Lynn, Jane and Roshel have equal number of coins. Lynn packs all her coins equally into 6 packets. Jane packs all her coins equally into 7 packets. Roshel packs all her coins equally into 3 packets. 5 packets of Lynn's coins, 4 packets of Jane's coins and 2 packets of Roshel's coins add up to 870 coins. How many coins do they have altogether?
| |
Lynn |
Jane |
Roshel |
| Number of packets |
6 |
7 |
3 |
| Number of coins |
42 u |
42 u |
42 u |
| Number of coins in each packet |
7 u |
6 u |
14 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 6, 7 and 3 = 42
Number of coins that each child has = 42 u
Number of coins in 1 packet of Lynn's coins = 42 u ÷ 6 = 7 u
Number of coins in 1 packet of Jane's coins = 42 u ÷ 7 = 6 u
Number of coins in 1 packet of Roshel's coins = 42 u ÷ 3 = 14 u
Number of coins in 5 packets of Lynn's coins, 4 packets of Jane's coins and 2 packets of Roshel's coins
= (5 x 7 u) + (4 x 6 u) + (2 x 14 u)
= 35 u + 24 u + 28 u
= 87 u
87 u = 870
1 u = 870 ÷ 87 = 10
Total number of coins that they have
= 3 x 42 u
= 126 u
= 126 x 10
= 1260
Answer(s): 1260
