ratio and proportion, ipm, mtse

A house has dogs, cats and parrot in the ratio of 3:7:5. If the number of cats was more than the number of dogs by a multiple of both 9 and 7 then what is the minimum of pets in the house possible?

Answer:

If total number of pets = p say

Then (7/15)p – (3/15)p = 63k

Where k is any natural number.

Or, p = 63k x (15/4)

For minimum value of p, k = 4

So, p = 63 x 4 x (15/4) = 945