Eva, Jaslyn and Pamela have equal number of marbles. Eva packs all her marbles equally into 10 packets. Jaslyn packs all her marbles equally into 4 packets. Pamela packs all her marbles equally into 8 packets. 8 packets of Eva's marbles, 2 packets of Jaslyn's marbles and 3 packets of Pamela's marbles add up to 201 marbles. How many marbles do they have altogether?
| |
Eva |
Jaslyn |
Pamela |
| Number of packets |
10 |
4 |
8 |
| Number of marbles |
40 u |
40 u |
40 u |
| Number of marbles in each packet |
4 u |
10 u |
5 u |
All the marbles can be put into the packets without remainder.
All the children have equal numbers of marbles.
Make the number of marbles that each child has the same. LCM of 10, 4 and 8 = 40
Number of marbles that each child has = 40 u
Number of marbles in 1 packet of Eva's marbles = 40 u ÷ 10 = 4 u
Number of marbles in 1 packet of Jaslyn's marbles = 40 u ÷ 4 = 10 u
Number of marbles in 1 packet of Pamela's marbles = 40 u ÷ 8 = 5 u
Number of marbles in 8 packets of Eva's marbles, 2 packets of Jaslyn's marbles and 3 packets of Pamela's marbles
= (8 x 4 u) + (2 x 10 u) + (3 x 5 u)
= 32 u + 20 u + 15 u
= 67 u
67 u = 201
1 u = 201 ÷ 67 = 3
Total number of marbles that they have
= 3 x 40 u
= 120 u
= 120 x 3
= 360
Answer(s): 360
