Question

Rs.6500 is divided equally among a certain number of persons .Had there been 15 more person each would have got rs.30 less . Find the orignal number of person

Answer 1

$\bigstar{\pmb{\huge{\underline{\mathfrak{\red{Answer}}}}}}$

Let the original number of persons be x.

Let the original number of persons be x. Then the amount received by each person = Rs. $\sf{\frac{6500}{x}}$

When 15 more persons are added, amount received by each person = Rs. $\sf{\frac{6500}{x+15}}$

A/C,

$\longrightarrow$$\sf{\frac{6500}{x} - \frac{6500}{x + 15} = 30}$

$\longrightarrow$$\sf{650( \frac{(x + 15) - x}{x(x + 15)} ) = 3}$

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

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

Here, a = 1, b = 15 and c = -3250

$\therefore$ $\sf{d = {b}^{2} - 4ac}$

$\longrightarrow$$\sf{ {(15)}^{2} - 4 \times 1 \times ( - 3250)}$

$\longrightarrow$$\sf{225 + 13000}$

$\longrightarrow$$\sf{13225 > 0}$

So, the real roots exist. Using the quadratic formula,

$\longrightarrow$$\sf{x = \frac{ - b ± \sqrt{d} }{2a}}$

$\longrightarrow$$\sf{\frac{ - 15 ± \sqrt{13225} }{2 \times 1}}$

$\longrightarrow$$\sf{ \frac{ - 15 ± 115}{2}}$

$\longrightarrow$$\sf{50 \: or \: - 65}$

As the number of persons cannot be negative, x ≠ -65, x = 50

As the number of persons cannot be negative, x ≠ -65, x = 50Hence, the original number of persons = 50

Answer 2

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,

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

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

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

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