The Premier Pie Stall problem

In the Hancock episode The Chef That Died of Shame, Hancock's pie stall owner sells one customer a cup of tea and a ham roll for 7½d, and second one a cup of coffee and a pie for 10½d.

He then sells Bill Kerr's character an unknown number of pies (at least two), and a cup of coffee, for 2s 6d (half a crown).

So how many pies were sold in the last transaction?

Let c be the price of coffee and p the price of a pie, and n the number of pies.
2s 6d is 30 pence, so np+c=30
Additionally we know p+c=10.5
so, (30-(p+c))/p = n-1
ie, if coffee+pie is 10½d, the remaining pies cost 19½d, so n is whatever integer 19.5 can be divided by that yields a coffee price which is a plausible penny/halfpenny price and allows c+p to be 10.5.

The only answer that fits is 3 remaining pies, yielding a pie price of 6½d.

So the number of pies is 4.

All four pies cost 26d and the price of coffee must be 4d. Check, 4+6.5=10.5.

This fits with the price quoted in The Last Bus Home for a cup of tea in Fred's cafe - 3d. You'd expect coffee to be slightly more expensive than tea, and if tea at the Premier Pie Stall is also 3d, then the ham rolls cost 4½d, and again you'd expect a ham roll to cost less than a hot pie.