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The problem statement:

The expenses of a tuition class are partly fixed and partly variable with the number of students.The charge is $40\$$ per head when there are $25$ students and $60\$$ per head when there are $50$ students.Find the charge per head when there are $100$ students.

I am looking for some hints/ideas about approaching this problem.

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    It may be worthwhile to check whether any numbers have been transposed.2011-08-21
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    @André Nicolas:Sorry,I don't understand.2011-08-21
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    @FoolForMath as written the problem seems unsolvable, so he is asking if you have written it incorrectly, or perhaps if we had more context we could make an assumption to allow it to be solved (as in nerdbert's answer)2011-08-21
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    If there are fixed costs (rent?) and costs per student (food?) one would expect the charge per head to fall as the number of heads increases. The obvious model, $C=F+ an$, where $C$ is total cost, $F$ is fixed costs, $a$ additional cost per head, $n$ number of students yields, with your numbers, $F=-1000$, so negative fixed costs!2011-08-21
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    I am pretty sure that I have copied it correctly.2011-08-21
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    André Nicolas:Yes,I noticed that,however the answer given is $70\$$.2011-08-21
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    I think there is no need for an answer from me. @nerbert has given the "formula" to use, essentially same as in my comment above. I get $1000=F+25a$, $3000=F+50a$, so solving the two linear equations in two unknowns we get $a=80$, $F=-1000$. Thus cost with $100$ students is $-1000+8000$, $70$ dollars a head. In real world terms, $-1000$ fixed cost makes no sense. The person who made up the problem should have taken the trouble to use simplified but more realistic numbers.2011-08-21
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    @André Nicolas:I know,but sometimes it's formal to give the answer as the problem is not stated properly.Thanks for the inputs :-)2011-08-21

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If $x$ is the number of students, and $y$ is the tuition costs, you have

$$y = ax + b$$

in which $a$ controls the variable part and $b$ controls the fixed part of the total... if you knew $a$ and $b$ you could calculate the costs for any given number of students. To find them, just use the two known values for $x$ and $y$, which gives you two equations (for your two unknowns $a$ and $b$).

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    but how do you know this is a linear function with respect to the part that is "varying with the number of students"; I would read it rather as $y = f(x) + b$ where $f(x)$ is some function of $x$.2011-08-21
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    You are right... without further information the problem as stated is unsolvable! But it looks like a question from an introductory algebra textbook, so linearity is the most likely assumption.2011-08-21
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    It's not much of a stretch to interpret "varying with the number of students" as "varying directly with the number of students," which is another way to say "proportional to the number of students." And if you don't make some stretch, there's no way to solve the problem, so presumably this is what was intended.2011-08-21