Moderate Ratios & Proportion Solved QuestionAptitude Discussion

 Q. 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

Solution:
Option(A) is correct

If three kinds of pets are taken be $3k, 7k$ and $5k$ respectively, then $7k−3k=63p$ (where $p$ is any positive integer).

As the number is a multiple of both 9 and 7, it has to be multiple of 63.

$\Rightarrow k=\dfrac{63p}{4}$

Minimum value of $p$ for which $k$ is a natural number is 4.
Thus, $k=63.$

Hence, the number of pets =$15k$= 945

(2) Comment(s)

Venkata Siva Sai Chandigarh Atchuta
()

It can also be solved by taking common multiple of 7 and 9

It is 63

Multiply the 63 for given ratio 3:7:5 so that

189:441:315

By adding all we get 945

Jayaram
()

pls i cant understand the problem explain it