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

Answer:

$\huge\text{Question}$

$Rs \: 9000 \: were \: divided \: equally \: among \: a \: certain \: number \: of \: persons. \: Had \: there \: been \: 20 \: more \: persons, \: each \: \: would \: have \: got \: Rs \: 160 less \: ? \: Find \: the \: original \: number \: of \: persons.</em></p><p><em>$

$\huge\text{Answer}</p><p>$

{ According to the given condition,

$\: \: \: \: \: \: = > \: \: y = x + 20....(i)$

{Total amount = ₹9000}

original share of each person - share of each of the increased persons = ₹ 160

$\: \: = > \: \frac{9000}{x} - \frac{9000}{y} = 160$

$</p><p>\fbox{each person get an amount = total amount \ number of person}$

{on dividing both sides by 40, we get}

$\: \: \: \: \: \frac{225}{x} - \frac{225}{y} = 4$

{ on putting y= x + 20 from Eq. (i) in Eq. (ii), we get}

$\frac{225}{x} - \frac{225}{x + 20} = 4$

$= > \frac{225x \: + \: 4500 \: - \: 225x}{x(x \: + \: 20} ) \: = 4$

$= > \frac{4500}{x(x \: + \: 20)} = 4$

$= > 4 {x}^{2} + 80x = 4500$

$= > {x}^{2} \: + \: 20x \: = 1125 \\ \\ \: \: \: \: \: \: = > {x}^{2} + 20x \: - \: 1125 = 0$

$= > {x}^{2} - 45x \: - 25x \: - 1125 \: = 0$

$\fbox{by \: splitting \: the \: middle \: term \: }$

$= > x(x \: + \: 45) - 25 \: (x \: + \: 45) = 0$

$= > (x \: - \: 25)(x \: + \: 45) = 0$

$\fbox{ x = 25 or x = - 45}</p><p>$

But number of persons cannot be negative.

Hence, the original number of person = 25