The total number of balls in Bag A, Bag B and Bag C was 61.
78 of the balls from Bag A and 17 balls from Bag B were removed. More balls were then added into Bag C until the number of balls in it was quadrupled. The ratio of the number of balls in Bag A to Bag B to Bag C became 1 : 2 : 4.
- How many less balls were there in Bag B than Bag A at first?
- Find the total number of balls in Bag B and Bag C in the end.
|
Bag A |
Bag B |
Bag C |
Total |
Before |
8 u |
2 u + 17 |
1 u |
61 |
Change |
- 7 u |
- 17 |
+ 3 u |
|
After |
1 u |
|
4 u |
|
Comparing the balls in the end |
1 u |
2 u |
4 u |
|
(a)
Total number of balls at first
= 8 u + 2 u + 17 + 1 u
= 11 u + 17
11 u + 17 = 61
11 u = 61 - 17
11 u = 44
1 u = 44 ÷ 11 = 4
Number of less balls in Bag B than Bag A at first
= 8 u - (2 u + 17)
= 8 u - 2 u - 17
= 6 u - 17
= 6 x 4 - 17
= 24 - 17
= 7
(b)
Total number of balls in Bag B and Bag C in the end
= 2 u + 4 u
= 6 u
= 6 x 4
= 24
Answer(s): (a) 7; (b) 24