Fred and Zeph had some coins in the ratio of 6 : 1. Zeph and Flynn had some coins in the ratio of 10 : 9. Fred sold 660 coins to Flynn and ended up with half the number of coins Zeph had. How many coins did Flynn have in the end?
| Fred |
Zeph |
Flynn |
| 6x10 |
1x10 |
|
| |
10x1 |
9x1 |
| 60 u |
10 u |
9 u |
The number of coins that Zeph had at first is repeated. Make the number of coins that Zeph had at first the same. LCM of 1 and 10 is 10.
| |
Fred |
Zeph |
Flynn |
| Before |
60 u |
10 u |
9 u |
| Change |
- 660 |
|
+ 660 |
| After |
5 u |
10 u |
9 u + 660 |
Number of coins that Fred had in the end
= 10 u ÷ 2
= 5 u
Number of coins that Fred sold to Flynn
= 60 u - 5 u
= 55 u
55 u = 660
1 u = 660 ÷ 55 = 12
Number of coins that Flynn had in the end
= 9 u + 660
= 9 x 12 + 660
= 108 + 660
= 768
Answer(s): 768
