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I am currently practising UKMT questions and this question was from the 1998 IMC. enter image description here

I am struggling to picture how this works. Does the $x$ amount of people behind Wallace equal the $y$ amount of places in front of Gromit? I don't really see the differences between $x,y$ and $n$. Aren't they all the same? I know which one's the answer but it would really help if someone explained how this works with maybe a diagram? Thank you.

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Draw the queue like this, where $W$ is wallace and $G$ is gromit:

$$ \begin{array}{ccccc} \cdots & G & \cdots & W & \cdots \end{array} $$

Let's call $a,b,c$ the number of people in line in each of the segments labelled "$\cdots$''. So there are $a$ people behind Gromit, $b$ people in between them, and $c$ people after Wallace. Let's think about what each statement in the original problem says:

  1. "There are $x$ people behind Wallace". This means $a+b+1 = x$
  2. "There are $n$ people in front of Gromit". This means $b+c+1 = n$.
  3. "Wallace is $y$ places in front of Gromit". This means $b=y-1$.

You should be able to use these equations to solve for $a$ and $c$. Finally, the total number of people in line is $a+b+c+2$.

Alternate Way: Add all the people behind Wallace plus all the people in front of Gromit. This gives $x+n$. But this double-counts all the people in between, of which there are $y-1$. So the total is $x+n-(y-1)$.

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  • $\begingroup$ Sorry, i'm confused. If there are c people after Wallace then doesn't c=x? $\endgroup$
    – yt.
    Nov 7 '18 at 19:01
  • $\begingroup$ Okay i think i understand the alternative way. But its quite confusing with the diagram ...G...W... above- are we assuming the queue is starting from the right so that Gromit is behind Wallace? I think that is where the confusion is coming from. $\endgroup$
    – yt.
    Nov 7 '18 at 19:06
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    $\begingroup$ I meant the "front" of the line is at the right and the "back" is the left. So for example, $c$ is the number of people "in front of Wallace" and $a$ is the number of people "behind" Gromit. $\endgroup$
    – Nick
    Nov 7 '18 at 20:05

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