Accedmychevron_right11thchevron_rightmathchevron_rightTheory Of Quadratic Functionschevron_rightExercise 3.2

Solution

Let the number of sheep bought be xx.

Price per sheep:

9000x\frac{9000}{x}

If he paid Rs. 100 less per sheep, new price per sheep is:

9000x100\frac{9000}{x}-100

Then he would get x+3x+3 sheep for Rs. 9000:

(x+3)(9000x100)=9000(x+3)\left(\frac{9000}{x}-100\right)=9000

Expand:

9000+27000x100x300=90009000+\frac{27000}{x}-100x-300=9000 27000x100x300=0\frac{27000}{x}-100x-300=0

Multiply by xx:

27000100x2300x=027000-100x^2-300x=0

Divide by 100-100:

x2+3x270=0x^2+3x-270=0 (x+18)(x15)=0(x+18)(x-15)=0

So x=18x=-18 or x=15x=15. Reject x=18x=-18.

x=15\boxed{x=15} The farmer bought 15 sheep.\boxed{\text{The farmer bought 15 sheep.}}