Question

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

Answer 1

Answer:

Step-by-step explanation:

Given :-

₹9000 were divided equally among a certain number of persons.

Had there been 20 more persons each would have got ₹ 160 less.

To Find :-

Original number of persons.

Solution :-

Let there be n persons and each get x rupees.

$\sf\implies nx = 9000$

$\sf\implies x=\dfrac{9000}{n}$

According to the Question,

$\sf\implies (n+20)(x-160)=9000$

$\sf\implies (n+20)(\dfrac{9000}{n} -160)=9000$

$\sf\implies 9000+\dfrac{20\times9000}{n}-160n-160\times20=9000$

$\sf\implies -160n^2+20\times9000-160\times20n=0$

$\bf\implies n^{2}+20n-1125=0$

$\sf\implies n^{2}+25n-45n-1125=0$

$\sf\implies (n+45)(n-25)=0$

$\bf\implies n=25,-45$

Hence, the number of persons are 25.

Answer 2

AnswEr :

25.

$\bf{\Large{\green{\underline{\underline{\tt{Given\::}}}}}}$

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

$\bf{\Large{\red{\underline{\underline{\tt{To\:find\::}}}}}}$

The original number of person.

$\bf{\Large{\blue{\underline{\underline{\rm{Explanation\::}}}}}}$

Let the original number of person be R

We have Rs.9000 were equally divided among a certain number of person.

Each person = $\bf{\dfrac{9000}{R} }$

if there 20 more person added so, we get;

Each person = $\bf{\dfrac{9000}{R+20} }$

$\bf{\large{\underline{\underline{\tt{\red{A.T.Q\::}}}}}}$

Now, we have found a equation :

$\longrightarrow\tt{\dfrac{9000}{R+20} =\dfrac{9000}{R} -160}\\\\\\\\\longrightarrow\tt{\dfrac{9000}{R} -\dfrac{9000}{R+20} =160}\\\\\\\\\longrightarrow\tt{\dfrac{9000(R+20)-9000R}{R(R+20)}=160}\\\\\\\\\longrightarrow\tt{\dfrac{\cancel{9000R}+180000\cancel{-9000R}}{R^{2}+20R } =160}\\\\\\\\\longrightarrow\tt{18000=160(R^{2} +20R)}\\\\\\\\\longrightarrow\tt{180000=160R^{2} +3200R}\\\\\\\\\longrightarrow\tt{160R^{2} +3200R-180000=0}\\\\\\\\\longrightarrow\tt{160(R^{2} +20R-1125)=0}}$

$\longrightarrow\tt{R^{2} +20R-1125=\cancel{\dfrac{0}{160} }}\\\\\\\\\longrightarrow\tt{R^{2} +20R-1125=0}\\\\\\\\\longrightarrow\tt{R^{2} +45R-25R-1125=0}\\\\\\\\\longrightarrow\tt{R(R+45)-25(R+45)=0}\\\\\\\\\longrightarrow\tt{(R+45)(R-25)=0}\\\\\\\\\longrightarrow\tt{R+45=0\:\:\:\:\:Or\:\:\:\:\:\:R-25=0}\\\\\\\\\longrightarrow\tt{\red{R\:=\:-45\:\:\:\:\:\:\:Or\:\:\:\:\:\:\:R=25}}$

∴ We know that negative value isn't acceptable.

Thus,

$\bf{\red{\underline{\sf{The\:original\:number\:of\:person\:is\:25.}}}}$