Cole and Lee had the same number of chocolate bars. Each of them packed his own chocolate bars into packets. Cole packed 8 chocolate bars in each packet and had 5 chocolate bars left. Lee packed 12 chocolate bars in each packet and was short of 3 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?
| |
Cole |
Lee |
| Number |
1 u |
1 u |
| Value |
8 |
12 |
| Total value |
8 u + 5 |
12 u - 3 |
The total number of chocolate bars that Cole and Lee each had is the same.
12 u - 3 = 8 u + 5
12 u - 8 u = 3 + 5
4 u = 8
1 u = 8 ÷ 4 = 2
Number of packets that each had if they have used the same number of packets = 2
(b)
The number of packets that each had is different.
Multiples of 8: 8, 16, 24, 32, 40
Multiples of 8 (+5): 13, 21, 29, 37, 45
Multiples of 12: 12, 24, 36, 48
Multiples of 12 (-3): 9, 21, 33, 45
Smallest common number: 45
Cole needs 5 packets of 8 chocolate bars and Lee needs 3 packets of 12 chocolate bars.
Smallest possible number of chocolate bars each of them had if they used different number of packets = 45
Answer(s): (a) 2; (b) 45
