A stationery kiosk sells pens in packs of 3 and 4. At first, there were 6 times as many packs of 3 as packs of 4. After selling half of the packs of 3 and some packs of 4, Mr Chua packs 10 additional packs of 4. How many packs of 3 and 4 are sold if there are 6 times as many packs of 3 as packs of 4 and there is a total of 44 unsold pens?
|
Packs of 3 |
Packs of 4 |
Comparing the number of packs at first |
6x2 = 12 u |
1x2 = 2 u |
Before |
2x6 = 12 u |
|
Change 1 |
- 1x6 = - 6 u |
- ? |
Change 2 |
|
+ 10 |
After |
1x6 = 6 u |
|
Comparing the number of packs in the end |
6 u |
1 u |
The number of packs of 3 in the end is repeated. Make the number of packs of 3 in the end the same. LCM of 1 and 6 is 6.
The number of packs of 3 at first is repeated. Make the number of packs of 3 at first the same. LCM of 12 and 6 is 12.
Number of pens left unsold |
Packs of 3
|
Packs of 4
|
Total
|
Number |
6 u |
1 u |
|
Value |
3 |
4 |
|
Total value |
18 u |
4 u |
22 u |
Total number of pens left unsold
= (6 u x 3) + (1 u x 4)
= 18 u + 4 u
= 22 u
22 u = 44
1 u = 44 ÷ 22 = 2
Number of packs of 3 and 4 sold
= 6 u + (2 u - 1 u) + 10
= 6 u + 1 u + 10
= 7 u + 10
= 7 x 2 + 10
= 14 + 10
= 24
Answer(s): 24