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

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   [tex]\huge\text{Question}[/tex]

   [tex]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.

   [/tex]

   [tex]\huge\text{Answer}

   [/tex]

   { According to the given condition,

   [tex] \: \: \: \: \: \: = > \: \: y = x + 20….(i)[/tex]

   {Total amount = ₹9000}

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

   [tex] \: \: = > \: \frac{9000}{x} – \frac{9000}{y} = 160 \\ [/tex]

   [tex]

   \fbox{each person get an amount = total amount \ number of person}[/tex]

   {on dividing both sides by 40, we get}

   [tex] \: \: \: \: \: \frac{225}{x} – \frac{225}{y} = 4 \\ [/tex]

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

   [tex] \frac{225}{x} – \frac{225}{x + 20} = 4 \\ [/tex]

   [tex] = > \frac{225x \: + \: 4500 \: – \: 225x}{x(x \: + \: 20} ) \: = 4 \\ [/tex]

   [tex] = > \frac{4500}{x(x \: + \: 20)} = 4 \\ [/tex]

   [tex] = > 4 {x}^{2} + 80x = 4500[/tex]

   [tex]= > {x}^{2} \: + \: 20x \: = 1125 \\ \\ \: \: \: \: \: \: = > {x}^{2} + 20x \: – \: 1125 = 0[/tex]

   [tex] = > {x}^{2} – 45x \: – 25x \: – 1125 \: = 0 \\ [/tex]

   [tex]\fbox{by \: splitting \: the \: middle \: term \: }[/tex]

   [tex] = > x(x \: + \: 45) – 25 \: (x \: + \: 45) = 0 \\ [/tex]

   [tex] = > (x \: – \: 25)(x \: + \: 45) = 0 \\ [/tex]

   [tex]\fbox{ x = 25 or x = – 45}

   \\ [/tex]

   But number of persons cannot be negative.

   Hence, the original number of person = 25

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