Bag H and Bag J have a total of 68 lollipops. After 4 lollipops were transferred from Bag H to Bag J, there were 3 times as many lollipops in Bag H as Bag J. How many lollipops were there in each bag at first?
- Bag H
- Bag J
|
Bag H |
Bag J |
Total |
Before |
3 u + 4 |
1 u - 4 |
68 |
Change |
- 4 |
+ 4 |
|
After |
3 u |
1 u |
68 |
(a)
The total number of lollipops at first and in the end remains the same.
Total number of lollipops in the end
= 3 u + 1 u
= 4 u
4 u = 68
1 u = 68 ÷ 4 = 17
Number of lollipops in Bag H at first
= 3 u + 4
= 3 x 17 + 4
= 51 + 4
= 55
(b)
Number of lollipops in Bag J at first
= 1 u - 4
= 17 - 4
= 13
Answer(s): (a) 55; (b) 13