Eva had
58 as many chocolate bars as Brandon. Brandon had
35 as many chocolate bars as Eric. If Eric had 225 more chocolate bars than Eva, find the total number of chocolate bars that was shared among these 3 children at first.
Eva |
Brandon |
Eric |
5x3 |
8x3 |
|
|
3x8 |
5x8 |
15 u |
24 u |
40 u |
Brandon is the repeated identity.
LCM of 8 and 3 = 24
Number of more chocolate bars that Eric had more than Eva
= 40 u - 15 u
= 25 u
25 u = 225
1 u = 225 ÷ 25 = 9
Total number of chocolate bars
= 15 u + 24 u + 40 u
= 79 u
= 79 x 9
= 711
Answer(s): 711