Samuel had 3 piles of coins, I, II and III. Altogether there was 72p. Pile II had twice as much as pile I. Pile III had three times as much as pile II. How much money was in Pile III?
- First let coins in pile I equal x. Then we the question tells us II has twice pile A so we can now write this as 2x. Finally, we can write pile III as 6x - reason is that III has three times pile II which is 2x so 3*2x= 6x. 2) Therefore add all these together to get: x + 2x + 6x = 9xand we're given there is altogether 72p so 9x = 723) Divide through by 9 on both sides to get 1x = 84) Then to find number of coins in pile III we multiply by 6 from earlier to get 48p and that is the correct answer
SR
