The total number of balls in Bag D, Bag E and Bag F was 135.
38 of the balls from Bag D and 27 balls from Bag E were removed. More balls were then added into Bag F until the number of balls in it was doubled. The ratio of the number of balls in Bag D to Bag E to Bag F became 5 : 3 : 2.
- How many less balls were there in Bag E than Bag D at first?
- Find the total number of balls in Bag E and Bag F in the end.
|
Bag D |
Bag E |
Bag F |
Total |
Before |
8 u |
3 u + 27 |
1 u |
135 |
Change |
- 3 u |
- 27 |
+ 1 u |
|
After |
5 u |
|
2 u |
|
Comparing the balls in the end |
5 u |
3 u |
2 u |
|
(a)
Total number of balls at first
= 8 u + 3 u + 27 + 1 u
= 12 u + 27
12 u + 27 = 135
12 u = 135 - 27
12 u = 108
1 u = 108 ÷ 12 = 9
Number of less balls in Bag E than Bag D at first
= 8 u - (3 u + 27)
= 8 u - 3 u - 27
= 5 u - 27
= 5 x 9 - 27
= 45 - 27
= 18
(b)
Total number of balls in Bag E and Bag F in the end
= 3 u + 2 u
= 5 u
= 5 x 9
= 45
Answer(s): (a) 18; (b) 45