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Re: Jim and Beth are in a ticket line along with 43 other people, and Jim [#permalink]
2
Kudos
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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Re: Jim and Beth are in a ticket line along with 43 other people, and Jim [#permalink]
I think x + y + z should be 43 . Jim and beth should not be included . Bunuel MartyMurray
gmatophobia wrote:
AnkurGMAT20 wrote:
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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Re: Jim and Beth are in a ticket line along with 43 other people, and Jim [#permalink]
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 wrote:
AnkurGMAT20 wrote:
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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Re: Jim and Beth are in a ticket line along with 43 other people, and Jim [#permalink]
Expert Reply
sayan640 wrote:
I think x + y + z should be 43 . Jim and beth should not be included . Bunuel MartyMurray
gmatophobia wrote:
AnkurGMAT20 wrote:
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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Re: Jim and Beth are in a ticket line along with 43 other people, and Jim [#permalink]
1
Kudos
Expert Reply
sayan640 wrote:
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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Jim and Beth are in a ticket line along with 43 other people, and Jim [#permalink]
1
Kudos
Expert Reply
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Jim and Beth are in a ticket line along with 43 other people, and Jim [#permalink]
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