Question

ranjana wants to distribute 540 oranges among some students.if 30 students were more each would get 3 oranges less.find the number of student's.

Answer 1

Let the number of students be x

and the number of oranges distributed per one student be y.

By first condition

xy=540..................(1)
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By second condition,

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

But from (1), 540 = xy

thus

(x+30)(y-3)=xy

Multiply on the L.H.S

x(y-3)+30(y-3)=xy

xy-3x+30y-90= xy

Subtracting xy from both sides,

-3x+30y-90=0

Dividing each term by 3

-x+10y-30=0. (Anything divides 0 becomes zero)

-x-30= -10y. ( Moving 10y to R.H.S. Yeah it's sign
changes)

-1(x+30)= -1(10y). ( Making -1 as common on both sides)

x+30= 10y. ( Say Multiplying by -1 on both sides)
------------------
x=10y-30. (Moving 30 to R.H.S and got value of
x)

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Let's substitute value of x in equation (1) ( Hope you didn't forgot equation (1) that was above)

Equation (1) was

xy = 540

(10y-30)(y)=540

Now, I have to change keyboard.

Multiplying the terms
$10 {y}^{2} - 30y = 540$

Dividing each term by 10
${y}^{2} - 3y = 54$
Moving the constant to right

${y}^{2} - 3y - 54 = 0$

Oh! It's a quadratic equation, there are various ways to solve it

Let us go for the beginners method i.e "FACTORISATION"

SPLITING -3y as a sum or difference

${y}^{2} + 6y - 9y - 54 = 0$
In first two terms, getting common as y and in next two getting common as 9

$y(y + 6) - 9(y + 6) = 0$
Again common as (y+6)
$(y - 9)(y + 6) = 0$
Now

y-9 = 0. or. y+6 = 0

y = 9. or. y = -6

These are the roots of the quadratic equation

Let's confirm which one is useful to us

y means the number of oranges distributed per one. So it is not going to be negative

So

$y = 9$

is the required solution

OH BUT WE WANT THE NUMBER OF STUDENTS

From (1)

xy=540

x(9)=540

x=540/9

x=60

ANS:- THE NUMBER OF STUDENTS IS 60




Hope I've helped you. Best wishes from me for your SSC exams

Answer 2

Hope it helps and all the best for your SSC boards☺️☺️