Xavier had 6 times as many coins as Fabian at first. Then each of them bought an equal number of new coins. In the end, Fabian had 10 coins and Xavier had 4 times as many coins as Fabian. How many coins did both of them buy altogether?
| |
Xavier |
Fabian |
Difference |
| Before |
6x3 =
18 u |
1x3 =
3 u |
5x3 =
15 u |
| Change |
+ ? |
+ ? |
|
| After |
4x5 =
20 u |
1x5 =
5 u |
3x5 =
15 u |
The difference in the number of coins at first and in the end remains unchanged. Make the difference in the number of coins at first and in the end the same. LCM of 5 and 3 is 15.
Number of coins that Fabian had in the end = 5 u
5 u = 10
1 u = 10 ÷ 5 = 2
Number of coins that Xavier bought
= 20 u - 18 u
= 2 u
Number of coins that both of them bought altogether
= 2 x 2 u
= 4 u
= 4 x 2
= 8
Answer(s): 8
