Jaslyn had
45 as many sweets as Lee. Lee had
310 as many sweets as Gabriel. If Gabriel had 228 more sweets than Jaslyn, find the total number of sweets that was shared among these 3 children at first.
Jaslyn |
Lee |
Gabriel |
4x3 |
5x3 |
|
|
3x5 |
10x5 |
12 u |
15 u |
50 u |
Lee is the repeated identity.
LCM of 5 and 3 = 15
Number of more sweets that Gabriel had more than Jaslyn
= 50 u - 12 u
= 38 u
38 u = 228
1 u = 228 ÷ 38 = 6
Total number of sweets
= 12 u + 15 u + 50 u
= 77 u
= 77 x 6
= 462
Answer(s): 462