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manolas89
as you said (2) can cleary answer the question but i think you are wrong for (1) .(1) is suff too because we know it;s about 6 teams and the rule says even*even=even , even*odd=even .

Yes, even*even=even and even*odd=even but this is not applicable to this problem.

A line of people waiting to enter a theater consists of 7 separate and successive groups. The first person in each group purchases one ticket for each person in the group and for no one else. If n is the total number of tickets sold for the first 6 groups, is n an even group?

(1) There are no more than 4 people in each group --> n is the sum of 6 positive integers each of which is at most 4. From this we cannot say whether n is even or odd. Consider {1, 1, 1, 1, 1, 1} and {1, 1, 1, 1, 1, 2}. Not sufficient.

(2) The 19th person in line purchases the tickets for the 7th group. This implies that there are 18 people in the first 6 groups, thus n=18=even. Sufficient.

Answer: B.

Hope it's clear.
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manolas89
as you said (2) can cleary answer the question but i think you are wrong for (1) .(1) is suff too because we know it;s about 6 teams and the rule says even*even=even , even*odd=even .

Yes, even*even=even and even*odd=even but this is not applicable to this problem.

A line of people waiting to enter a theater consists of 7 separate and successive groups. The first person in each group purchases one ticket for each person in the group and for no one else. If n is the total number of tickets sold for the first 6 groups, is n an even group?

(1) There are no more than 4 people in each group --> n is the sum of 6 positive integers each of which is at most 4. From this we cannot say whether n is even or odd. Consider {1, 1, 1, 1, 1, 1} and {1, 1, 1, 1, 1, 2}. Not sufficient.

(2) The 19th person in line purchases the tickets for the 7th group. This implies that there are 18 people in the first 6 groups, thus n=18=even. Sufficient.

Answer: B.

Hope it's clear.

I didn't think that the question was clear ' If n is the total number of tickets sold for the first 6 groups, is n an even group? '

n is the number of tickets not a group, so I think the questions should be is 'n' even just like that.

Any comments?

Cheers
J :)
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manolas89
as you said (2) can cleary answer the question but i think you are wrong for (1) .(1) is suff too because we know it;s about 6 teams and the rule says even*even=even , even*odd=even .

Yes, even*even=even and even*odd=even but this is not applicable to this problem.

A line of people waiting to enter a theater consists of 7 separate and successive groups. The first person in each group purchases one ticket for each person in the group and for no one else. If n is the total number of tickets sold for the first 6 groups, is n an even group?

(1) There are no more than 4 people in each group --> n is the sum of 6 positive integers each of which is at most 4. From this we cannot say whether n is even or odd. Consider {1, 1, 1, 1, 1, 1} and {1, 1, 1, 1, 1, 2}. Not sufficient.

(2) The 19th person in line purchases the tickets for the 7th group. This implies that there are 18 people in the first 6 groups, thus n=18=even. Sufficient.

Answer: B.

Hope it's clear.

I didn't think that the question was clear ' If n is the total number of tickets sold for the first 6 groups, is n an even group? '

n is the number of tickets not a group, so I think the questions should be is 'n' even just like that.

Any comments?

Cheers
J :)


I chose B.. Bt m thinking one thing that ..N will always b even .. Because..6 groups* X(no of persons in each group)= N .. So n will never b odd integer..

I think question shud have asked that.. THE NUMBER OF MEMBERS IN EACH GROUP IS EVEN OR ODD? then Choice B wud be No. because 18 total members will contain 3 members in each group..so 6*3= 18..

correct me if m wrong
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sanjoo
I think question shud have asked that.. THE NUMBER OF MEMBERS IN EACH GROUP IS EVEN OR ODD? then Choice B wud be No. because 18 total members will contain 3 members in each group..so 6*3= 18..

correct me if m wrong

The question asks whether n is even. n does not equal to 6*(the # of people in each group) because the # of people in each group is not necessarily the same. For example, if the # of people in 6 groups are {1, 1, 1, 1, 1, 2} then n=7=odd.

Hope it's clear.
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Jamesk486
A line of people waiting to enter a theater consists of 7 separate and successive groups. The first person in each group purchases one ticket for each person in the group and for no one else. If n is the total number of tickets sold for the first 6 groups, is n an even group?

(1) There are no more than 4 people in each group
(2) The 19th person in line purchases the tickets for the 7th group

we have to figure out the number of ticket sold (thus by logical extension the number of people in the group) ; we have to exclude the 7th group
(1) There are no more than 4 people in each group
6 group * (1 person or 2 person or 3 person or max 4 person)
total people can be 6 to 24
NOT SUFFICIENT


(2) The 19th person in line purchases the tickets for the 7th group
The beginning of 7th group is 19, so it means the end of 6th group is 18
18 is even ; so the number of people is even
SUFFICIENT
ANSWER B
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Bunuel
manolas89
as you said (2) can cleary answer the question but i think you are wrong for (1) .(1) is suff too because we know it;s about 6 teams and the rule says even*even=even , even*odd=even .

Yes, even*even=even and even*odd=even but this is not applicable to this problem.

A line of people waiting to enter a theater consists of 7 separate and successive groups. The first person in each group purchases one ticket for each person in the group and for no one else. If n is the total number of tickets sold for the first 6 groups, is n an even group?

(1) There are no more than 4 people in each group --> n is the sum of 6 positive integers each of which is at most 4. From this we cannot say whether n is even or odd. Consider {1, 1, 1, 1, 1, 1} and {1, 1, 1, 1, 1, 2}. Not sufficient.

(2) The 19th person in line purchases the tickets for the 7th group. This implies that there are 18 people in the first 6 groups, thus n=18=even. Sufficient.

Answer: B.

Hope it's clear.

Was it just me that got thrown off by the wording of this question? "Is n an even group?", is this just a weird way of saying is n an even number? Because the question statement is clearly talking about 7 groups then says that n is the total number of tickets sold for the first 6 groups. I don't know, I just thought this wording was weird.
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Bunuel
manolas89
as you said (2) can cleary answer the question but i think you are wrong for (1) .(1) is suff too because we know it;s about 6 teams and the rule says even*even=even , even*odd=even .

Yes, even*even=even and even*odd=even but this is not applicable to this problem.

A line of people waiting to enter a theater consists of 7 separate and successive groups. The first person in each group purchases one ticket for each person in the group and for no one else. If n is the total number of tickets sold for the first 6 groups, is n an even group?

(1) There are no more than 4 people in each group --> n is the sum of 6 positive integers each of which is at most 4. From this we cannot say whether n is even or odd. Consider {1, 1, 1, 1, 1, 1} and {1, 1, 1, 1, 1, 2}. Not sufficient.

(2) The 19th person in line purchases the tickets for the 7th group. This implies that there are 18 people in the first 6 groups, thus n=18=even. Sufficient.

Answer: B.

Hope it's clear.

Was it just me that got thrown off by the wording of this question? "Is n an even group?", is this just a weird way of saying is n an even number? Because the question statement is clearly talking about 7 groups then says that n is the total number of tickets sold for the first 6 groups. I don't know, I just thought this wording was weird.

Same query here ... Bunuel can you clarify?
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Bunuel
manolas89
as you said (2) can cleary answer the question but i think you are wrong for (1) .(1) is suff too because we know it;s about 6 teams and the rule says even*even=even , even*odd=even .

Yes, even*even=even and even*odd=even but this is not applicable to this problem.

A line of people waiting to enter a theater consists of 7 separate and successive groups. The first person in each group purchases one ticket for each person in the group and for no one else. If n is the total number of tickets sold for the first 6 groups, is n an even group?

(1) There are no more than 4 people in each group --> n is the sum of 6 positive integers each of which is at most 4. From this we cannot say whether n is even or odd. Consider {1, 1, 1, 1, 1, 1} and {1, 1, 1, 1, 1, 2}. Not sufficient.

(2) The 19th person in line purchases the tickets for the 7th group. This implies that there are 18 people in the first 6 groups, thus n=18=even. Sufficient.

Answer: B.

Hope it's clear.

I didn't think that the question was clear ' If n is the total number of tickets sold for the first 6 groups, is n an even group? '

n is the number of tickets not a group, so I think the questions should be is 'n' even just like that.

Any comments?

Cheers
J :)


+1 .....Can someone please confirm
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from the question we can say that there are 7 total groups
target
if n is the total number of tickets sold for the first 6 groups, is n an even group

#1
(1) There are no more than 4 people in each group

this can be yes or no as multiples of all 6 groups can be ( 5 even and 1 odd ) or all even
#2
The 19th person in line purchases the tickets for the 7th group
which means that for 6 groups total count of tickets were 18 , even
sufficient

Jamesk486
A line of people waiting to enter a theater consists of 7 separate and successive groups. The first person in each group purchases one ticket for each person in the group and for no one else. If n is the total number of tickets sold for the first 6 groups, is n an even group?

(1) There are no more than 4 people in each group
(2) The 19th person in line purchases the tickets for the 7th group


DS21246
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disagree and this is a **** question. the 19th person in line purchases tickets for the 7th group - we have no idea whether they are first, middle, or last of the y7th group.

Could very well be the last person of the 7th group meaning the start could even perhaps be at person 2.
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