Math, asked by ameer487, 1 year ago

6500/- were divided equally among a certain number of persons. had there been 15 more persons, each would have got 30/- less. find the original number of persons?

Answers

Answered by abhi178
11
let x is no of person and y is amount taje by each person.

6500/x=y ----------------(1)
when 15 more person appear then
6500/(x+15)=y-30 ------------(2)

solve both (1) and (2) equation
subtracting (1) to (2)
6500{15/(x^2+15x)}=30
3250=x^2+15x

x^2+15x-3250 =0
solve this by quadratic formula ,
x=(-15+_root (225+13000))/2
=50,-65
but negative not possible
hence 50 number of person.
Answered by mathdude500
0

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 = 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 = 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 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 \\  \\

\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

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