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Updated on: 06 Jul 2019, 05:23
Originally posted by jamifahad on 14 Sep 2011, 12:27.
Last edited by Sajjad1994 on 06 Jul 2019, 05:23, edited 1 time in total.
Last edited by Sajjad1994 on 06 Jul 2019, 05:23, edited 1 time in total.
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Question Stats:
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47%
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Suppose five circles, each 4 inches in diameter, are cut from a rectangular strip of paper 12 inches long. If the least amount of paper is to be wasted, what is the width of the paper strip?
(A) 5
(B) 4+2\(\sqrt{3}\)
(C) 8
(D) 4(1+\(\sqrt{3}\))
(E) 12
(A) 5
(B) 4+2\(\sqrt{3}\)
(C) 8
(D) 4(1+\(\sqrt{3}\))
(E) 12
Source: Nova GMAT
Difficulty Level: 700
Difficulty Level: 700
11 Sep 2014, 22:35
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jamifahad
First, think what they mean by 'least amount of paper is to be wasted'? Since the dimension of the circles is given, perhaps what they mean is that the width of the paper used must be minimum since the leftover paper will be wasted. So how will you place the 5 circles? The diameter of the circles is 4 each and the length of the paper is 12 so you can place 3 circles neatly across the length. The other two must be placed in a way which reduces width of the paper, which means between two circles rather than edge to edge.
Attachment:
Ques4.jpg [ 23.97 KiB | Viewed 10809 times ]
So width \(= 2 + 2*\sqrt{3} + 2 = 4 + 2*\sqrt{3}\)
hiimmunish
Joined: 10 May 2011
Last visit: 03 Nov 2012
Posts: 14
Location: India
Concentration: Strategy, General Management
GMAT 1: 690 Q50 V32
GMAT 1: 690 Q50 V32
Posts: 14
14 Sep 2011, 19:49
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Well, don't know how to draw diagrams here but let me see if i can explain through text.
First Approach (Traditional): First place 3 circles back-to-back (touching each other) to make it for 12 inches length. Now for least wastage of paper, you need to put 4th and 5th circles above 3 circles. 4th between 1st and 2nd circles and 5th between 2nd and 3rd. Now you can make equilateral triangle by joining centers of 1st, 2nd, and 4th circles. So total width of the strip will be radius of 4th cricle + height of equilateral triangle + radius of 1st (or 2nd) circle. It equals 2 + (sqrt3/2)4 + 2 = 4 + 2sqrt3
Second approach (quicker): Just think, if we place 4th circle exactly on the top of the 1st circle width of the strip will be sum of diameters of 2 circles 1.e 8. But that is not the case. Since 4th circle is placed between 1st and 2nd, some part of their diameters will overlap. It means our answer should be less than 8. You can eliminate options (C), (D), and (E). Option (A) is too less a value (it is suggesting that 4th circle is actually fitting 3/4th of its diameter between 1st and 2nd circles, which is not feasible). So you are left with only option and that's your answer.
I hope it helps!!!
First Approach (Traditional): First place 3 circles back-to-back (touching each other) to make it for 12 inches length. Now for least wastage of paper, you need to put 4th and 5th circles above 3 circles. 4th between 1st and 2nd circles and 5th between 2nd and 3rd. Now you can make equilateral triangle by joining centers of 1st, 2nd, and 4th circles. So total width of the strip will be radius of 4th cricle + height of equilateral triangle + radius of 1st (or 2nd) circle. It equals 2 + (sqrt3/2)4 + 2 = 4 + 2sqrt3
Second approach (quicker): Just think, if we place 4th circle exactly on the top of the 1st circle width of the strip will be sum of diameters of 2 circles 1.e 8. But that is not the case. Since 4th circle is placed between 1st and 2nd, some part of their diameters will overlap. It means our answer should be less than 8. You can eliminate options (C), (D), and (E). Option (A) is too less a value (it is suggesting that 4th circle is actually fitting 3/4th of its diameter between 1st and 2nd circles, which is not feasible). So you are left with only option and that's your answer.
I hope it helps!!!
