Question

ALEXA , on her birthday, distributed 540 oranges equally among the students.
If there would have been 30 students more, each would have received 3 oranges less.
Find the number of students.​

Answer 1

Step-by-step explanation:

Let no of students be x

Thus, no of oranges received by each student is = (540/x)

If there were 30 more students no of students would be (x + 30)

Now, no of oranges received by each student is (540/(x+30))

According to question,

(540/x) - (540/(x+ 30)) = 3

(180/x) - (180/(x + 30)) = 1 [DIVIDING BOTH SIDES BY 3]

(180x +5400 - 180x)/x(x+30) = 1

5400 = x² + 30 x

0 = x² + 30x - 5400

0 = x² - 60x + 90x - 5400

0 = x(x - 60) + 90(x - 60)

0 = (x - 60) (x + 90)

Thus, No of students cannot be negative

Therefore No of students = 60

THERE WERE 60 STUDENTS

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

➢ ALEXA , on her birthday, distributed 540 oranges equally among the students.

Let assume that number of students be x.

Number of oranges to be distributed = 540

So,

$\red{\rm :\longmapsto\:Each \: student \: share, \: s_1 = \dfrac{540}{x} - - - (1)}$

According to second condition,

Number of students = x + 30

Number of oranges to be distributed = 540

Thus,

$\red{\rm :\longmapsto\:Each \: student \: share, \: s_2 = \dfrac{540}{x + 30} - - - (2)}$

Now,

If there would have been 30 students more, each would have received 3 oranges less.

Thus,

$\rm :\longmapsto\: \: s_1 - s_2 = 3$

On substituting the values from equation (1) and (2), we get

$\rm :\longmapsto\:\dfrac{540}{x} - \dfrac{540}{x + 30} = 3$

$\rm :\longmapsto\:540\bigg[\dfrac{1}{x} - \dfrac{1}{x + 30}\bigg] = 3$

$\rm :\longmapsto\:180\bigg[\dfrac{x + 30 - x}{x(x + 30)}\bigg] = 1$

$\rm :\longmapsto\:180\bigg[\dfrac{30}{x(x + 30)}\bigg] = 1$

$\rm :\longmapsto\:x(x + 30) = 5400$

$\rm :\longmapsto\: {x}^{2} + 30x = 5400$

$\rm :\longmapsto\: {x}^{2} + 30x - 5400 = 0$

Now, its a quadratic equation in x, so to get the values of x, we use method of splitting of middle terms.

So,

$\rm :\longmapsto\: {x}^{2} + 90x - 60- 5400 = 0$

$\rm :\longmapsto\:x(x + 90) - 60(x + 90) = 0$

$\rm :\longmapsto\:(x + 90)(x - 60) = 0$

$\bf\implies \:x = 60 \: \: \: or \: \: \: x = - 90 \: \{rejected \}$

Hence,

• Number of students = 60.

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.

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