Tiffany had
45 as many mochi balls as Sean. Sean had
38 as many mochi balls as Bobby. If Bobby had 308 more mochi balls than Tiffany, find the total number of mochi balls that was shared among these 3 children at first.
Tiffany |
Sean |
Bobby |
4x3 |
5x3 |
|
|
3x5 |
8x5 |
12 u |
15 u |
40 u |
Sean is the repeated identity.
LCM of 5 and 3 = 15
Number of more mochi balls that Bobby had more than Tiffany
= 40 u - 12 u
= 28 u
28 u = 308
1 u = 308 ÷ 28 = 11
Total number of mochi balls
= 12 u + 15 u + 40 u
= 67 u
= 67 x 11
= 737
Answer(s): 737