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saurabhkhatrinitk
Bunuel
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.

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"
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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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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.
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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

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

(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!
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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.
(2) No 2 of the members sold the same number of raffle tickets yesterday.
\(A,B,C\, \ge 0\,\,{\rm{ints}}\,\,\,\left( * \right)\)

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

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

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


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

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


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

Regards,
Fabio.
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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.
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So, we need to find whether any one of the 3 members sell at least 2 raffle tickets or not i.e., 2 or more raffle tickets

Statement 1.
It’s given that 3 members sold a total of 6 tickets yesterday.
If we assume that if each of the 3 members sold only less than 2 tickets, the maximum sum of tickets they can sell is (1,1,1) combination i.e. 3 tickets.
But in the statement its given that they sold a total of 6 tickets. That means at least one of the 3 members should definitely sell more than 2 tickets.
Hence Statement 1 is sufficient.
Option B ,C and E are eliminated. The answer could be either A or D.

Statement 2.
It means each of the 3 members sold different no of tickets. That means no of tickets each sold should be 3 different non negative integers.
So, the minimum combination you can form is (0,1,2)
Hence , we can confirm at least one of the members sold at least 2 tickets
Statement 2 is also sufficient .
Hence Option D


Thanks,
Clifin J Francis,
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I find the wording of this question confusing. If two people sold at least two raffle tickets, then the answer to the question ‘Did one person sell…?’ is no, because more than one person sold tickets.
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I find the wording of this question confusing. If two people sold at least two raffle tickets, then the answer to the question ‘Did one person sell…?’ is no, because more than one person sold tickets.

That's not the correct interpretation of the question. The question actually means 'Did any of the team members sell two or more tickets?' This includes the possibility that one, two, or all three members could have sold at least two tickets each. It is asking if at least one member sold two or more tickets. It doesn't matter if only one, two, or all three members did so; as long as at least one person sold two or more, the answer is yes.
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Question Stem Analysis:
  • There are 3 members in a team who sell raffle tickets.
  • We are asked: Did at least one member sell ≥ 2 tickets? This is a YES/NO question -- we need a definite answer

Let the number of tickets sold by the three members be a, b, c.
We need to check whether it is guaranteed that at least one of a, b, c ≥ 2?

Statement Analysis:

Statement (1): Total tickets sold = 6
  • a + b + c = 6
  • If we try to avoid ≥ 2 for all three of a, b, and c, we will only reach a maximum of 1 + 1 + 1 = 3, which is too small.
  • So, to reach 6, at least one must be ≥ 2

Since we get a definite Yes, statement (1) is sufficient alone. Eliminate B, C, and E.

Statement (2): No two members sold the same number.
This means that a, b, c are all distinct.
  • To check whether it’s possible that no one sold ≥ 2, we try to keep all values as small as possible.
  • The three smallest distinct non-negative integers are: 0, 1, 2
  • Even in this minimum case, one member sold 2 tickets. So, it is not possible to have all three values less than 2.

So, at least one must be ≥2
Since we get a definite Yes, statement (2) is sufficient alone. Eliminate A.

Correct Answer: D

Shweta Koshija
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