The total number of balls in Bag N, Bag P and Bag Q was 104.
38 of the balls from Bag N and 8 balls from Bag P were removed. More balls were then added into Bag Q until the number of balls in it was quadrupled. The ratio of the number of balls in Bag N to Bag P to Bag Q became 5 : 3 : 4.
- How many more balls were there in Bag N than Bag P at first?
- Find the total number of balls in Bag N and Bag Q in the end.
|
Bag N |
Bag P |
Bag Q |
Total |
Before |
8 u |
3 u + 8 |
1 u |
104 |
Change |
- 3 u |
- 8 |
+ 3 u |
|
After |
5 u |
|
4 u |
|
Comparing the balls in the end |
5 u |
3 u |
4 u |
|
(a)
Total number of balls at first
= 8 u + 3 u + 8 + 1 u
= 12 u + 8
12 u + 8 = 104
12 u = 104 - 8
12 u = 96
1 u = 96 ÷ 12 = 8
Number of more balls in Bag N than Bag P at first
= 8 u - (3 u + 8)
= 8 u - 3 u - 8
= 5 u - 8
= 5 x 8 - 8
= 40 - 8
= 32
(b)
Total number of balls in Bag N and Bag Q in the end
= 5 u + 4 u
= 9 u
= 9 x 8
= 72
Answer(s): (a) 32; (b) 72