Jeremy, Ken and Sean had a total of 307 coins at first. After a week, the number of Jeremy's coins became 4 times the number of coins he had at first. The number of Ken's coins decreased by 86. Sean had half as many coins as he had at first. In the end, the three boys had the same number of coins.
- How many coins did Sean have at first?
- What was the total number of coins that the three boys had in the end?
|
Jeremy |
Ken |
Sean |
Total |
Before |
1 u |
4 u + 86 |
8 u |
|
Change |
x 4 |
- 86 |
x 12 |
|
After |
4 u |
4 u |
4 u |
12 u |
(a)
Total number of coins at first
=
1 u + 4 u + 86 + 8 u
= 13 u + 86
13 u + 86 = 30713 u = 307 - 8613 u = 221
1 u = 221 ÷ 13 = 17 Number of coins that Sean had at first
= 4 u ÷
12= 4 u x
21= 8 u = 8 x 17 = 136(b)Total number of coins that the three boys had at the end = 12 u = 12 x 17 = 204 Answer(s): (a) 136; (b) 204