Aptitude 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**

**Venkata Siva Sai Chandigarh Atchuta**

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**Jayaram**

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pls i cant understand the problem explain it

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