Did one of the 3 members of a certain team sell at least 2 raffle tick

Did one of the 3 members of a certain team sell at least 2 raffle tickets yesterday?

The question basically asks whether there is a member who sold at least 2 tickets (so 2 or more).

(1) The 3 members sold a total of 6 raffle tickets yesterday. If each of the 3 members sold less than 2 tickets, then the total # of tickets sold cannot be 6, hence at least one member sold at least 2 tickets. Sufficient.

Or: can we split 6 tickets so that ALL 3 members to have sold less than 2 tickets? No: (6,0,0); (5,1,0), (4,1,1); (4,2,0); (3,3,0),(3,2,1), (2,2,2). Sufficient.

(2) No 2 of the members sold the same number of raffle tickets yesterday. If one member sold 0 tickets and another sold 1 ticket (the least possible numbers), then the third one must have sold more than 1, so 2 or more. Sufficient.

Answer: D.

Hope it's clear.
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If the Q asks "whether there is a member who sold at least 2 tickets (so 2 or more)" then it should have been framed as atleast one NOT one which indicates Exactly one. Hence the doubt. Infact, i would say the Q should use the word Atleast or Exactly to make it clear.

Not so (even though I do see why you are confused).

Actually it's opposite, if the question meant exactly (only) one, then it would say so.
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what does this mean

" No 2 members sold the same number of tickets"

It means that each of the 3 members sold different number of tickets.

Hope it's clear.
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The question looks kind of tricky to me,
However deep diving I think- the bottomline of the question is did any of the 3 members sold more than 1 ticket.

Now total no of tickets sold in 6 and tickets cannot be negative or fractions hence yes statement 1 is true.

Similarly statement 2 is also true if 3 of them sold different number of tickets then they can sell (0,1, 2) tickets .

Thus both the statements are sufficient to prove it correct.

Hope this helps.

statement (1):
there's a statement called the pigeonhole principle, which basically says the following two things:
* if the AVERAGE of a set of integers is an INTEGER n, then at least one element of the set is > n.
* if the AVERAGE of a set of integers is a NON-INTEGER n, then at least one element of the set is > the next integer above n.
this principle is easy to prove: if you assume the contrary, then you get the absurd situation in which every element of a set is below the average of the set. that is of course impossible.

specifically, statement (1) is a case of the first part of the principle: the average of the set is 6/3 = 2, so at least one element of the set must be 2 or more.


statement (2):
there are only two ways not to sell at least 2 tickets: sell 0 tickets, and sell 1 ticket.
if everyone sells a different # of tickets, then you can't fit three people into these two categories.
therefore, someone must have sold at least 2 tickets.

Hence D.

(1) 3 members sold a total of 6 raffle tickets - Combinations are 2,2,2, 4,1,1, 5,0,1, and so on. So in each case at least one would've sold more than 2. - Sufficient

(2) No 2 of the members sold the same number of raffle tickets yesterday - We don't know how many tickets were sold yesterday but we know no 2 person sold the same no. of tickets. Since they are tickets they must be 3 different non negative integers. The 2 different smallest non negative integers are 0, and 1. Hence the other one has to be either 2 or greater than 2. So either way we can tell one person has sold at least 2 raffle tickets. - Sufficient

Hope this explains.

Cheers!

Sure. This is a good alternative solution.

But will you really remember this principle on the test day ?

In my opinion, it's better to go intuitive about questions. But that's what I think. Everyone's got his/her way of doing things !

I agree with some people who posted their doubts about the questions earlier. I chose option E.

If the question had clearly mentioned "Did atleast one of the 3 members of a certain team sell at least 2 raffle tickets yesterday." then it would've caused much less confusion.

But at the end of the day it's an official question, so we can't argue much. :-/

\(A,B,C\, \ge 0\,\,{\rm{ints}}\,\,\,\left( * \right)\)

\(?\,\,\,:\,\,\,\,A \ge 2\,\,\,{\rm{or}}\,\,\,B \ge 2\,\,\,{\rm{or}}\,\,\,C \ge 2\)

\(\left( 1 \right)\,\,A + B + C = 6\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)

\(\left( {**} \right)\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\left\{ \matrix{\\
\,A \le \,\,1 \hfill \cr \\
\,B \le \,\,1 \hfill \cr \\
\,C \le \,\,1 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,A + B + C\,\, \le \,\,3\,\,\,\,\, \Rightarrow \,\,\,\,\left( 1 \right)\,\,\,{\rm{contradicted}}\)


\(\left( 2 \right)\,\,A,B,C\,\,{\rm{different}}\,\,\,\,\mathop \Rightarrow \limits^{\left( {***} \right)} \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)

\(\left( {***} \right)\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{\\
\,A \le \,\,1\,\,\,\, \Rightarrow \,\,\,A \in \left\{ {0,1} \right\} \hfill \cr \\
\,B \le \,\,1\,\,\,\, \Rightarrow \,\,\,B \in \left\{ {0,1} \right\} \hfill \cr \\
\,C \le \,\,1\,\,\,\, \Rightarrow \,\,\,C \in \left\{ {0,1} \right\} \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( 2 \right)\,\,\,{\rm{contradicted}}\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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(1) The 3 members sold a total of 6 raffle tickets yesterday.
Minimum case = 2 each by 3 members
other cases -0,0,6 or 1,0,5 and many more in all at least one has sold 2 tickets


(2) No 2 of the members sold the same number of raffle tickets yesterday.
Minimum case - 0,1,5 or 1,2,3 or 0,2,4 0r any other Tickets will always be at least 2 for one of member as case
2,2,2 is not valid here

So answer is D

(2) No 2 of the members sold the same number of raffle tickets yesterday.

I still confuse with (2)

what if yesterday the total of sell was 1
No 1 : 0
No 2 : 1
No 3 : 0

and what if yesterday the total sell was 3

No 1 : 2
No 2 : 1
No 3 : 0

So how could this statement sufficient

Did one of the 3 members of a certain team sell at least 2 raffle tickets yesterday?

(1) The 3 members sold a total of 6 raffle tickets yesterday. sufficient as even if two members 1,1 the third one will sell 4
(2) No 2 of the members sold the same number of raffle tickets yesterday. sufficient since members can sell 0,1,2 (a all three sold diff nos so atleast one will sell 2 or more

1) If one person sold all tickets => Yes
If all of them sold equal amount of tickets, then each of them sold 2 tickets => Yes
Sufficient

2) Minimum number of tickets
1st person: 0
2nd person: 1
3rd person: 2 => Yes
Sufficient => D

The question is asking whether one of the three members sell at least 2 tickets.

The way I interpreted this was: if there is a case where more than one member sold at least 2, then the statement is not true. Hence, the statement is true only when exactly one sells at least 2 raffle tickets

Can anyone explain which topic this would fall under. apart from it being a plain word problem?

This is a Yes-No type of DS question. Any data that helps us answer the question with a definite YES or a definite NO will be sufficient data.
We are trying to ascertain if one of the 3 members of a team sold at least 2 i.e. a minimum of 2 raffle tickets yesterday.

From statement I alone, the 3 members sold a total of 6 raffle tickets yesterday. If we take the number of tickets sold by these persons to be a, b and c, we can say,
a + b + c = 6.

Considering the worst case scenario, if we say a=0 and b = 0, c=6. Did one of the members sell a minimum of 2 tickets? YES, the person who sold 6 did.
Let’s take a different case now. If a = 1 and b = 2, c =3. Did one of the 3 members sell a minimum of 2 tickets? I think we get a more resounding YES here.
We can also consider the case where a=b=c=2. In this case also, the answer to the main question is a YES.

Statement I alone is sufficient. Answer options B, C and E can be eliminated. Possible answer options are A or D.

From statement II alone, no two of the members sold the same raffle tickets yesterday.
This means a≠b≠c.

In the worst case scenario, a=0, b = 1 and c=2. Did one of the two members sell a minimum of 2 tickets? YES.

Remember that the case above is the worst case possible. In other cases, the total number of tickets sold is bound to be more and hence, there will be this ONE person who will definitely sell at least 2 raffle tickets.
Statement II alone is sufficient. Answer option A can be eliminated.

The correct answer option is D.

In a question like this, where the values involved are smaller, it makes a lof sense to adopt the plugging in strategy because it helps you to clearly identify your YESs and NOs which is very important.

Hope that helps!
Arvind.

Ha - I actually interpreted this statement as : Number two member sold the same number of tickets

thank you for clarifying!