Nick and Japheth had some coins in the ratio of 11 : 4. Japheth and Xavier had some coins in the ratio of 7 : 1. Nick sold 126 coins to Xavier and ended up with half the number of coins Japheth had. How many coins did Xavier have in the end?
| Nick |
Japheth |
Xavier |
| 11x7 |
4x7 |
|
| |
7x4 |
1x4 |
| 77 u |
28 u |
4 u |
The number of coins that Japheth had at first is repeated. Make the number of coins that Japheth had at first the same. LCM of 4 and 7 is 28.
| |
Nick |
Japheth |
Xavier |
| Before |
77 u |
28 u |
4 u |
| Change |
- 126 |
|
+ 126 |
| After |
14 u |
28 u |
4 u + 126 |
Number of coins that Nick had in the end
= 28 u ÷ 2
= 14 u
Number of coins that Nick sold to Xavier
= 77 u - 14 u
= 63 u
63 u = 126
1 u = 126 ÷ 63 = 2
Number of coins that Xavier had in the end
= 4 u + 126
= 4 x 2 + 126
= 8 + 126
= 134
Answer(s): 134
