12 Days of Christmas GMAT Competition 2025 - Day 7: At a customer

7 agents to be given total 28 tickets. Each ticket to 1 agent only.

I: If each agent given different no of tickets = we can have 1,2,3,4,5,6,7 or 0,1,2,3,4,5,13. Thus we can not say that each agent receive at least 1 ticket. NOT SUFFICIENT.

II. An agent was assigned 7 ticket (max for group).
So we can have 7,0,0,6,5,5,5 or 7,1,1,4,5,5,5. Thus this statement alone is also NOT SUFFICIENT.

I&II: If 1 agent gets 7 which is max and all rest get different no, we have 7,6,5,4,3,2,1. It is only possible outcome. Thus each agent received at least 1 ticket can be assessed. Hence both together are sufficient. C.

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23 Dec 2025, 14:50

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Bunuel

Stmnt 1-- tickets may be assigned as per following
1,2,3,4,5,6,7
or
0,2,3,4,5,6,8
Not sufficient
Stmnt 2
the assigned tickets could be like this
1,2,3,4,5,6,7
or
0,2,3,4,5,7,7
combing both statements now we cannot repeat ticket value the only possible arrangement is
1,2,3,4,5,6,7
sufficient
Ans-C

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(1) Different number of tickets by each agent. This is not sufficient to identify if each has at least one ticket, as it could be one of them have 0 tickets and rest on get different digit adding up to 28.

(2) Max = 7. This is also not sufficient alone.

Considering together, the only possible scenario is:
7+6+5+4+3+2+1 = 28.
which implies together it is sufficient.

(C) is the answer.

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23 Dec 2025, 19:19

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Statement 1 - Assined 0 ticket to one agent rest 6 can be assigned different nos similarly all seven can be assigned different nos without assigning 0 to anyone hence not sufficient.
Statement 2 - 7, 0, 4, 4, 4, 4, 5 or 7, 1, 4, 4, 4, 4, 4 hence not sufficient

Combining both only one option possible - 7, 6, 5, 4, 3, 2, 1

Hence answer "C"

Bunuel

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23 Dec 2025, 20:54

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There are 7 agents and 28 tickets assigned to them in that particular shift

From the statement 1 : if each agent is assigned different number of tickets which means 0,1,23,4,5,6,7,8 can be assigned to the agents available. Therefore the agent may be get 0 ticket as well l. This says that only A isn't enough to answer the question.

From statement 2 : max number of tickets assigned to one agent is 7, if this is the case one of the agent just get 1 and not 0 from the 0-8 probable cases of getting tickets. And the tickets that get assigned to the agents will be from 1-7

That means to answer the question both statements are required hence C is the answer

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7 agents. 28 total tickets.

Did each agent get one ?

Statement 1:
each agent got different ticket.

one possible scenario:
1, 2, 3, 4, 5, 6, 7

another possible scenario:
20, 8, 0, 0, 0, 0, 0

insufficient

Statement 2:

max is 7

one possible scenario
1, 2, 3, 4, 5, 6, 7

another possible scenario
7, 7, 7, 7, 0, 0, 0
or
7, 6, 6, 6, 3, 0, 0

insufficient

Combined:

max is 7 and each one has unique number

only one solution:
1, 2, 3, 4, 5, 6, 7

sufficient

ans: option C

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Bunuel

This is a yes or no question, so any statement that gives the result "maybe" won't work.

Statement 1: We could potentially have an order such as $13+5+4+3+2+1+0=28$, or we could have an order where everyone gets a ticket. Answer to the stem's question is maybe, therefore insufficient.

Statement 2: We could potentially have a setup where: $7+6+6+6+3+0+0=28$, or we could have an order where everyone gets a ticket. Answer to the stem is "maybe", therefore insufficient.

Statement 1+2: In this situation we are forced to have only one setup: $7+6+5+4+3+2+1=28$. Answer to the stem is a definitive "yes". Sufficient, answer is C.

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imo C, both statements required.
1. For this to not work, the last person could still have 0 tickets or atleast 1 while still keeping all 7 different
2. setting the max to 7 allows multiple agents to hold/not hold 7 tickets, i.e. altering the number of agents with 0 tickets
Combining both, you end up with only 1 possible arrangement of tickets where everyone gets atleast 1 ticket

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(1) Each agent was assigned a different number of tickets
We can have {0,1,2,3,4,5,13} or {1,2,3,4,5,6,7} so this is NOT SUFFICIENT.
(2) From this, we can also have {0,4,4,4,4,5,7} or {1,4,4,4,4,4,7} as long as the maximum number of the set is 7 and total = 28. NOT SUFFICIENT.

(1)+(2) => Each agent had different number of tickets + Max = 7 so we only have {1,2,3,4,5,6,7}. SUFFICIENT.

Bunuel

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Bunuel

1)

Case 1 : The first agent gets 20 tickets and the second agent gets the 8 tickets. In this case, not every agent receives a ticket.
Case 2 : The distribution is as follows -

1 2 3 4 5 6 7

As we have two posssible scenarios, we can eliminate A and D.

In this case, every agent receives a ticket.

2)

Case 1: The distribution is as follows

7 7 7 7

In this case, not every agent receives a ticket.

Case 2: The distrbution is as follow

1 2 3 4 5 6 7

In this case, every agent receives a ticket.

Combining :

Only case 2 is common. Hence, we can conclude that each agent received a ticket.

Option C

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23 Dec 2025, 23:35

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Bunuel

7 agents are working at a customer centre, and 28 tickets are assigned to these agents.

No two agents work on the same ticket.

Did each recieve at least 1 ticket ?

Statement 1:

Each agent was assigned a different number of tickets.

28 tickets divide among 7 agents, so each gets 4 tickets.

Each gets a different number.

One of the possible combination is : (1,2,3,4,5,6,7).

Another possible combination is : (0,1,2,3,4,5,13).

Each gets a different amount. Since, we get two possible outcomes.

Insufficient.

Statement 2:

The agent who was assigned the most tickets was assigned 7 tickets.

There are many cases: (0,0,0,7,7,7,7).

(0,0,1,6,7,7,7) and we can keep a case 0, and change the combinations.

Many values occurs, Insufficient

Combining both statements 1 and 2, we get

Max value an agent can get is 7.

All have unique values.

Only one possible combination prevails amongst the rest.

(1,2,3,4,5,6,7)

Option C

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participating in 12 days of charismatic in GMAT competitions 2025- 7 days. At a Customer really highlighted how customer experience matter across industry. for example Asna Jewellery is the best jewellery shop in dubai, customer expert personalised attention.

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Given that , 7 agents worked a particular evening shift. And there are 28 customer tickets which were assigned to them. Each ticket was assigned to exactly one agent. We need to find out whether each agent receives at least 1 ticket.

(1) Each agent was assigned a unique number of tickets. There can be several possibilities : 1 to 7 , or 0 ,1,2,3,4,5,13 etc. So we cannot tell using this information. So , Statement 1 is insufficient.
(2) Max number of tickets assigned to agent is 7. But here also there can be several possibilities : 1 to 7 or 7,7,7,7,0,0,0 etc. So Statement 2 is insufficient.

Now combining both statements :

Only one possibility : 1 to 7. It means each agent receives atleast one ticket.

So yes C is the answer.

Bunuel

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S1 That means we could have 1234567 which is sufficient but also 0234568 which is insufficient hence insufficient
S2 We could have something like 7777000 so insufficient
S1+S2 = Means we can only have 1234567 as the only order that fulfils the constraints hence sufficient
Ans C

Bunuel

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(1) It could be 0, 1, 2, and so on, or 1, 2, and so on. Not sufficient.

(2) So remaining 21 tickets, it could be assigned anyway. Say, 3 people get 6, one gets 3, and the rest 2 people get no ticket, and so on. Not sufficient.

(1) + (2) Let's see the worst-case possible. 7(this is the maximum), 6(next biggest number), 5, 4, 3, 2, 1, this equals 28. So each agent gets at least 1 ticket for sure, as we can see that is the only possible case. Sufficient.
Option C.

Bunuel

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I am now wiser for how tricky the DS questions in this competition are, and not trusting these to be anything but tricky, I believe, really helped me with this one!

So, we have 7 agents working during an evening shift. 28 customer tickets were assigned - this number, is the total such tickets assigned, divided among 7 agents (an average of 4 tickets, if you may).

Also, each ticket to exactly one agent just means there's no overlaps in tickets. Noted.

We need to see if did each agent receive at least 1 ticket - or did, even one of then, spend a day chilling without any assignments?

Simple stem, tricky question, let's go!

Statement 1: Each agent was assigned a different number of tickets.

Key realization: 0 is also a number.

Now, if we assign each agent the lowest possible different numbers without considering the 0, we have exactly 28.

1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.

But if, say, 1 of the agents takes 0 tickets, then too 28 is achievable.

0 + 1 + 2 + 3 + 4 + 5 + 13 = 28.

Two possible, results, hence Statement I alone IS NOT SUFFICIENT.

Statement II: The agent assigned the highest number of tickets was assigned 7 tickets.

Key realization: We no longer know that each agent was assigned a different number of tickets.

This means, plenty of combinations of 28 are possible. It could be 0 + 0 + 3 + 6 + 6 + 6 + 7 = 28; or 0 + 0 + 5 + 5 + 5 + 5 + 7; anything really.

Multiple results. Hence Statement I alone IS NOT SUFFICIENT.

Let's take two statements together:

We know that each individual is assigned a different number, and the largest number of these is 7. So, we can only go backwards. We saw above that 1 + 2.... + 7 = 28, so we have that as a lock-in.

Now, here's the thing, if we assume any of these numbers to be a 0, the other 6, while being different, will never add up to 28. 0 + 2 + 3 + 4 + 5 + 6 + 7 = 27; now, if you replace any other number, we'll just derive a number that's lesser than lesser than 28, never exactly 28, which the number needs to be.

Hence, Both Statement I and Statement II - Answer is C

Bunuel

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Statement 1: Even if each agent received different number of tickets, there is no guarantee that all the agents were assigned the tickets (eg: 8, 6, 5, 4, 3, 2, 0)
Statement 2: If the max #tickets assigned were 7, there can still be agents who were not assigned any ticket (eg: 7, 7, 5, 4, 3, 2, 0)

However, a combination of these statements assures that each agent was assigned atleast 1 ticket.

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Bunuel