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. Which value of depicted?

Answer 1

Answer:

$\boxed{\sf \: Number\:of\:persons = 50 \: }$

Step-by-step explanation:

Let assume that the number of persons be x.

Case :- 1

Amount to be distributed = Rs 6500

Number of persons = x

So, Each person share is

$\boxed{ \sf{ \:\sf \: S_1 = \dfrac{6500}{x} \: }} - - - (1)$

Case :- 2

Amount to be distributed = Rs 6500

Number of persons = x + 15

So, Each person share is

$\boxed{ \sf{ \:\sf \: S_2 = \dfrac{6500}{x + 15} \: }} - - - (2)$

According to statement, it is given that had there been 15 persons more, each would get Rs 30 less.

$\bf\implies \:S_1 - S_2 = 30$

$\sf \: \dfrac{6500}{x} - \dfrac{6500}{x + 15} = 30$

$\sf \:6500\bigg( \dfrac{1}{x} - \dfrac{1}{x + 15}\bigg) = 30$

$\sf \:650\bigg(\dfrac{x + 15 - x}{x(x + 15)}\bigg) = 3$

$\sf \:\dfrac{650 \times 15}{x(x + 15)} = 3$

$\sf \: {x}^{2} + 15x = 3250$

$\sf \: {x}^{2} + 15x - 3250 = 0$

$\sf \: {x}^{2} - 50x + 65x - 3250 = 0$

$\sf \: x(x - 50) + 65(x - 50) = 0$

$\sf \: (x - 50) \: (x + 65) = 0$

$\bf\implies \:x = 50\: \: \: \: \: or \: \: \: \: \: x = - 65 \ \: \: \: \{rejected \}$

Hence,

$\implies\sf \: Number\:of\:persons = 50$

This shows how sharing affect the share of each person.

$\rule{190pt}{2pt}$

${{ \mathfrak{Additional\:Information}}}$

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

Three cases arises :

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

Step-by-step explanation:

$hey \: there \: buddy$!!

$given \: rupees = 6500$

$let \: the \: original \: no \: of \: persons \: be \: x$

$then \: the \: money \: given \: to \: each \: person \: would \: be \: = \frac{6500}{x}$

$the \: question \: says \: that \: 15 \: more \: people \: are \: added$

$so \: now \: the \: total \: no \: people \: became = (x + 15)$

$the \: money \: divided \: \: in \: (x + 15) \: persons \: = \frac{6500}{x + 15}$

$now \: here \: according \: to \: the \: question \: the \: money \: that \: is \: divided \: in \: (x + 15) \: persons = \frac{6500}{x} - 13$

$so \: now \: we \: can \: equate \: both \: the \: equations$

$\frac{6500}{x + 15} = \frac{6500}{x} - 13$

$by \: cross \: multiplication$

$6500x = 500x + 6500(15) - 30x(x + 15)$

$we \: know \: that \: ax + by + c = 0$

$30 {x}^{2} + 450x - 6500(15) = 0$

$x = \frac{ - 15± \sqrt{225 + 4(3250)} }{2}$

$x = \frac{ - 15±115}{2}$

$x = - 65 \: (rejected)$

$x = 50$

$hence \: the \: original \: no \: of \: persons \: = 50$

$always \: here \: to \: help \: you$