Ratios & Proportion
Aptitude

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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

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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