Elijah and Luke had the same number of chocolate bars. Each of them packed his own chocolate bars into packets. Elijah packed 8 chocolate bars in each packet and had 2 chocolate bars left. Luke packed 10 chocolate bars in each packet and was short of 8 chocolate bars.
- How many packets did each of them have if they have used the same number of packets?
- What was the smallest possible number of chocolate bars each of them had if they used different number of packets?
| |
Elijah |
Luke |
| Number |
1 u |
1 u |
| Value |
8 |
10 |
| Total value |
8 u + 2 |
10 u - 8 |
The total number of chocolate bars that Elijah and Luke each had is the same.
10 u - 8 = 8 u + 2
10 u - 8 u = 8 + 2
2 u = 10
1 u = 10 ÷ 2 = 5
Number of packets that each had if they have used the same number of packets = 5
(b)
The number of packets that each had is different.
Multiples of 8: 8, 16, 24, 32, 40
Multiples of 8 (+2): 10, 18, 26, 34, 42
Multiples of 10: 10, 20, 30, 40, 50
Multiples of 10 (-8): 2, 12, 22, 32, 42
Smallest common number: 42
Elijah needs 5 packets of 8 chocolate bars and Luke needs 4 packets of 10 chocolate bars.
Smallest possible number of chocolate bars each of them had if they used different number of packets = 42
Answer(s): (a) 5; (b) 42
