Xavier, Ivan and Luis shared a bag of coins. Ivan took 35 as many coins as Luis. Xavier took twice as many coins as the total Ivan and Luis took. After Xavier had given 45 to Ivan and 19 to Luis, Ivan gave 5 to Luis. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Xavier and Ivan had at first.
| |
Ivan |
Luis |
Xavier |
Total |
| Before |
3 u |
5 u |
16 u |
24 u |
| Change 1 |
+ 45 |
|
- 45 |
|
| Change 2 |
|
+ 19 |
- 19 |
|
| Change 3 |
- 5 |
+ 5 |
|
|
| After |
1x8 =
8 u |
1x8 =
8 u |
1x8 =
8 u |
3x8 =
24 u |
Total number of coins that Ivan and Luis had at first
= 3 u + 5 u
= 8 u
Number of coins that Xavier had at first
= 2 x 8 u
= 16 u
The total number of coins remains unchanged. Make the total number of coins the same. LCM of 3 and 24 is 24.
Number of coins that Luis received from Ivan and Xavier
= 8 u - 5 u
= 3 u
3 u = 19 + 5
3 u = 24
1 u = 24 ÷ 3 = 8
Difference between the number coins that Xavier and Ivan had at first
= 16 u - 3 u
= 13 u
= 13 x 8
= 104
Answer(s): 104
