Irene, Tiffany and Hazel have equal number of marbles. Irene packs all her marbles equally into 5 packets. Tiffany packs all her marbles equally into 10 packets. Hazel packs all her marbles equally into 8 packets. 4 packets of Irene's marbles, 6 packets of Tiffany's marbles and 3 packets of Hazel's marbles add up to 497 marbles. How many marbles do they have altogether?
| |
Irene |
Tiffany |
Hazel |
| Number of packets |
5 |
10 |
8 |
| Number of marbles |
40 u |
40 u |
40 u |
| Number of marbles in each packet |
8 u |
4 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 5, 10 and 8 = 40
Number of marbles that each child has = 40 u
Number of marbles in 1 packet of Irene's marbles = 40 u ÷ 5 = 8 u
Number of marbles in 1 packet of Tiffany's marbles = 40 u ÷ 10 = 4 u
Number of marbles in 1 packet of Hazel's marbles = 40 u ÷ 8 = 5 u
Number of marbles in 4 packets of Irene's marbles, 6 packets of Tiffany's marbles and 3 packets of Hazel's marbles
= (4 x 8 u) + (6 x 4 u) + (3 x 5 u)
= 32 u + 24 u + 15 u
= 71 u
71 u = 497
1 u = 497 ÷ 71 = 7
Total number of marbles that they have
= 3 x 40 u
= 120 u
= 120 x 7
= 840
Answer(s): 840
