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# When a class of n students is divided into groups of 6 students each,

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Intern
Joined: 27 Feb 2018
Posts: 3
When a class of n students is divided into groups of 6 students each,  [#permalink]

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Updated on: 15 Mar 2018, 08:39
1
00:00

Difficulty:

45% (medium)

Question Stats:

75% (02:08) correct 25% (02:21) wrong based on 66 sessions

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When a class of n students is divided into groups of 6 students each, 2 students are left without a group. When the class is divided into groups of 8 students each, 4 students are left without a group. What is the smallest number of students that can be added to or removed from the class so that the resulting number of students can be equally divided into groups of 12 students each?

(A) 2
(B) 4
(C) 8
(D) 10
(E) 12

Originally posted by jwin125 on 14 Mar 2018, 18:54.
Last edited by Bunuel on 15 Mar 2018, 08:39, edited 3 times in total.
Math Expert
Joined: 02 Aug 2009
Posts: 7960
Re: When a class of n students is divided into groups of 6 students each,  [#permalink]

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14 Mar 2018, 19:22
1
jwin125 wrote:
When a class of n students is divided into groups of 6 students each, 2 students are left without a group. When the class is divided into groups of 8 students each, 4 students are left without a group. What is the smallest number of students that can be added to or removed from the class so that the resulting number of students can be equally divided into groups of 12 students each?

Please follow the rules and so not miss out on giving choices..

Our requirement is that the final number has to be MULTIPLE of 3 and 4

Given..
$$n=6x+2$$... so n is a multiple of 2 but not of 3... Example 2,8,14,20,26,32,38,44,50
$$n=8y+4$$... n is surely a multiple of 4.. example 4,12,20,28,36,44,52

So numbers are $$20+24a...$$
Also we can get this from (FIRSTCommon Number+LCM of 6,8)

When you divide (20+24a) by 12, 24a is div by 12 and 20 gives a remainder of 8..
So we can add 4 or remove 8 to get n div by 12..
Least is 4..
Ans is 4
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Re: When a class of n students is divided into groups of 6 students each,   [#permalink] 14 Mar 2018, 19:22
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