Bag C and Bag D have a total of 114 candy canes. After 5 candy canes were transferred from Bag C to Bag D, there were 5 times as many candy canes in Bag C as Bag D. How many candy canes were there in each bag in the end?
- Bag C
- Bag D
|
Bag C |
Bag D |
Total |
Before |
5 u + 5 |
1 u - 5 |
114 |
Change |
- 5 |
+ 5 |
|
After |
5 u |
1 u |
114 |
(a)
The total number of candy canes at first and in the end remains the same.
Total number of candy canes in the end
= 5 u + 1 u
= 6 u
6 u = 114
1 u = 114 ÷ 6 = 19
Number of candy canes in Bag C in the end
= 5 u
= 5 x 19
= 95
(b)
Number of candy canes in Bag D in the end
= 1 u
= 19
Answer(s): (a) 95; (b) 19