Hi All,

We're told that a shop sells two variants of chocolates - one that costs $3 and the other that costs $5 and the shop sold $108 of chocolates on a given day. We're asked for the number of different combinations of (number of $3 sold, number of $5 sold) that could have been sold. This question comes down to basic Arithmetic, but there's a built-in pattern that you can use to save some time.

To start, $108 is NOT a multiple of $5, so you can't hit that total with just $5 chocolates. $105 IS a multiple of $5 though, so we could have....
One $3 and twenty-one $5 chocolates

At this point, it's worth noting that five $3 chocolates and three $5 chocolates both cost $15, so we can use this pattern to 'exchange' groups of chocolates and quickly determine all of the possibilities....

Four $3s and eighteen $5s
Nine $3s and fifteen $5s
Fourteen $3s and twelve $5s
Nineteen $3s and nine $5s
Twenty four $3s and six $5s
Twenty nine $3s and three $5s
Thirty four $3s and zero $5s

Total options = 8

Final Answer: C

GMAT assassins aren't born, they're made,
Rich