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