Yoko, Joelle and Min have equal number of coins. Yoko packs all her coins equally into 4 packets. Joelle packs all her coins equally into 6 packets. Min packs all her coins equally into 10 packets. 3 packets of Yoko's coins, 5 packets of Joelle's coins and 4 packets of Min's coins add up to 238 coins. How many coins do they have altogether?
| |
Yoko |
Joelle |
Min |
| Number of packets |
4 |
6 |
10 |
| Number of coins |
60 u |
60 u |
60 u |
| Number of coins in each packet |
15 u |
10 u |
6 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 4, 6 and 10 = 60
Number of coins that each child has = 60 u
Number of coins in 1 packet of Yoko's coins = 60 u ÷ 4 = 15 u
Number of coins in 1 packet of Joelle's coins = 60 u ÷ 6 = 10 u
Number of coins in 1 packet of Min's coins = 60 u ÷ 10 = 6 u
Number of coins in 3 packets of Yoko's coins, 5 packets of Joelle's coins and 4 packets of Min's coins
= (3 x 15 u) + (5 x 10 u) + (4 x 6 u)
= 45 u + 50 u + 24 u
= 119 u
119 u = 238
1 u = 238 ÷ 119 = 2
Total number of coins that they have
= 3 x 60 u
= 180 u
= 180 x 2
= 360
Answer(s): 360
