Question

Rs 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs 160 less. Find the original number of persons.

Answer 1

Step 1:

Form a quadratic equation:

Let original number of persons be $x$.

Given that total rupees is $\mathrm{Rs}.9000$.

Money given to each person will be $\mathrm{Rs}.\frac{9000}{\mathrm{x}}$

Now $20$ more people are added, so total number of people became $x+20$.

Dividing the amount among $x+20$ people, each person will get $\mathrm{Rs}.\frac{9000}{\mathrm{x}+20}$.

But according to given condition, on dividing the amount among $x+20$ people, each person get $\mathrm{Rs}.160$ less i.e., $\mathrm{Rs}.\frac{9000}{\mathrm{x}}-160$

Therefore,

$\begin{array}{rcl}\frac{9000}{\mathrm{x}+20}& =& \frac{9000}{x}-160\\ & \Rightarrow & \frac{9000}{x+20}=\frac{9000-160x}{x}\\ & \Rightarrow & 9000x=\left(x+20\right)\left(9000-160x\right)\\ & \Rightarrow & 9000x=9000x+9000\left(20\right)-160x^2-3200x\\ & \Rightarrow & 160x^2+3200x-9000\left(20\right)=0\\ & \Rightarrow & x^2+20x-1125=0\end{array}$

Step 2:

Solve the quadratic equation:

Use quadratic formula to solve the equation:

$\begin{array}{rcl}x& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & \Rightarrow & x=\frac{-20\pm \sqrt{{\left(20\right)}^2-4\times 1\times \left(-1125\right)}}{2\times 1}\\ & \Rightarrow & x=\frac{-20\pm \sqrt{400+4500}}{2}\\ & \Rightarrow & x=\frac{-20\pm \sqrt{4900}}{2}\\ & \Rightarrow & x=\frac{-20\pm 70}{2}\\ & \Rightarrow & x=\frac{-20+70}{2}\mathrm{or}x=\frac{-20-70}{2}\\ & \Rightarrow & x=\frac{50}{2}\mathrm{or}x=\frac{-90}{2}\\ & \Rightarrow & x=25\mathrm{or}x=-45\end{array}$

But the number of persons cannot be negative. Rejecting $x=-45$.

So, take the value $x=25$

Final answer:

Hence, the original number of persons are $25$.