Pamela, Tina and Nicole have equal number of coins. Pamela packs all her coins equally into 4 packets. Tina packs all her coins equally into 3 packets. Nicole packs all her coins equally into 6 packets. 3 packets of Pamela's coins, 2 packets of Tina's coins and 4 packets of Nicole's coins add up to 175 coins. How many coins do they have altogether?
| |
Pamela |
Tina |
Nicole |
| Number of packets |
4 |
3 |
6 |
| Number of coins |
12 u |
12 u |
12 u |
| Number of coins in each packet |
3 u |
4 u |
2 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, 3 and 6 = 12
Number of coins that each child has = 12 u
Number of coins in 1 packet of Pamela's coins = 12 u ÷ 4 = 3 u
Number of coins in 1 packet of Tina's coins = 12 u ÷ 3 = 4 u
Number of coins in 1 packet of Nicole's coins = 12 u ÷ 6 = 2 u
Number of coins in 3 packets of Pamela's coins, 2 packets of Tina's coins and 4 packets of Nicole's coins
= (3 x 3 u) + (2 x 4 u) + (4 x 2 u)
= 9 u + 8 u + 8 u
= 25 u
25 u = 175
1 u = 175 ÷ 25 = 7
Total number of coins that they have
= 3 x 12 u
= 36 u
= 36 x 7
= 252
Answer(s): 252
