Question

In a class of 48 students the number of regular students is more than the number of irregular students. If two irregular students had been regular, the product of the number of two types of students would be 380. Find the number of each type of student.

Answer 1

Hint: Let us say we x regular students and y irregular students. Then we have the total students is 48. Hence we get the first equation in x and y. Now we are given that if two irregular students had been regular, the product of the number of two types of students would be 380. Hence we have a second equation in x and y. now we will solve the equations simultaneously and find x and y.

Complete step by step answer:
Now we are given there are 48 students
Let us say there are x regular students and y irregular students. Then we have x > y.
Now we can also say that x + y = 48
Rearranging the terms we get x = 48 – y ………………….. (1)
Now according to the second condition we have if two irregular students had been regular, the product of the number of two types of students would be 380.
Now if two irregular students are regular then number of irregular students = (y – 2) and number of regular students = (x + 2).
Hence we get
$\begin{align}
  & \left( x+2 \right)\left( y-2 \right)=380 \\
 & xy-2x+2y-4=380 \\
\end{align}$
Substituting the value of x from equation (1) we get
$\left( 48-y \right)y-2\left( 48-y \right)+2y-4=380$
$\begin{align}
  & 48y-{{y}^{2}}-96+2y+2y-4=380 \\
 & \Rightarrow 52y-{{y}^{2}}-100=380 \\
 & \Rightarrow {{y}^{2}}-52y+380+100=0 \\
 & \Rightarrow {{y}^{2}}-52y+480=0 \\
\end{align}$
Now splitting the middle term we get
$\begin{align}
  & {{y}^{2}}-40y-12y+480=0 \\
 & \Rightarrow y\left( y-40 \right)-12\left( y-40 \right)=0 \\
 & \Rightarrow \left( y-12 \right)\left( y-40 \right)=0 \\
\end{align}$
Therefore we have y = 12 or y = 40
Now x = 48 – y
Hence if y = 12 we get x = 48 – 12 = 36
And if y = 40 we get x = 48 – 40 = 8
But it is given that x > y.
Hence y = 12 and x = 36.

Hence there are 36 regular students and 12 irregular students.

Note: Note that while solving equations there was a term of xy. Hence we could not multiply the equation with constant and add the equations to find x or y. hence we used substitution. Now to solve the quadratic equation we can also use formulas for roots. The roots of quadratic equation $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .