Question

Ramrsh divided Rs 6500 equally among a certain no. of children in an orphanage to buy gifts for Christmas. Had there been 15 more children, each would have go Rs 30 less. Find the original no. of students.

Answer 1

Let the no. of students be x
Then,
CASE 1: When the amount was shared to x students, i.e.
6500÷x

CASE 2:When the amount was shared to 15 more students i.e.,
6500÷x+15
Now,
According to question,
6500÷x - 6500÷x+15=30
6500[1÷x - 1÷x+15]=30
So,
6500[x+15 - x÷x²-15x]=30
6500[15÷x²+15x]=30
=>97500=30(x²+15x)
97500÷30=x²+15x
3250=x²+15x
x²+15x-3250=0

On performing the quadratic formula with this we get
x=Rs.50 (or)Rs.-65
But Rupees or cash will not be in negative
So, x=Rs.50

Answer 2

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

Let the original number of persons be x.

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

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