Aditya from Bristol Grammar has been very busy this month and also gave us the reasoning behind this answer:

If we call the numbers of pennies in each bag a, b, c and d, we know that a+b+c+d=15, otherwise the problem would have stated that Ram can pay any sum of money from 1 to more than 15.
Just for convenience, let a < b < c < d.
The smallest value, a, must equal 1.
As there is no use in having two bags with one penny in them, b must equal 2.
If Ram was to pay 3p, he would hand over bag a and b.
But if he were to pay 4p (this is the same as if he had to pay 2p) it would be pointless repeating another bag of 2p. So bag c has 4p in it.
As a+b+c+d = 15, and a+b+c = 7, d therefore is 8.

Matthew, James and Yuji from Moorfield Junior School, and Sarah L and Joe from Tattingstone School agreed that the money should be split in this way:

1p

2p

4p

8p

They all then went on to show how each sum of money from 1p to 15p can be made with these four bags:

1p
2p
1p+2p=3p
4p
4p+1p=5p
4p+2p=6p
4p+1p+2p=7p
8p
8p+1p=9p
8p+2p=10p
8p+2p+1p=11p
8p+4p=12p
8p+1p+4p=13p
8p+4p+2p=14p
8p+4p+2p+1p=15p