Shannon, Kimberly and Sarah have equal number of coins. Shannon packs all her coins equally into 8 packets. Kimberly packs all her coins equally into 10 packets. Sarah packs all her coins equally into 4 packets. 7 packets of Shannon's coins, 4 packets of Kimberly's coins and 2 packets of Sarah's coins add up to 426 coins. How many coins do they have altogether?
| |
Shannon |
Kimberly |
Sarah |
| Number of packets |
8 |
10 |
4 |
| Number of coins |
40 u |
40 u |
40 u |
| Number of coins in each packet |
5 u |
4 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 8, 10 and 4 = 40
Number of coins that each child has = 40 u
Number of coins in 1 packet of Shannon's coins = 40 u ÷ 8 = 5 u
Number of coins in 1 packet of Kimberly's coins = 40 u ÷ 10 = 4 u
Number of coins in 1 packet of Sarah's coins = 40 u ÷ 4 = 10 u
Number of coins in 7 packets of Shannon's coins, 4 packets of Kimberly's coins and 2 packets of Sarah's coins
= (7 x 5 u) + (4 x 4 u) + (2 x 10 u)
= 35 u + 16 u + 20 u
= 71 u
71 u = 426
1 u = 426 ÷ 71 = 6
Total number of coins that they have
= 3 x 40 u
= 120 u
= 120 x 6
= 720
Answer(s): 720
