Question

Rs $6500$ was divided equally among a certain number of persons. Had there been $15$ persons more , each would have got Rs $30$ less . Find the original number of persons.

Answer 1

Hint: It is given that Rs $6500$ was divided equally among a certain number of persons hence each person get initial is = $\dfrac{{6500}}{x}$ Rs where $x$ is the number of person initial , Now when we increase the $15$ person then amount will be $30$ less then initial amount now in this case the amount will be $\dfrac{{6500}}{{x + 15}}$ or $\dfrac{{6500}}{x}$ Rs this amount is $30$ more than this $\dfrac{{6500}}{{x + 15}}$ Rs, hence equation become $\dfrac{{6500}}{x}$ $ - $ $\dfrac{{6500}}{{x + 15}}$ $ = $ $30$.

Complete step-by-step answer:
As first we try to understand that what question will say , In the question it is given that : Rs $6500$ was divided equally among a certain number of persons ,
Now the persons will increase by $15$ then the amount will each person get is Rs $30$ less than previous amount ,
So for now let us assume the initial number of people is $x$ .
So the amount will given to each person initial is = $\dfrac{{6500}}{x}$ Rs
After that in the question it is given that the person is increase by $15$
Hence number of person now is = $x + 15$
So now the amount will given to each person is = $\dfrac{{6500}}{{x + 15}}$ Rs
Now in the question it is given that this amount is Rs $30$ less than the previous amount .
or we can say that when we subtract the initial amount we get $30$ Rs remaining .
or $\dfrac{{6500}}{x}$ this amount is $30$ more than $\dfrac{{6500}}{{x + 15}}$ .
As
$\dfrac{{6500}}{x}$ $ - $ $\dfrac{{6500}}{{x + 15}}$ $ = $ $30$
$6500\left( {\dfrac{1}{x} - \dfrac{1}{{x + 15}}} \right) = 30$
Divide by $10$ in whole equation we get ,
$650\left( {\dfrac{1}{x} - \dfrac{1}{{x + 15}}} \right) = 3$
Now take LCM of denominator and change numerator according to that ,
$650\left( {\dfrac{{x + 15 - x}}{{x(x + 15)}}} \right) = 3$
Now further solving ,
$650\left( {\dfrac{{15}}{{x(x + 15)}}} \right) = 3$
On dividing $3$ in whole equation we get ,
$650\left( {\dfrac{5}{{x(x + 15)}}} \right) = 1$
On cross multiplication we get ,
$650 \times 5 = x(x + 15)$
$3250 = x(x + 15)$
${x^2} + 15x - 3250 = 0$
So for the factor we find that $65$ and $ - 50$ its sum is $15$ and when we multiply it we get $ - 3250$
Hence the equation become ,
${x^2} + 65x - 50x - 3250 = 0$
Now take common $x$ from first two term and $ - 50$ from second two term ,
$x(x + 65) - 50(x + 65) = 0$
$(x - 50)(x + 65) = 0$
Hence $x = 50, - 65$
So $x = - 65$ is not possible because the number of people is not negative .
Hence there is $50$ person in initial or original number of persons

Note: This question we will also solve by two variables as , let the amount given to each person initially is y . Hence $y = \dfrac{{6500}}{x}$ and the other will be $y - 30 = \dfrac{{6500}}{{x + 15}}$ because we increase the $15$ person then amount will be $30$ less then initial amount . Now solve both equations and get the value of x .

In the step $3250 = x(x + 15)$ we will solve it directly without converting it into quadratic equation as the above equation will be written as $50 \times 65 = x(x + 15)$ or $50 \times (50 + 15) = x(x + 15)$ on comparing we get $x = 50$