Answer: $4$ more five-pence coins than two-pence coins

Forming an equation

Let the number of five-pence coins be $x$

So the number of two-pence coins is $50 - x$

Then $5x + 2(50 - x) = 181$, that is $3x = 81$, that is $x = 27$.

So there are $27$ five-pence coins and $23$ two-pence coins, so there are $4$ more five-pence coins than two-pence coins.

Starting from 25 of each coin

25$\times$2p + 25$\times$5p = 175p

Swap $n\times$2p for $n\times$5p to get 181p

175p $-n\times$2p + $n\times$5p = 181p

$\Rightarrow$ 175p + $n($5p$-$2p$)$ = 181p

$\Rightarrow$ 175p + $n\times$3p = 181p

$\Rightarrow n\times$3p = 3p

$\Rightarrow n$ = 2

After swapping 2 2ps for 2 5ps there are 4 more 5ps than 2ps