Raeann, Lynn and Dana have equal number of coins. Raeann packs all her coins equally into 10 packets. Lynn packs all her coins equally into 8 packets. Dana packs all her coins equally into 4 packets. 7 packets of Raeann's coins, 2 packets of Lynn's coins and 3 packets of Dana's coins add up to 272 coins. How many coins do they have altogether?
| |
Raeann |
Lynn |
Dana |
| Number of packets |
10 |
8 |
4 |
| Number of coins |
40 u |
40 u |
40 u |
| Number of coins in each packet |
4 u |
5 u |
10 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 10, 8 and 4 = 40
Number of coins that each child has = 40 u
Number of coins in 1 packet of Raeann's coins = 40 u ÷ 10 = 4 u
Number of coins in 1 packet of Lynn's coins = 40 u ÷ 8 = 5 u
Number of coins in 1 packet of Dana's coins = 40 u ÷ 4 = 10 u
Number of coins in 7 packets of Raeann's coins, 2 packets of Lynn's coins and 3 packets of Dana's coins
= (7 x 4 u) + (2 x 5 u) + (3 x 10 u)
= 28 u + 10 u + 30 u
= 68 u
68 u = 272
1 u = 272 ÷ 68 = 4
Total number of coins that they have
= 3 x 40 u
= 120 u
= 120 x 4
= 480
Answer(s): 480
