Fred and Vaidev had some coins in the ratio of 3 : 2. Vaidev and Xavier had some coins in the ratio of 5 : 2. Fred sold 100 coins to Xavier and ended up with half the number of coins Vaidev had. How many coins did Xavier have in the end?
| Fred |
Vaidev |
Xavier |
| 3x5 |
2x5 |
|
| |
5x2 |
2x2 |
| 15 u |
10 u |
4 u |
The number of coins that Vaidev had at first is repeated. Make the number of coins that Vaidev had at first the same. LCM of 2 and 5 is 10.
| |
Fred |
Vaidev |
Xavier |
| Before |
15 u |
10 u |
4 u |
| Change |
- 100 |
|
+ 100 |
| After |
5 u |
10 u |
4 u + 100 |
Number of coins that Fred had in the end
= 10 u ÷ 2
= 5 u
Number of coins that Fred sold to Xavier
= 15 u - 5 u
= 10 u
10 u = 100
1 u = 100 ÷ 10 = 10
Number of coins that Xavier had in the end
= 4 u + 100
= 4 x 10 + 100
= 40 + 100
= 140
Answer(s): 140
