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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
nguyendinhtuong wrote:
rahuldev31 wrote:
Of the 20 people who each purchased 2 tickets to a concert, some used both tickets, some used only 1 ticket, and some
used neither ticket. What percent of the tickets that were purchased by the 20 people were used by those people?

(1) Of the 20 people, 10 used only 1 ticket.

(2) Of the 20 people, 4 used neither ticket.


Can this be solved using Overlapping sets instead of double matrix?



20 people who each purchased 2 tickets, so there are total 40 tickets.

Of 20 people, some used both tickets, some used only 1 ticket, and some used 0 ticket.

(1) 10 people used only 1 ticket, but we can't know how many people who used both and how many people who used neither. Insufficient.

(2) 4 people used neither, so 16 remainings used both or only 1. We still don't know how many people who used only 1 and how many people who used both. Insufficient.

Combine (1) and (2) we have 4 used neither, 10 used only 1 so remaining 6 used both. Since that, we can deduce the number of used ticket. Sufficient.



Thank you. The only confusing part is " What percent of the tickets that were purchased by the 20 people were used by those people?"

So I interpret (1 ticket used + Both) /Total. Since option (2) gives us number of neither used, remaining must have been used.
But obviously I'm misinterpreting something cuz C is the right answer
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
Bunuel, whats your take on this?
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
rahuldev31 wrote:
Of the 20 people who each purchased 2 tickets to a concert, some used both tickets, some used only 1 ticket, and some
used neither ticket. What percent of the tickets that were purchased by the 20 people were used by those people?

(1) Of the 20 people, 10 used only 1 ticket.

(2) Of the 20 people, 4 used neither ticket.


Can this be solved using Overlapping sets instead of double matrix?



S1: 10 used 1 ticket but we don't know how many tickets each of the other 10 used.....INSUFFICIENT

S2: 4 used neither ticket but we don't know how many tickets each of the other 16 used.....INSUFFICIENT


S1+S2: 10 buy 1 = `10*1 =10
4 buy 0 =0*4=0
therefore, 6 buy 2 = 6*2 =12.....SUFFICIENT...SO C
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
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rahuldev31 wrote:
Thank you. The only confusing part is " What percent of the tickets that were purchased by the 20 people were used by those people?"

So I interpret (1 ticket used + Both) /Total. Since option (2) gives us number of neither used, remaining must have been used.
But obviously I'm misinterpreting something cuz C is the right answer


The question is like this:
20 people purchased total 40 tickets. What percent of these 40 tickets that were used by those 20 people.

In (2), we know the number of people who used neither, then yes we know the total number of people who used only 1 or both. However, the question is asking the percent of tickets, not the percent of people. Hence, we need to know the specific number of people who used only 1 ticket and the specific number who used both.

Hope this helps.
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
rahuldev31 wrote:
Of the 20 people who each purchased 2 tickets to a concert, some used both tickets, some used only 1 ticket, and some
used neither ticket. What percent of the tickets that were purchased by the 20 people were used by those people?

(1) Of the 20 people, 10 used only 1 ticket.

(2) Of the 20 people, 4 used neither ticket.


Can this be solved using Overlapping sets instead of double matrix?



misread the question and thought that not all purchased 2 ticket rather some have purchased ...so landed to E

its C...
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
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rahuldev31 wrote:
Of the 20 people who each purchased 2 tickets to a concert, some used both tickets, some used only 1 ticket, and some
used neither ticket. What percent of the tickets that were purchased by the 20 people were used by those people?

(1) Of the 20 people, 10 used only 1 ticket.

(2) Of the 20 people, 4 used neither ticket.


We can create the following formula:

Total people = # who only used only 1 ticket + # who used both tickets + # who used neither ticket

20 = # who only used only 1 ticket + # who used both tickets + # who used neither ticket

Since each of the 20 people bought 2 tickets, a total of 40 tickets were bought. Of course, not all tickets were used since some used only 1 ticket and some used neither ticket. We need to determine what percentage of the tickets that were purchased by the 20 people were used by those people.

Statement One Alone:

Of the 20 people, 10 used only 1 ticket.

We see that the # who only used only 1 ticket is 10. However, since we know nothing about the number of people who used neither ticket, we still cannot answer the question. Statement one alone is not sufficient.

Statement Two Alone:

Of the 20 people, 4 used neither ticket.

We see that the # who used neither ticket is 4. However, since we don’t know the number of people who used only 1 ticket, we still cannot answer the question. Statement two alone is not sufficient.

Statements One and Two Together:

Using the information from statements one and two, we see that:

20 = 10 + # who used both tickets +4

Thus, 20 - 14 = 6 people used both tickets.

Since a total of 40 tickets were purchased and only (6 x 2) + (10 x 1) = 22 were used, the percentage of tickets purchased by the 20 people that were used is 22/40 x 100 = 11/20 x 100 = 55 percent.

Answer: C
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
rahuldev31 wrote:
nguyendinhtuong wrote:
rahuldev31 wrote:
Of the 20 people who each purchased 2 tickets to a concert, some used both tickets, some used only 1 ticket, and some
used neither ticket. What percent of the tickets that were purchased by the 20 people were used by those people?

(1) Of the 20 people, 10 used only 1 ticket.

(2) Of the 20 people, 4 used neither ticket.


Can this be solved using Overlapping sets instead of double matrix?



20 people who each purchased 2 tickets, so there are total 40 tickets.

Of 20 people, some used both tickets, some used only 1 ticket, and some used 0 ticket.

(1) 10 people used only 1 ticket, but we can't know how many people who used both and how many people who used neither. Insufficient.

(2) 4 people used neither, so 16 remainings used both or only 1. We still don't know how many people who used only 1 and how many people who used both. Insufficient.

Combine (1) and (2) we have 4 used neither, 10 used only 1 so remaining 6 used both. Since that, we can deduce the number of used ticket. Sufficient.



Thank you. The only confusing part is " What percent of the tickets that were purchased by the 20 people were used by those people?"

So I interpret (1 ticket used + Both) /Total. Since option (2) gives us number of neither used, remaining must have been used.
But obviously I'm misinterpreting something cuz C is the right answer



Hi Rahul,


For finding percentage you require the count of tickets utilised and not the count of people. Option 2 gives us the count of people who have utilised the tickets. Out of 16, the breakup of number of people who utilized 2 tickets and the number of people who utilised 1 ticket is not provided.The total number of tickets used depends on this. Only after getting this value can we calculate the percantage of used tickets using formula(Total used tickets/40).
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
On combining both the statements we get 4 used neither, 10 used only one and remaining 6 used both. Hence Sufficient.
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
rahuldev31

This ques can be solved using sets but since there are only three cases - 1 ticket , 2 ticket , Neither
The intersection part would not be there
So there could be two circles , one for 1 ticket other for 2 tickets without any overlapping and outside region of these two circles will be neither case.

In such case also C would be answer
As the sum of all three would be 20 = 1st circle + 2nd circle + Outside region
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
Bunuel ScottTargetTestPrep

I am unable to infer why option B alone doesnt suffice.
We are asked to find out percent of the tickets which have been used i.e ( used 1 ticket+ Used both the tickets )/40.

From option B , we know 4 people dint use the tickets i.e each of these 4people havent used their 2 tickets =>> 40-(2*4)=32 tickets are remaining . 32 =used 1 ticket+ Used both the tickets. Using this information , i can directly find the percentage of tickets used.

Pls correct me where my understanding is wrong?
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
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Niveditha28 wrote:
Bunuel ScottTargetTestPrep

I am unable to infer why option B alone doesnt suffice.
We are asked to find out percent of the tickets which have been used i.e ( used 1 ticket+ Used both the tickets )/40.

From option B , we know 4 people dint use the tickets i.e each of these 4people havent used their 2 tickets =>> 40-(2*4)=32 tickets are remaining . 32 =used 1 ticket+ Used both the tickets. Using this information , i can directly find the percentage of tickets used.

Pls correct me where my understanding is wrong?

Response:
You are incorrectly assuming that all of the remaining 32 tickets are used; however, some of those 32 tickets won’t be used since some people use only one of their tickets. In other words, unused tickets = 2 * the number of people who use neither ticket + the number of people who use only one ticket.

Let’s use actual numbers to make it clear. First, suppose 4 people used neither ticket, 10 people used only one ticket and 6 people used both tickets. In this case, 10 + 6 x 2 = 22 tickets are used (and 10 + 4 x 2 = 18 tickets are not used). However, if 4 people used neither ticket, 5 people used only one ticket and 11 people used both tickets, then 5 + 11 x 2 = 27 tickets are used (and 5 + 4 x 2 = 13 tickets are not used). As we can see, there are multiple possibilities for the percent of tickets that were used. Also, notice that simply subtracting twice the number of people who use neither ticket from 40 does not give us the number of tickets that were actually used.
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Re: Of the 20 people who each purchased 2 tickets to a concert, some used [#permalink]
ScottTargetTestPrep Thank you for the explanation. I now understand where i went wrong.

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