Question

In a class of 48 students, the number of regular students is more than the irregular students. Had two irregular students 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:
Number of Regular students + Number of Irregular/ non-regular students = the total number of students in the class. Quadratic equations can be solved easily by using a middle term splitting method.

Complete step by step solution:
Let the number of regular students is x.
The number of irregular students = (48-x).
Also given, \[x > \left( {48 - x} \right)\]…… Eq.01
According to the condition;
\[ \Rightarrow (x + 2)\left( {48 - x - 2} \right) = 380\]
Solving above equation we get;
$\Rightarrow (x + 2)\left( {46 - x} \right) = 380 $
On opening the brackets and multiplying terms we get,
$\Rightarrow 46x - {x^2} + 92 - 2x = 380 $
trying to make a quadratic equation in standard from,
$\Rightarrow - {x^2} + 44x = 380 - 92 $
Multiplying equation by $-1$ we get,
$\Rightarrow {x^2} - 44x + 288 = 0 $
Now, we’ll solve this equation by factoring method,
$\Rightarrow {x^2} - 36x - 8x + 288 = 0 $
taking $x$ common from the first pair and $-8$ common from the second pair.
$\Rightarrow x(x - 36) - 8(x - 36) = 0 $
Taking $x-36$ common from both terms
$\Rightarrow (x - 36)(x - 8) = 0 $
Equating one by one to zero
$\Rightarrow x = 36; x = 8 $
Here, the value of x can’t be 8. It does not follow the condition prescribed in Eq. 01.
\[ \Rightarrow x = 36\]

Number of Regular students=36
Number of Irregular students=48-36=12.

Note:
There is one another way to solve a quadratic equation which is a hit and trial method. In this method by looking at the equation we try to guess the roots Then Using factor theorem we find another root. SInce it is just a guess method and there is no thumb rule for guessing, it’s not recommended usually.