He has six more 50c coins than 20c coins

And all the money is $24 (2400c).

The extra six more 50c coins translates to an extra $3.

Remove the six coins and you have an equal number of 50c and 20c coins which has a total value of $21.

Thanks, this is the step that I am stuck.

Pretend that one 50c coin and one 20c coin is now one ‘70c’ coin (two coins become one).

Since the total value of all the ‘70c’ coins is $21 (2100c), there must have been 2100/70 = 30 of such ‘70c’ coins.

This means that the $21 consists of 30 of the 50c coins and 30 of the 20c coins. At the start there were 36 of the 50c coins and 30 of the 20c coins.

There must have been 66 coins altogether.

I can’t send all the working in one post because on some occasions the screen does not allow my to type beyond a certain length.

Oh I got it now, all thanks to your detailed explanation! Thank you!

Thanks Eric! Appreciate it!