Question

a man sold A certain number of cows for 12000 had he sold 5 cows less for the same sum he would have received 80 rupees more per cow find the number of cows he sold

Answer 1

let the no. of cows be x and the price of each be y
x*y=12000
y=12000/x   (1)

(x-5)(y+80)=12000
xy+80x-5y-400=12000
12000+80x-5y-400=12000
80x-5y=400
16x-y=80
Substituting for y,
16x-(12000/x)=80
(16x²-12000)/x=80
16x²-12000=80x
16x²-80x-12000=0
16(x²-5x-750)
x²-30x+25x-750=0
x(x-30)+25(x-30)=0
(x+25)(x-30)=0
x=-25   (not possible)
x=30

30 cows has been sold.

Answer 2

Let the man have sold x number of cows
So cost of each cow=12000/x
Now if he had sold 5 cows less
Number of cows sold=x-4
Cost of each cow now=(12000/x)+80
Therefore
(x-4)(12000/x +80)=12000
(x-4)(12000+80x/x)=12000
12000x+80x^2-48000-320x=12000x
80x^2-320x-48000=0
8x^2-32x-4800=0
x^2-4x-600=0
Now after solving this quadratic equation you will get the value of x