The students of a class contributed for a programmer. Each student contributed the same amount. Had there been 15 more in the class and each student had contributed Rs.40 less, the total amount contributed would have increased from Rs.3000 to Rs.3200. Find the strength of the class.

Hint: Let’s suppose strength of class be x and amount each student contributed to be y, and then write the equation as per question.

Complete step-by-step answer:
Total amount contributed is
xy = 3000 ⇒$y = \dfrac{{3000}}{x}$
When student in class is 15 more amounts raised to 3200
(x + 15) (y – 40) = 3200
⇒ xy + 15y – 40x – 600 = 3200
$ \Rightarrow 15\left( {\dfrac{{3000}}{x}} \right) - 40x - 800 = 0$
⇒ $x^2$ + 20x –1125 = 0
By factoring we get,
⇒ (x + 45) (x – 25) = 0
x = 25
x = – 45 (strength of student cannot be negative)
So, we accept only x = 25.
The strength of class = 25.

Note: Since strength cannot be –ve so, we neglect the negative value.
The solution of linear equations in two variables, ax + by = c, is a particular point in the graph, such that when x-coordinate is multiplied by a and y-coordinate is multiplied by b, then the sum of these two values will be equal to c.
Basically, for a linear equation in two variables, there are infinitely many solutions.