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If there are B boys and G girls in a club, can the girls be

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If there are B boys and G girls in a club, can the girls be [#permalink]

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12 Mar 2013, 03:18
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If there are B boys and G girls in a club, can the girls be divided equally among 6 teams with no girls left over?

(1) If there were 4 fewer girls then the number of girls would be twice the number of boys
(2) If the number of boys were 2 less than twice the actual number of boys, then the boys could be divided equally among 6 teams with no boys left over.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 12 Mar 2013, 03:42, edited 1 time in total.
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Re: If there are B boys and G girls in a club, can the girls be [#permalink]

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12 Mar 2013, 03:54
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If there are B boys and G girls in a club, can the girls be divided equally among 6 teams with no girls left over?

The question basically asks whether G is a multiple of 6.

(1) If there were 4 fewer girls then the number of girls would be twice the number of boys --> G-4=2B. If B=1, then G=6 and the answer is YES but if B=2, then G=8 and the answer is NO. Not sufficient.

(2) If the number of boys were 2 less than twice the actual number of boys, then the boys could be divided equally among 6 teams with no boys left over. This statement implies that 2B-2 is a multiple of 6. No info about G. Not sufficient.

(1)+(2) From (2) we have that 2B-2=6k --> B=3k+1. Thus from (1) we have that G-4=2(3k+1) --> G=6k+6=6(k+1) --> G IS a multiple of 6. Sufficient.

Hope it's clear.
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Re: If there are B boys and G girls in a club, can the girls be [#permalink]

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27 Jul 2014, 08:15
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Re: If there are B boys and G girls in a club, can the girls be [#permalink]

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16 Jul 2015, 10:13
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Question: "Is G/6 = int?"

(1) Insufficient
G – 4 = 2B

(2) Insufficient.
(2B – 2)/6 = int
You are not telling me anything about girls!

(1 & 2) Using Substitution
(G – 4 – 2)/6 = int
(G – 6)/6 = int

Split fraction --> G/6 – 1 = integer --> G/6 = integer + 1.
And since integer + 1 = another integer, then G/6 = another integer
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Re: If there are B boys and G girls in a club, can the girls be [#permalink]

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23 Dec 2015, 11:08
Bunuel wrote:
If there are B boys and G girls in a club, can the girls be divided equally among 6 teams with no girls left over?

The question basically asks whether G is a multiple of 6.

(1) If there were 4 fewer girls then the number of girls would be twice the number of boys --> G-4=2B. If B=1, then G=6 and the answer is YES but if B=2, then G=8 and the answer is NO. Not sufficient.

(2) If the number of boys were 2 less than twice the actual number of boys, then the boys could be divided equally among 6 teams with no boys left over. This statement implies that 2B-2 is a multiple of 6. No info about G. Not sufficient.

(1)+(2) From (2) we have that 2B-2=6k --> B=3k+1. Thus from (1) we have that G-4=2(3k+1) --> G=6k+6=6(k+1) --> G IS a multiple of 6. Sufficient.

Hope it's clear.

For statement 2, where does the 2B come from exactly? why is not b-2 = 6k
Re: If there are B boys and G girls in a club, can the girls be   [#permalink] 23 Dec 2015, 11:08
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