Description : In a house, there are dogs, cats and parrot in the ratio 3:7:5. If the number of cats was more than the number of dogs by a multiple of both 9 and 7, what is the minimum of pets in the house? A) 945 B) 630 C) 252 D) 238 E) NONE
Last Answer : Answer: A If three kinds of pets are taken be 3k,7k and 5k respectively, then 7k−3k=63p (where pp is any positive integer). As the number is a multiple of both 9 and 7, it has to be multiple of 63. ... pp for which kk is a natural number is 4. Thus, k =63 Hence, the number of pets =15k= 945