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?
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.
Minimum value of $p$ for which $k$ is a natural number is 4.
Hence, the number of pets =$15k$= 945
A fort has provisions for 60 days. If after 15 days 500 men strengthen them and the food lasts 40 days longer, how many men are there in the fort?
Solution:Option(B) is correct
Let there be $'x'$ men in the beginning so that after 15 days the food for them is left for 45 days.
After adding 500 men the food lasts for only 40 days.
Now $(x+500)$ men can have the same food for 40 days.
Therefore by equating the amount of food we get,
Therefore there are 4,000 men in the fort.
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