Fanny, Jean and Victoria have equal number of pens. Fanny packs all her pens equally into 5 packets. Jean packs all her pens equally into 3 packets. Victoria packs all her pens equally into 6 packets. 4 packets of Fanny's pens, 2 packets of Jean's pens and 3 packets of Victoria's pens add up to 177 pens. How many pens do they have altogether?
| |
Fanny |
Jean |
Victoria |
| Number of packets |
5 |
3 |
6 |
| Number of pens |
30 u |
30 u |
30 u |
| Number of pens in each packet |
6 u |
10 u |
5 u |
All the pens can be put into the packets without remainder.
All the children have equal numbers of pens.
Make the number of pens that each child has the same. LCM of 5, 3 and 6 = 30
Number of pens that each child has = 30 u
Number of pens in 1 packet of Fanny's pens = 30 u ÷ 5 = 6 u
Number of pens in 1 packet of Jean's pens = 30 u ÷ 3 = 10 u
Number of pens in 1 packet of Victoria's pens = 30 u ÷ 6 = 5 u
Number of pens in 4 packets of Fanny's pens, 2 packets of Jean's pens and 3 packets of Victoria's pens
= (4 x 6 u) + (2 x 10 u) + (3 x 5 u)
= 24 u + 20 u + 15 u
= 59 u
59 u = 177
1 u = 177 ÷ 59 = 3
Total number of pens that they have
= 3 x 30 u
= 90 u
= 90 x 3
= 270
Answer(s): 270
