The total number of balls in Bag R, Bag S and Bag T was 83.
25 of the balls from Bag R and 11 balls from Bag S were removed. More balls were then added into Bag T until the number of balls in it was doubled. The ratio of the number of balls in Bag R to Bag S to Bag T became 3 : 2 : 2.
- How many more balls were there in Bag R than Bag S at first?
- Find the total number of balls in Bag R and Bag T in the end.
|
Bag R |
Bag S |
Bag T |
Total |
Before |
5 u |
2 u + 11 |
1 u |
83 |
Change |
- 2 u |
- 11 |
+ 1 u |
|
After |
3 u |
|
2 u |
|
Comparing the balls in the end |
3 u |
2 u |
2 u |
|
(a)
Total number of balls at first
= 5 u + 2 u + 11 + 1 u
= 8 u + 11
8 u + 11 = 83
8 u = 83 - 11
8 u = 72
1 u = 72 ÷ 8 = 9
Number of more balls in Bag R than Bag S at first
= 5 u - (2 u + 11)
= 5 u - 2 u - 11
= 3 u - 11
= 3 x 9 - 11
= 27 - 11
= 16
(b)
Total number of balls in Bag R and Bag T in the end
= 3 u + 2 u
= 5 u
= 5 x 9
= 45
Answer(s): (a) 16; (b) 45