The total number of balls in Bag S, Bag T and Bag U was 147.
58 of the balls from Bag S and 48 balls from Bag T were removed. More balls were then added into Bag U until the number of balls in it was tripled. The ratio of the number of balls in Bag S to Bag T to Bag U became 3 : 2 : 3.
- How many less balls were there in Bag T than Bag S at first?
- Find the total number of balls in Bag T and Bag U in the end.
|
Bag S |
Bag T |
Bag U |
Total |
Before |
8 u |
2 u + 48 |
1 u |
147 |
Change |
- 5 u |
- 48 |
+ 2 u |
|
After |
3 u |
|
3 u |
|
Comparing the balls in the end |
3 u |
2 u |
3 u |
|
(a)
Total number of balls at first
= 8 u + 2 u + 48 + 1 u
= 11 u + 48
11 u + 48 = 147
11 u = 147 - 48
11 u = 99
1 u = 99 ÷ 11 = 9
Number of less balls in Bag T than Bag S at first
= 8 u - (2 u + 48)
= 8 u - 2 u - 48
= 6 u - 48
= 6 x 9 - 48
= 54 - 48
= 6
(b)
Total number of balls in Bag T and Bag U in the end
= 2 u + 3 u
= 5 u
= 5 x 9
= 45
Answer(s): (a) 6; (b) 45