Xavier, Bobby and Pierre had a total of 329 buttons at first. After a week, the number of Xavier's buttons became 3 times the number of buttons he had at first. The number of Bobby's buttons decreased by 169. Pierre had one-quarter as many buttons as he had at first. In the end, the three boys had the same number of buttons.
- How many buttons did Pierre have at first?
- What was the total number of buttons that the three boys had in the end?
|
Xavier |
Bobby |
Pierre |
Total |
Before |
1 u |
3 u + 169 |
12 u |
|
Change |
x 3 |
- 169 |
x 14 |
|
After |
3 u |
3 u |
3 u |
9 u |
(a)
Total number of buttons at first
=
1 u + 3 u + 169 + 12 u
= 16 u + 169
16 u + 169 = 32916 u = 329 - 16916 u = 160
1 u = 160 ÷ 16 = 10 Number of buttons that Pierre had at first
= 3 u ÷
14= 3 u x
41= 12 u = 12 x 10 = 120(b)Total number of buttons that the three boys had at the end = 9 u = 9 x 10 = 90 Answer(s): (a) 120; (b) 90