The total number of balls in Bag K, Bag L and Bag M was 25.
58 of the balls from Bag K and 3 balls from Bag L were removed. More balls were then added into Bag M until the number of balls in it was quadrupled. The ratio of the number of balls in Bag K to Bag L to Bag M became 3 : 2 : 4.
- How many more balls were there in Bag K than Bag L at first?
- Find the total number of balls in Bag K and Bag M in the end.
|
Bag K |
Bag L |
Bag M |
Total |
Before |
8 u |
2 u + 3 |
1 u |
25 |
Change |
- 5 u |
- 3 |
+ 3 u |
|
After |
3 u |
|
4 u |
|
Comparing the balls in the end |
3 u |
2 u |
4 u |
|
(a)
Total number of balls at first
= 8 u + 2 u + 3 + 1 u
= 11 u + 3
11 u + 3 = 25
11 u = 25 - 3
11 u = 22
1 u = 22 ÷ 11 = 2
Number of more balls in Bag K than Bag L at first
= 8 u - (2 u + 3)
= 8 u - 2 u - 3
= 6 u - 3
= 6 x 2 - 3
= 12 - 3
= 9
(b)
Total number of balls in Bag K and Bag M in the end
= 3 u + 4 u
= 7 u
= 7 x 2
= 14
Answer(s): (a) 9; (b) 14