The total number of balls in Bag L, Bag M and Bag N was 191.
58 of the balls from Bag L and 35 balls from Bag M were removed. More balls were then added into Bag N until the number of balls in it was doubled. The ratio of the number of balls in Bag L to Bag M to Bag N became 3 : 4 : 2.
- How many less balls were there in Bag M than Bag L at first?
- Find the total number of balls in Bag M and Bag N in the end.
|
Bag L |
Bag M |
Bag N |
Total |
Before |
8 u |
4 u + 35 |
1 u |
191 |
Change |
- 5 u |
- 35 |
+ 1 u |
|
After |
3 u |
|
2 u |
|
Comparing the balls in the end |
3 u |
4 u |
2 u |
|
(a)
Total number of balls at first
= 8 u + 4 u + 35 + 1 u
= 13 u + 35
13 u + 35 = 191
13 u = 191 - 35
13 u = 156
1 u = 156 ÷ 13 = 12
Number of less balls in Bag M than Bag L at first
= 8 u - (4 u + 35)
= 8 u - 4 u - 35
= 4 u - 35
= 4 x 12 - 35
= 48 - 35
= 13
(b)
Total number of balls in Bag M and Bag N in the end
= 4 u + 2 u
= 6 u
= 6 x 12
= 72
Answer(s): (a) 13; (b) 72