The total number of balls in Box L, Box M and Box N was 191.
58 of the balls from Box L and 47 balls from Box M were removed. More balls were then added into Box N until the number of balls in it was doubled. The ratio of the number of balls in Box L to Box M to Box N became 3 : 3 : 2.
- How many less balls were there in Box M than Box L at first?
- Find the total number of balls in Box M and Box N in the end.
|
Box L |
Box M |
Box N |
Total |
Before |
8 u |
3 u + 47 |
1 u |
191 |
Change |
- 5 u |
- 47 |
+ 1 u |
|
After |
3 u |
|
2 u |
|
Comparing the balls in the end |
3 u |
3 u |
2 u |
|
(a)
Total number of balls at first
= 8 u + 3 u + 47 + 1 u
= 12 u + 47
12 u + 47 = 191
12 u = 191 - 47
12 u = 144
1 u = 144 ÷ 12 = 12
Number of less balls in Box M than Box L at first
= 8 u - (3 u + 47)
= 8 u - 3 u - 47
= 5 u - 47
= 5 x 12 - 47
= 60 - 47
= 13
(b)
Total number of balls in Box M and Box N in the end
= 3 u + 2 u
= 5 u
= 5 x 12
= 60
Answer(s): (a) 13; (b) 60