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Four cylindrical cans each with a radius of 2 inches are placed on th
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18 Feb 2018, 22:54
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29% (01:34) correct 71% (01:37) wrong based on 68 sessions
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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
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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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18 Feb 2018, 23:37
Bunuel wrote: 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 nonsymmetrical 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.
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Four cylindrical cans each with a radius of 2 inches are placed on th
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19 Feb 2018, 00:02
Bunuel wrote: 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 diameter of can = 4 so if side of the box buiges there is one can at the end of each of the box/facing each side and for it to bulge the box should be able to accommodate 8 inches in length atleast imo 8 is the side of the square and perimeter = 32 (B) looks good but considering bulge of the box i.e takes more area a side of 7.5 can suffice ps  considering the bulge is caused by the exterior of a can went it C at first i thought it'd be B



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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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19 Feb 2018, 00:03
DavidTutorexamPAL wrote: Bunuel wrote: 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 nonsymmetrical 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?



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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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19 Feb 2018, 00:06
DavidTutorexamPAL wrote: 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



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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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19 Feb 2018, 00:08
Hatakekakashi wrote: 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
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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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19 Feb 2018, 00:15
DavidTutorexamPAL wrote: Hatakekakashi wrote: 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



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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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19 Feb 2018, 00:47
Hatakekakashi wrote: DavidTutorexamPAL wrote: 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 30inchperimeter 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
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Four cylindrical cans each with a radius of 2 inches are placed on th
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19 Feb 2018, 01:51
DavidTutorexamPAL wrote: Hatakekakashi wrote: DavidTutorexamPAL wrote: 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 30inchperimeter 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?



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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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19 Feb 2018, 02:00
Hatakekakashi wrote: 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.
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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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19 Feb 2018, 02:08
DavidTutorexamPAL wrote: 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



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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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21 Feb 2018, 14:21
Bunuel wrote: 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 2by2 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
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Re: Four cylindrical cans each with a radius of 2 inches are placed on th
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