05:10
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
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,811
Own Kudos:
810,956
 [1]
Given Kudos: 105,869
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,811
Kudos: 810,956
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
User avatar
DavidTutorexamPAL
User avatar
examPAL Representative
Joined: 07 Dec 2017
Last visit: 09 Sep 2020
Posts: 1,002
Own Kudos:
Given Kudos: 26
Posts: 1,002
Kudos: 2,042
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Hatakekakashi
Joined: 07 Jan 2016
Last visit: 22 Feb 2025
Posts: 1,228
Own Kudos:
Given Kudos: 126
Location: United States (MO)
GMAT 1: 710 Q49 V36
Products:
GMAT 1: 710 Q49 V36
Posts: 1,228
Kudos: 483
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Hatakekakashi
Joined: 07 Jan 2016
Last visit: 22 Feb 2025
Posts: 1,228
Own Kudos:
Given Kudos: 126
Location: United States (MO)
GMAT 1: 710 Q49 V36
Products:
GMAT 1: 710 Q49 V36
Posts: 1,228
Kudos: 483
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DavidTutorexamPAL
Bunuel
Four cylindrical cans each with a radius of 2 inches are placed on their bases inside an open square pasteboard box. If the four sides of the box bulge slightly, which of the following could be the internal perimeter of the base of the box, expressed in inches?

(A) 64

(B) 32

(C) 30

(D) 20

(E) 16

'the four sides of the box bulge slightly' means that the cans take up a slightly larger area than that of the box.
So, we'll look at the extreme of the range - a Logical approach.

4 cans arranged in a box on their base would be arranged in 2 rows of 2
(if they were arranged in 1 row of 4 or another non-symmetrical arrangement then there would be one side that didn't 'bulge').
Since each can has a radius of 2, this means that the cans takes up (2 cans * diameter of 4) = 8 inches.
Then, if the cans fit perfectly, the perimeter of the box would be 8*4 =32 inches.
But we know that the cans don't quite fit (the sides 'bulge') and therefore our answer is a bit less than 32.

(C), 30, is our answer.

Bunuel, just for the sake of nitpicking, why can't the cans be arranged in a triangle with another one on top? They would still all be 'on their bases', as required.

I went with C as well initially .. but does the bulge definitetly mean it's going to take more area?
User avatar
Hatakekakashi
Joined: 07 Jan 2016
Last visit: 22 Feb 2025
Posts: 1,228
Own Kudos:
Given Kudos: 126
Location: United States (MO)
GMAT 1: 710 Q49 V36
Products:
GMAT 1: 710 Q49 V36
Posts: 1,228
Kudos: 483
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DavidTutorexamPAL

Bunuel, just for the sake of nitpicking, why can't the cans be arranged in a triangle with another one on top? They would still all be 'on their bases', as required.

even though the question is not directed towards me I think if it's a triangle all 4 sides of the box won't bulge
User avatar
DavidTutorexamPAL
User avatar
examPAL Representative
Joined: 07 Dec 2017
Last visit: 09 Sep 2020
Posts: 1,002
Own Kudos:
Given Kudos: 26
Posts: 1,002
Kudos: 2,042
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hatakekakashi


I went with C as well initially .. but does the bulge definitetly mean it's going to take more area?

I think so... if the sides 'bulge' e.g are pressed outward it is implied that the cans need more space then they have.
To be fair this question is more about English than about math, not a standard question IMO
User avatar
Hatakekakashi
Joined: 07 Jan 2016
Last visit: 22 Feb 2025
Posts: 1,228
Own Kudos:
483
 [1]
Given Kudos: 126
Location: United States (MO)
GMAT 1: 710 Q49 V36
Products:
GMAT 1: 710 Q49 V36
Posts: 1,228
Kudos: 483
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DavidTutorexamPAL
Hatakekakashi


I went with C as well initially .. but does the bulge definitetly mean it's going to take more area?

I think so... if the sides 'bulge' e.g are pressed outward it is implied that the cans need more space then they have.
To be fair this question is more about English than about math, not a standard question IMO

I understand.. I think the word bulge makes all the difference


Regards,
HK
User avatar
DavidTutorexamPAL
User avatar
examPAL Representative
Joined: 07 Dec 2017
Last visit: 09 Sep 2020
Posts: 1,002
Own Kudos:
2,042
 [1]
Given Kudos: 26
Posts: 1,002
Kudos: 2,042
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hatakekakashi
DavidTutorexamPAL

Bunuel, just for the sake of nitpicking, why can't the cans be arranged in a triangle with another one on top? They would still all be 'on their bases', as required.

even though the question is not directed towards me I think if it's a triangle all 4 sides of the box won't bulge

So I did the math and actually you would lose very little height by arranging in a triangle, the answer would still be (C).
Basically if you put two cans on the base of the box = the base of the triangle they still need 8 inches of space for width.
And if you put the third right above them in the middle then you lose about 0.54 inches in total length giving a length of about 7.46.
So a 30-inch-perimeter box with a side of 7.5 inches would still likely bulge on all sides and at any rate is the only relevant answer.

Since we're on the topic, consider an arrangement where two cans are in opposite corners and are tangent to each other.
The other two cans are on top of them.
We have no information on the total height of the box vs. height of the cans so this could work.
Then the side of the minimal square is 4 + 2*sqrt(2) and the perimieter is about 27.3 inches
This could also be made to bulge on 4 sides quite easily by decreasing the perimeter slightly (but there is no relevant answer).

Anyways, for anyone reading please note that the above is not really relevant for your exam, feel free to skip over it :)
User avatar
Hatakekakashi
Joined: 07 Jan 2016
Last visit: 22 Feb 2025
Posts: 1,228
Own Kudos:
Given Kudos: 126
Location: United States (MO)
GMAT 1: 710 Q49 V36
Products:
GMAT 1: 710 Q49 V36
Posts: 1,228
Kudos: 483
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DavidTutorexamPAL
Hatakekakashi
DavidTutorexamPAL

Bunuel, just for the sake of nitpicking, why can't the cans be arranged in a triangle with another one on top? They would still all be 'on their bases', as required.

even though the question is not directed towards me I think if it's a triangle all 4 sides of the box won't bulge

So I did the math and actually you would lose very little height by arranging in a triangle, the answer would still be (C).
Basically if you put two cans on the base of the box = the base of the triangle they still need 8 inches of space for width.
And if you put the third right above them in the middle then you lose about 0.54 inches in total length giving a length of about 7.46.
So a 30-inch-perimeter box with a side of 7.5 inches would still likely bulge on all sides and at any rate is the only relevant answer.

Since we're on the topic, consider an arrangement where two cans are in opposite corners and are tangent to each other.
The other two cans are on top of them.
We have no information on the total height of the box vs. height of the cans so this could work.
Then the side of the minimal square is 4 + 2*sqrt(2) and the perimieter is about 27.3 inches
This could also be made to bulge on 4 sides quite easily by decreasing the perimeter slightly (but there is no relevant answer).

Anyways, for anyone reading please note that the above is not really relevant for your exam, feel free to skip over it :)


Thanks for taking your time out to do the math. in any case 30 should be fine then

but as you've mentioned that the height of the can is not mentioned and considering the cans resting on the lateral surface i.e like this - instead of | then the answer could also not be determined ( discussion not related to the question)

don't you think so?
User avatar
DavidTutorexamPAL
User avatar
examPAL Representative
Joined: 07 Dec 2017
Last visit: 09 Sep 2020
Posts: 1,002
Own Kudos:
Given Kudos: 26
Posts: 1,002
Kudos: 2,042
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hatakekakashi

Thanks for taking your time out to do the math. in any case 30 should be fine then

but as you've mentioned that the height of the can is not mentioned and considering the cans resting on the lateral surface i.e like this - instead of | then the answer could also not be determined ( discussion not related to the question)

don't you think so?

WRT to the question then the cans are placed on their bases so I don't think they can be placed on their lateral surface.
In general I agree with you, without information on the can's height the problem is unsolvable.
There are potentially all kinds of different solutions including a mix of 'on the side' and 'on the base'

If any of you are incredibly bored, consider this a spare time project to try at home!
Take a bunch of cans and play around with them, see how many you can cram into a box.
User avatar
Hatakekakashi
Joined: 07 Jan 2016
Last visit: 22 Feb 2025
Posts: 1,228
Own Kudos:
Given Kudos: 126
Location: United States (MO)
GMAT 1: 710 Q49 V36
Products:
GMAT 1: 710 Q49 V36
Posts: 1,228
Kudos: 483
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DavidTutorexamPAL


If any of you are incredibly bored, consider this a spare time project to try at home!
Take a bunch of cans and play around with them, see how many you can cram into a box.

hahaha :D I weigh 270 pounds so i guess can cram a lot of them ;)

Regards,
HK
User avatar
JeffTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,974
Own Kudos:
Given Kudos: 1,646
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert
Expert reply
Posts: 2,974
Kudos: 8,710
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Four cylindrical cans each with a radius of 2 inches are placed on their bases inside an open square pasteboard box. If the four sides of the box bulge slightly, which of the following could be the internal perimeter of the base of the box, expressed in inches?

(A) 64

(B) 32

(C) 30

(D) 20

(E) 16

The 4 cans will be placed on the square base of the box in the 2-by-2 formation. Since the total length of the diameters of 2 cans is 4 + 4 = 8, the cans would fit in a square with perimeter 8 x 4 = 32 without any bulging. Since all the sides of the box bulge slightly, the perimeter of the base of the box should be slightly less than 32. A possible value is 30.

Answer: C
Moderators:
Math Expert
109811 posts
Tuck School Moderator
853 posts