Comparing arithmetic complexity of FFT radix-2 and convolution
Let the input signal $x[n]$ has a legth of $N=50$ samples, and the filter $h[n]$ has a length of $M=10$ samples, then the output $y[n]$ (by linear conv) will have a length of $L = N+M-1=59$ samples.
If you use time-domain convolution, then the number of real MACS can be seen to be  $N \cdot M = 50 \times 10 = 500$.
If you want to use radix-2 FFT to implement the linear convolution result, then you should select a length of $R = 64$ for FFTs. And you will: 1- convert $x[n]$ and $h[n]$ into $X[k]$ and $H[k]$ by two $R$-point FFTS, 2- multiply the results to get $Y[k] = X[k]H[k]$, and 3- apply inverse FFT of $R$-point on $Y[k]$ to get the output $y[n]$.
Each $R$-point FFT (and IFFT) requires about $\frac{1}{2} R  \cdot \log_2(R)$ complex MACs. One complex MAC is 4 real MACs, therefore this is equivalent to $ 2 ~R ~ \log_2(R) = 128 \log_2(64) = 128 \times 6 = 768 $ real MACs. And the total of two FFTs and one inverse FFT requires about $3 \times 768 = 2304$ real MACs. The intermediate multiplication also requires $4 \times R = 256$ real MACs and hence FFT based implementation requires a total of $2560$ real MACs.
So in this case, time-domain convolution is more efficient than an FFT based implementation.