Pierre, Brandon and Simon had a total of 501 coins at first. After a week, the number of Pierre's coins became 3 times the number of coins he had at first. The number of Brandon's coins decreased by 381. Simon 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 Simon have at first?
- What was the total number of coins that the three boys had in the end?
|
Pierre |
Brandon |
Simon |
Total |
Before |
1 u |
3 u + 381 |
6 u |
|
Change |
x 3 |
- 381 |
x 12 |
|
After |
3 u |
3 u |
3 u |
9 u |
(a)
Total number of coins at first
=
1 u + 3 u + 381 + 6 u
= 10 u + 381
10 u + 381 = 50110 u = 501 - 38110 u = 120
1 u = 120 ÷ 10 = 12 Number of coins that Simon had at first
= 3 u ÷
12= 3 u x
21= 6 u = 6 x 12 = 72(b)Total number of coins that the three boys had at the end = 9 u = 9 x 12 = 108 Answer(s): (a) 72; (b) 108