Solution:


The most important this to pay attention to is that \(\frac{1}{6}\) show-dogs are poodles and \(\frac{1}{7}\) show-dogs are poodles are equal.
So let the total poodles be P & Show dogs be S

Thus, \(\frac{1}{6}\)P= \(\frac{1}{7}\)S
i.e. \(\frac{P}{S}\)= \(\frac{6}{7}\)

Hence, the minimum number of poodles and show dogs has to be a multiple of 7+6=13 and 13 being the lowest total number.

13 or any other multiple of 13>10
QTY A> QTY B


Your Answer is correct but the total would be 12 instead of 13.
You have probably counted a dog which is Poodle as well as Show-dog twice!