Marion, Fanny and Tiffany have equal number of coins. Marion packs all her coins equally into 5 packets. Fanny packs all her coins equally into 6 packets. Tiffany packs all her coins equally into 10 packets. 2 packets of Marion's coins, 5 packets of Fanny's coins and 8 packets of Tiffany's coins add up to 488 coins. How many coins do they have altogether?
| |
Marion |
Fanny |
Tiffany |
| Number of packets |
5 |
6 |
10 |
| Number of coins |
30 u |
30 u |
30 u |
| Number of coins in each packet |
6 u |
5 u |
3 u |
All the coins can be put into the packets without remainder.
All the children have equal numbers of coins.
Make the number of coins that each child has the same. LCM of 5, 6 and 10 = 30
Number of coins that each child has = 30 u
Number of coins in 1 packet of Marion's coins = 30 u ÷ 5 = 6 u
Number of coins in 1 packet of Fanny's coins = 30 u ÷ 6 = 5 u
Number of coins in 1 packet of Tiffany's coins = 30 u ÷ 10 = 3 u
Number of coins in 2 packets of Marion's coins, 5 packets of Fanny's coins and 8 packets of Tiffany's coins
= (2 x 6 u) + (5 x 5 u) + (8 x 3 u)
= 12 u + 25 u + 24 u
= 61 u
61 u = 488
1 u = 488 ÷ 61 = 8
Total number of coins that they have
= 3 x 30 u
= 90 u
= 90 x 8
= 720
Answer(s): 720
