Box M and Box N have a total of 48 candy canes. After 5 candy canes were transferred from Box M to Box N, there were 3 times as many candy canes in Box M as Box N. How many candy canes were there in each box at first?
- Box M
- Box N
|
Box M |
Box N |
Total |
Before |
3 u + 5 |
1 u - 5 |
48 |
Change |
- 5 |
+ 5 |
|
After |
3 u |
1 u |
48 |
(a)
The total number of candy canes at first and in the end remains the same.
Total number of candy canes in the end
= 3 u + 1 u
= 4 u
4 u = 48
1 u = 48 ÷ 4 = 12
Number of candy canes in Box M at first
= 3 u + 5
= 3 x 12 + 5
= 36 + 5
= 41
(b)
Number of candy canes in Box N at first
= 1 u - 5
= 12 - 5
= 7
Answer(s): (a) 41; (b) 7