Xylia had
45 as many chocolate bars as Sean. Sean had
47 as many chocolate bars as Carl. If Carl had 76 more chocolate bars than Xylia, find the total number of chocolate bars that was shared among these 3 children at first.
Xylia |
Sean |
Carl |
4x4 |
5x4 |
|
|
4x5 |
7x5 |
16 u |
20 u |
35 u |
Sean is the repeated identity.
LCM of 5 and 4 = 20
Number of more chocolate bars that Carl had more than Xylia
= 35 u - 16 u
= 19 u
19 u = 76
1 u = 76 ÷ 19 = 4
Total number of chocolate bars
= 16 u + 20 u + 35 u
= 71 u
= 71 x 4
= 284
Answer(s): 284