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Ranjan, on her birthday, distributed 540 oranges equally among the students. if there would have been 30 students more, it would have received 3 oranges less. find the number of students.
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Let the number of students be x
and the number of oranges distributed per one student be y.
By first condition
xy=540..................(1)
________________________________________________
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)
_______________________________________________
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

Dividing each term by 10

Moving the constant to right

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

In first two terms, getting common as y and in next two getting common as 9

Again common as (y+6)

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

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
and the number of oranges distributed per one student be y.
By first condition
xy=540..................(1)
________________________________________________
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)
_______________________________________________
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

Dividing each term by 10

Moving the constant to right

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

In first two terms, getting common as y and in next two getting common as 9

Again common as (y+6)

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

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
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