Question

Rs 6500 were divided equally among a certain number of persons. Had there been 15 more persons, each would have got Rs 30 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 amount is $\mathrm{Rs}.6500$.

Money given to each person will be $\mathrm{Rs}.\frac{6500}{\mathrm{x}}$

Now $15$ more persons are added, so total number of people became will be $x+15$.

Dividing the amount among $x+15$ people, each person get $\mathrm{Rs}.\frac{6500}{\mathrm{x}+15}$.

But according to given condition, on dividing the amount among $x+15$ people, each person will get $\mathrm{Rs}.30$ less than earlier

i.e each person gets $\mathrm{Rs}.\frac{6500}{\mathrm{x}}-30$

Therefore, we can write

$\begin{array}{rcl}\frac{6500}{\mathrm{x}+15}& =& \frac{6500}{x}-30\\ & \Rightarrow & \frac{6500}{x+15}=\frac{6500-30x}{x}\\ & \Rightarrow & 6500x=\left(x+15\right)\left(6500-30x\right)\\ & \Rightarrow & 6500x=6500x+6500\left(15\right)-30x^2-450x\\ & \Rightarrow & 30x^2+450x-6500\left(15\right)=0\\ & \Rightarrow & x^2+15x-3250=0\end{array}$

Step 2:

Solve the quadratic equation:

Use quadratic formula to solve the equation:

$\begin{array}{rcl}x& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & \Rightarrow & x=\frac{-15\pm \sqrt{{\left(15\right)}^2-4\times 1\times \left(-3250\right)}}{2\times 1}\\ & \Rightarrow & x=\frac{-15\pm \sqrt{225+13000}}{2}\\ & \Rightarrow & x=\frac{-15\pm \sqrt{13225}}{2}\\ & \Rightarrow & x=\frac{-15\pm 115}{2}\\ & \Rightarrow & x=\frac{-15+115}{2}\mathrm{or}x=\frac{-15-115}{2}\\ & \Rightarrow & x=\frac{100}{2}\mathrm{or}x=\frac{-130}{2}\\ & \Rightarrow & x=50\mathrm{or}x=-65\end{array}$

But the number of persons cannot be negative. Rejecting $x=-65$.

So, take the value $x=50$

Final answer:

Hence, the original number of persons are $50$.