Ranjana wants to distribute 540 oranges among students. 30 students were increased each will get 3 oranges less. Find no. Of students
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Secondary SchoolMath5 points
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.
Ask for details Follow Report by Danish1111510.05.2018
Answers

Kmg13teen Virtuoso
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
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
==========================
→→→→→→→→→→→→→→→→→→→→
=> 
1
Secondary SchoolMath5 points
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.
Ask for details Follow Report by Danish1111510.05.2018
Answers

Kmg13teen Virtuoso
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
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
==========================
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