CBSE Class 12-science Answered

Let w, x, y, z be the number of rats in the first, second, third and fourth section respectively.

And the cat catches A number of rats at a time.

So after she catches A number of rats in first section, the remaining rats i.e. w - A go to second section and the number of rats there doubles.

Hence, x + w - A = 2x.

After catching the same number of rats in second section, the remaining rats i.e. 2x - A go to third section, and the number of rats there triples.

Hence, y + 2x - A = 3y.

Similarly for fourth section,

z + 3y - A = 4z.

and finally the remaining rats i.e. 4z - A go to first section and cat catches A rats, there we exhaust of rats. 

Hence, 4z - A - A = 0.

So rearranging the four equations,

w - x = A

2x - 2y = A

3y - 3z = A

2z = A,

So we have FIVE VARIABLES  AND FOUR EQUATIONS. 

we can express w, x, y, z in terms of A,

(substitute the expression of z in third equation to obtain y, then the expression for y in second and so on.)

z = A/2, y = 5A/6, x = 4A/3, w = 7A/3. 

Let's PICK MINIMUM A i.e. NUMBER OF RATS THE CAT CATCHES AT A TIME AS 6,

then w = 14, x = 8, y = 5 and z = 3.

So TOTAL NUMBER OF RATS IN THE FIRST PORTION i.e. w = 14, and

TOTAL NUMBER OF RATS THE CAT CAUGHT = 6 X 5 = 30

Regards,

Team,

TopperLearning.