A man rared cows and buffalows. The number of buffalowes 7 more than one third of the number of cows. What was the total no. of animals if the number of buffalowes was 21?

Suppose the number of cows be x.
So the number of buffaloes = $\frac{x}{3}+7$
And given number of buffaloes = 21
So we have;
$$\begin{gathered} \frac{x}{3}+7=21 \\ \Rightarrow \frac{x}{3}=21-7 \\ \Rightarrow \frac{x}{3}=14 \\ \Rightarrow x=14\times 3=42 \end{gathered}$$
So the number of cows = 42.
Therefore number of animals = 42 + 21 = 63