"A chocolate bar is separated into several equal peices. If one person eats 1/4 of the pieces and the second person eats 1/2 of the remaining pieces, there are six pieces left over. Into how many peices was the chocolate bad originally divided?"
2007-10-04 09:53:36 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The two people together ate:
1/4 + 1/2 = 3/4
That means 1/4 of the bar remains. We are told that 1/4 of the bar is 6 pieces. So all together the bar must have had 24 pieces.
1/4n = 6
n = 24
To double-check, the first person eats 6 pieces (1/4) and the other person eats 12 pieces (1/2). That leaves 6 pieces which matches with the original problem.
2007-10-04 10:22:58 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
Let's call we have x pieces of chocolate.
- The first person eats (1/4)x, so we have (3/4)x pieces left.
- The second person eats half of the remaining (3/4)x pieces, meaning he eats (3/8)x pieces. The number of chocolate pieces left: (3/4)x-(3/8)x=(3/8)x pieces.
- (3/8)x=6 => x= 16.
- Check: 16-4=12, 12-6=6.
2007-10-04 17:23:14 · answer #2 · answered by misa_dep_trai 2 · 1⤊ 0⤋
16 p
half of the remain = 1/2 x3/4=3/8 of the total
2007-10-04 17:21:23 · answer #3 · answered by Anonymous · 1⤊ 0⤋