Xuan, Jade and Emma have equal number of coins. Xuan packs all her coins equally into 10 packets. Jade packs all her coins equally into 8 packets. Emma packs all her coins equally into 4 packets. 7 packets of Xuan's coins, 3 packets of Jade's coins and 2 packets of Emma's coins add up to 126 coins. How many coins do they have altogether?
| |
Xuan |
Jade |
Emma |
| 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 Xuan's coins = 40 u ÷ 10 = 4 u
Number of coins in 1 packet of Jade's coins = 40 u ÷ 8 = 5 u
Number of coins in 1 packet of Emma's coins = 40 u ÷ 4 = 10 u
Number of coins in 7 packets of Xuan's coins, 3 packets of Jade's coins and 2 packets of Emma's coins
= (7 x 4 u) + (3 x 5 u) + (2 x 10 u)
= 28 u + 15 u + 20 u
= 63 u
63 u = 126
1 u = 126 ÷ 63 = 2
Total number of coins that they have
= 3 x 40 u
= 120 u
= 120 x 2
= 240
Answer(s): 240
