How to solve this word problem on the topic of quadratic equations?

Question

The maths teacher of Mumbai is transferred to another school. The students of Class 10 decided to buy a book for 360 rupees(currency) as a gift for her. On the farewell day, 4 students did not turn up and so those present had to contribute an extra 1 rupee each. How many students were there in the class all together?

Progress

Let the number of students in the class be $x$. I got $360/x$ as the cost of one student in the class.

Answer 1

$x = $ number of students
$y = $ cost per student

$y=\frac{360}{x}$

$y+1=\frac{360}{x-4}$

Substitute $y=\frac{360}{x}$ into second equation:

$$\frac{360}{x}+1=\frac{360}{x-4}$$

$$(x-4)\frac{360}{x}+(x-4)=360$$

$$360-\frac{1440}{x}+x-4=360$$

$$-\frac{1440}{x}+x=4$$

$$-1440+x^2=4x$$

$$x^2-4x-1440=0$$

$$(x+36)(x-40)=0$$

$$x=40$$

We can ignore $x=-36$ because a class cannot have negative students. The class has $40$ students.

Answer 2

Suppose there are $x$ students in the class.

The full class per-student book cost is ${360 \over x}$. We are told that ${360 \over x-4}$ is one rupee more that the full class per-student cost. Appropriately equate the two and solve.

Answer 3

This is how I would do it. Let $n$ be the number of students and $x$ be the cost per student. We are given two pieces of information:$$nx=360$$ and $$(n-4)(x+1)=360$$

If we equate $nx=(n-4)(x+1)$ we can get rid of the awkward term in $nx$ and get a linear relationship between $n$ and $x$, and substitute into either of the other equations to obtain a quadratic.

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Namely $n=4x+4$, whence $nx=4x(x+1)=360$ and $x^2+x-90=0$ or $x(x+1)=90$

It depends how you want to solve the quadratic, but $x(x+1)$ is an increasing function for $x\gt 0$and it is clear that $x=9, n=40$ is the only positive solution.

Alternatively factorise as $(x+10)(x-9)=0$ or use the quadratic formula.