Abi had
45 as many sweets as Cody. Cody had
37 as many sweets as Eric. If Eric had 46 more sweets than Abi, find the total number of sweets that was shared among these 3 children at first.
Abi |
Cody |
Eric |
4x3 |
5x3 |
|
|
3x5 |
7x5 |
12 u |
15 u |
35 u |
Cody is the repeated identity.
LCM of 5 and 3 = 15
Number of more sweets that Eric had more than Abi
= 35 u - 12 u
= 23 u
23 u = 46
1 u = 46 ÷ 23 = 2
Total number of sweets
= 12 u + 15 u + 35 u
= 62 u
= 62 x 2
= 124
Answer(s): 124