Al, Bert, and Carl are the winners of a school drawing for a pile of

As they each come at different times thinking they are the first, they each assume that the amount of sweets present is the full amount and thus will take the fraction value of the amount present.

The ratio of the amount of sweets ­Al, Bert, and Carl will each get of 3 : 2 : 1 in terms of fractions will be: \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{6}\). However, as we want to work out what is left we should look at it rather as Al will leave \(\frac{1}{2}\) of wwhat is infront of him, Bert will leave \(\frac{2}{3}\) of what is infront of him and Carl will leave \(\frac{5}{6}\) of what he sees infront of him.

Multiplying these values together will yield the fractional value of the amount of sweets which remain, regardless of the order of who comes first, second or third.

\(\frac{1}{2}*\frac{2}{3}*\frac{5}{6} = \frac{5}{18}\)

ANSWER D

Alternatively one could assign a value to the number of sweets available. As we are dealing with the fractions of \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{6}\) a good number to pick is \(2*3*6 = 36\)

Once again, order is irrelevant. Let's assume Al comes first, he will take 18 and leave 18. Then Bert comes and he takes 6 and leaves 12. Finally Carl comes and takes 2 sweets and leaves 10.

The amount left over as a fraction will be: \(\frac{10}{36}\), simplified to \(\frac{5}{18}\)

ANSWER D­