Question

Rs. 7500 were divided equally among a certain number of children. Had there been 20 less children, each would have received Rs. 100 more. Find the original number of children.

Answer 1

Hint: We will apply the trick of solving a word problem. The trick is known as the unitary problem. This method is used to find the individual value of an object first and then finding the value of an object by multiplying the number of objects to its single object’s value.

Complete step-by-step solution -
Now, we will consider the number of students as x. Since Rs. 7500 is divided among x number of children so by applying a unitary method here we can write that Rs. 7500 is divided among x so x students got Rs. 7500. This results into the fact that 1 student got Rs. $\dfrac{7500}{x}$.
Now, according to the question we have been informed that there had been 20 children less. Therefore, we have that the number of children changes to x - 20. So, if Rs. 7500 is divided among x - 20 children therefore, one student gets Rs. $\dfrac{7500}{x-20}$. Now, we will form an equation for these conditions. As each student would have received 100 more rupees therefore, we have $\dfrac{7500}{x}+100$. This expression is equal to $\dfrac{7500}{x-20}$. Now, we can write
$\begin{align}
  & \dfrac{7500}{x-20}=\dfrac{7500}{x}+100 \\
 & \Rightarrow \dfrac{7500}{x-20}-\dfrac{7500}{x}=100 \\
 & \Rightarrow \dfrac{7500x-7500\left( x-20 \right)}{x\left( x-20 \right)}=100 \\
 & \Rightarrow 7500x-7500\left( x-20 \right)=100\left( x\left( x-20 \right) \right) \\
 & \Rightarrow 7500x-7500x+20\times 7500=100\left( {{x}^{2}}-20x \right) \\
 & \Rightarrow 20\times 7500=100{{x}^{2}}-20\times 100x \\
 & \Rightarrow 150000=100{{x}^{2}}-2000x \\
 & \Rightarrow 100{{x}^{2}}-2000x=150000 \\
 & \Rightarrow {{x}^{2}}-20x-1500=0 \\
\end{align}$
Now, we will apply a square root formula to find the roots of the equation ${{x}^{2}}-20x-1500=0$. The formula is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Therefore, we have
$\begin{align}
  & x=\dfrac{-\left( -20 \right)\pm \sqrt{{{\left( 20 \right)}^{2}}-4\left( 1 \right)\left( -1500 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{20\pm \sqrt{400+6000}}{2} \\
 & \Rightarrow x=\dfrac{20\pm \sqrt{6400}}{2} \\
\end{align}$
Since, we know that the square root of 6400 is 80.
$\begin{align}
  & \Rightarrow x=\dfrac{20\pm \sqrt{6400}}{2} \\
 & \Rightarrow x=\dfrac{20\pm 80}{2} \\
\end{align}$
Now, we can have two different roots here which are given by $x=\dfrac{20+80}{2}$ and $x=\dfrac{20-80}{2}$. Therefore, we have the roots as $x=\dfrac{100}{2}$ and $x=\dfrac{-60}{2}$. Or we can say that $x=50$ or $x=-30$. Since, the number of children cannot be negative.
Hence the number of children is 50.

Note: We need to take care of the minus and plus signs here otherwise, it will give the wrong number of students. Also, we never consider the number of students as negative. This is because it will create a huge contradiction on the counting of numbers. That is why we have selected the number as 50 and rejected -30.