Papercut parts

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Papercut parts

by harsh.champ » Sun Feb 07, 2010 11:45 am
I cut a piece of paper in four equal parts. Now, I cut exactly one out of these four parts into four equal parts. Now, again if I keep on repeating the same process for infinite number of times, then which of the following can be the number of parts of paper at any instant of time?

(A)2048
(B)2049
(C)2050
(D)2051
(E)2052

The correct answer is C.
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by shashank.ism » Sun Feb 07, 2010 11:57 am
harsh.champ wrote:I cut a piece of paper in four equal parts. Now, I cut exactly one out of these four parts into four equal parts. Now, again if I keep on repeating the same process for infinite number of times, then which of the following can be the number of parts of paper at any instant of time?

(A)2048
(B)2049
(C)2050
(D)2051
(E)2052

The correct answer is C.
in first cut no. of parts = 4 = 3x1 +1
in 2nd cut no. of parts = 3+4 = 3x2 +1
in 3rd cut no. of parts = 3+3+4 = 3x3 +1


similarly in nth cut no. of parts = 3n+1

solving for any of 2048,2049, 2050 , 2051 or 2052
2049 = 3n+1 --> n=2048/3=682.66
--> n=2050
The correct answer is C.[/quote]
Last edited by shashank.ism on Sun Feb 07, 2010 2:17 pm, edited 2 times in total.
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by ajith » Sun Feb 07, 2010 12:33 pm
harsh.champ wrote:I cut a piece of paper in four equal parts. Now, I cut exactly one out of these four parts into four equal parts. Now, again if I keep on repeating the same process for infinite number of times, then which of the following can be the number of parts of paper at any instant of time?

(A)2048
(B)2049
(C)2050
(D)2051
(E)2052

The correct answer is C.
after the first cut the no of parts 4
after the second cut num of pieces = 3+4
after the third cut num of pieces = 3+3+4

After nth cut no of pieces = 3n+1
Inthe given numbers only 2050 can be expressed 2050 hence C
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