The total number of balls in Basket R, Basket S and Basket T was 34.
45 of the balls from Basket R and 2 balls from Basket S were removed. More balls were then added into Basket T until the number of balls in it was tripled. The ratio of the number of balls in Basket R to Basket S to Basket T became 1 : 2 : 3.
- How many less balls were there in Basket S than Basket R at first?
- Find the total number of balls in Basket S and Basket T in the end.
|
Basket R |
Basket S |
Basket T |
Total |
Before |
5 u |
2 u + 2 |
1 u |
34 |
Change |
- 4 u |
- 2 |
+ 2 u |
|
After |
1 u |
|
3 u |
|
Comparing the balls in the end |
1 u |
2 u |
3 u |
|
(a)
Total number of balls at first
= 5 u + 2 u + 2 + 1 u
= 8 u + 2
8 u + 2 = 34
8 u = 34 - 2
8 u = 32
1 u = 32 ÷ 8 = 4
Number of less balls in Basket S than Basket R at first
= 5 u - (2 u + 2)
= 5 u - 2 u - 2
= 3 u - 2
= 3 x 4 - 2
= 12 - 2
= 10
(b)
Total number of balls in Basket S and Basket T in the end
= 2 u + 3 u
= 5 u
= 5 x 4
= 20
Answer(s): (a) 10; (b) 20