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AnkurGMAT20
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I think MartyMurray's solution makes better sense than gmatophobia's.

By definition:
As X is not supposed to be 25 but 24
Z is not supposed to be 15 but 14

But these mistakes are covered up by letting Y be 5 instead of 7.

If the question ask further how many people are in front of Jim, the right answer is X=24 not X=25.

J must be 1 and B must be 1

X+J+Y+B+Z=45
X+1+Y+1+Z=45
Not
X+Y+Z=45

Posted from my mobile device
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I think x + y + z should be 43 . Jim and beth should not be included . Bunuel MartyMurray
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Jim and Beth are in a ticket line along with 43 other people, and Jim is ahead of Beth in the line. If there are 20 people behind Jim and 30 people ahead of Beth, how many people in the line are between Beth and Jim?

a) 5
b) 7
c) 10
d) 13
e) 15
Jim and Beth are in a ticket line along with 43 other people

Inference: 45 people are standing in the ticket line

...If there are 20 people behind Jim and 30 people ahead of Beth...

Assume that there are \(x\) people ahead of Jim, \(y\) people between Jim and Beth, and \(z\) people behind Beth as shown below -

Attachment:
Screenshot 2024-01-14 234127.png

\(x + y + z = 45\) -- (1)

As there are 20 people behind Jim,

\(y + z = 20\) -- (2)

As there are 30 people ahead Beth,

\(y + x = 30\) -- (3)

(2) + (3)

\(x + y + z + y = 50\)

From (1)

\(45 + y = 50\)

\(y = 5\)

Hence, there are 5 people between Jim and Beth.

Option A
­
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Considering your current diagram ,

it should be ,

x + 1 +y +1 +z = 45
x +y +z = 43

y +1 +z = 20
y+z = 19---------(2)

x +1 +y = 30
x+y = 29--------------(3)
Adding (2) and (3) , we get , x +2y + z = 48


Solving we get, x + 2*y  + z - (x +y +z ) = 48 -43 = 5

Am I right in my approach ? KarishmaB MartyMurray
gmatophobia
AnkurGMAT20
Jim and Beth are in a ticket line along with 43 other people, and Jim is ahead of Beth in the line. If there are 20 people behind Jim and 30 people ahead of Beth, how many people in the line are between Beth and Jim?

a) 5
b) 7
c) 10
d) 13
e) 15
Jim and Beth are in a ticket line along with 43 other people

Inference: 45 people are standing in the ticket line

...If there are 20 people behind Jim and 30 people ahead of Beth...

Assume that there are \(x\) people ahead of Jim, \(y\) people between Jim and Beth, and \(z\) people behind Beth as shown below -

Attachment:
Screenshot 2024-01-14 234127.png

\(x + y + z = 45\) -- (1)

As there are 20 people behind Jim,

\(y + z = 20\) -- (2)

As there are 30 people ahead Beth,

\(y + x = 30\) -- (3)

(2) + (3)

\(x + y + z + y = 50\)

From (1)

\(45 + y = 50\)

\(y = 5\)

Hence, there are 5 people between Jim and Beth.

Option A
­
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I think x + y + z should be 43 . Jim and beth should not be included . Bunuel MartyMurray
gmatophobia
AnkurGMAT20
Jim and Beth are in a ticket line along with 43 other people, and Jim is ahead of Beth in the line. If there are 20 people behind Jim and 30 people ahead of Beth, how many people in the line are between Beth and Jim?

a) 5
b) 7
c) 10
d) 13
e) 15
Jim and Beth are in a ticket line along with 43 other people

Inference: 45 people are standing in the ticket line

...If there are 20 people behind Jim and 30 people ahead of Beth...

Assume that there are \(x\) people ahead of Jim, \(y\) people between Jim and Beth, and \(z\) people behind Beth as shown below -

Attachment:
Screenshot 2024-01-14 234127.png

\(x + y + z = 45\) -- (1)

 
Yes, exactly, x + y + z should be 43.­
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sayan640
Considering your current diagram ,

it should be ,

x + 1 +y +1 +z = 45
x +y +z = 43

y +1 +z = 20
y+z = 19---------(2)

x +1 +y = 30
x+y = 29--------------(3)
Adding (2) and (3) , we get , x +2y + z = 48


Solving we get, x + 2*y  + z - (x +y +z ) = 48 -43 = 5

Am I right in my approach ? KarishmaB MartyMurray
 
Yes, that approach is spot on.
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AnkurGMAT20
Jim and Beth are in a ticket line along with 43 other people, and Jim is ahead of Beth in the line. If there are 20 people behind Jim and 30 people ahead of Beth, how many people in the line are between Beth and Jim?

A. 5
B. 7
C. 10
D. 13
E. 15

Attachment:
The attachment Jim.png is no longer available
Always make a small diagram for such questions. 

Total 45 people so 1 to 45.
Since 20 people are behind Jim, it means he is at 25th place (26 to 45 behind him)
Since 30 people are ahead of Beth, it means she is at 31st place.
Attachment:
Screenshot 2024-06-23 at 4.56.13 PM.png
Screenshot 2024-06-23 at 4.56.13 PM.png [ 24.67 KiB | Viewed 7529 times ]
Between them are 26 - 30 i.e. 5 people. 

Answer (A)

 ­
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Note 1: Jim is ahead of Beth in the line.
20 people behind Jim indicates that Jim is at the 25th position in the line.
30 people ahead of Beth indicates that Beth is at the 31st position.

Which means that the number of people between them are positions - 26, 27, 28, 29 and 30 - 5 people.
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­Think about the questions in the below way --> 

Jim & Beth and there "with 43 people"
That makes it 45 people line in total. 

As the line is of 45 people,
&, the question talks about 20 and 30 people behind & ahead.
So the number of people between Jim and Beth has to be ending with digit "5". as -> ('X'5-'Y'0='Z'5)

This leads to options A and E. 

Now make a number line and use 5 and 15 between Jim and Beth. 
You will end up with "A"- the answer.
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Total are 43
But 30 ahead of beth and 20 behind Jim means there is an overlap of 7 people

But note, these 7 people also have J and B included
Therefore there are 5 people between J and B
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I solved it using set theory, where Total = A + B - Both.

Hence, 45 = 30 + 20 - Both
Both = 50 - 45
Both = 5

IMO A
AnkurGMAT20
Jim and Beth are in a ticket line along with 43 other people, and Jim is ahead of Beth in the line. If there are 20 people behind Jim and 30 people ahead of Beth, how many people in the line are between Beth and Jim?

A. 5
B. 7
C. 10
D. 13
E. 15

Attachment:
Jim.png
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I'm not sure if my method is correct, but I used a venn diagram to solve this question.

There are 45 people (include Jim and Beth)
Jim and number of people behind him= 21 people
Beth and number of people ahead of her= 31 people
x is the overlapping people (including Jim and Beth)

Applying the formula, we will get

45 = 21 + 31 - x
x = 7 (This means there are 7 people total in the overlap, including Jim and Beth.)

Therefore, the the number of people between J and B will be 7-2 = 5 people
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Done in 1 minute.

20+30-2-x=43
Then x=5

Answer A
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Jim and Beth are in a ticket line along with 43 other people, and Jim is ahead of Beth in the line. If there are 20 people behind Jim and 30 people ahead of Beth, how many people in the line are between Beth and Jim?
A. 5
B. 7
C. 10
D. 13
E. 15

I always try to avoid double counting as I usually forget subtracting 1 or 2 later at the end and fall for the trap answers.

What worked for me:

______X____ Beth ______Y_____ Jim _____Z______

X = people behind Beth
Y = people between Beth and Jim (This is what we need to find out)
Z = people ahead of Jim

Now, Total people in the line = Jim + Beth + 43 = 45.

People behind Jim = 20
X + Beth + Y = 20 ........ (1)

People ahead of Beth = 30
Y + Jim + Z = 30 .......... (2)

Adding (1) and (2)
X + Beth + Y + Y + Jim + Z = 50
Y + (X + Beth + Y + Jim + Z) = 50
Y + Total people = 50
Y + 45 = 50
Y = 5

Answer: A.
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Consider it as an "overlapping sets" problem:

Let A be the number of people (including Jim) ahead of Beth, and B the number of people behind Jim (including Beth). The intersection of the two sets is exactly the number of people between Jim and Beyh.

We have:

A + B -x = 45 (1)

solving for x:

x= 30+20-45 = 5.
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