Eric, Seth and Owen have some beads. The ratios of their beads are as follows:
Eric : (Eric + Seth + Owen) = 3 : 7
Seth : (Eric + Owen) = 1 : 3
Owen has 14 beads more than Seth. In the end, Owen gives away all he has to Seth and Eric so that both Seth and Eric end up with the same number of beads, how many beads has Seth received from Owen?
Seth |
Owen |
Eric |
Total |
4x4 |
3x4 |
7x4 |
1x7 |
3x7 |
4x7 |
7 u |
9 u |
12 u |
28 u |
The total number of beads is repeated. Make the total number of beads the same. LCM of 7 and 4 is 28.
Number of beads that Owen has more than Seth
= 9 u - 7 u
= 2 u
2 u = 14
1 u = 14 ÷ 2 = 7
|
Seth |
Owen |
Eric |
Before |
7 u |
9 u |
12 u |
Change |
+ 7 u |
- 9 u |
+ 2 u |
After |
14 u |
0 u |
14 u |
Number of beads that Seth has received from Owen
= 14 u - 7 u
= 7 u
= 7 x 7
= 49
Answer(s): 49