The total number of balls in Bag T, Bag U and Bag V was 136.
78 of the balls from Bag T and 37 balls from Bag U were removed. More balls were then added into Bag V until the number of balls in it was quadrupled. The ratio of the number of balls in Bag T to Bag U to Bag V became 1 : 2 : 4.
- How many less balls were there in Bag U than Bag T at first?
- Find the total number of balls in Bag U and Bag V in the end.
|
Bag T |
Bag U |
Bag V |
Total |
Before |
8 u |
2 u + 37 |
1 u |
136 |
Change |
- 7 u |
- 37 |
+ 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 + 37 + 1 u
= 11 u + 37
11 u + 37 = 136
11 u = 136 - 37
11 u = 99
1 u = 99 ÷ 11 = 9
Number of less balls in Bag U than Bag T at first
= 8 u - (2 u + 37)
= 8 u - 2 u - 37
= 6 u - 37
= 6 x 9 - 37
= 54 - 37
= 17
(b)
Total number of balls in Bag U and Bag V in the end
= 2 u + 4 u
= 6 u
= 6 x 9
= 54
Answer(s): (a) 17; (b) 54