Mark and Albert had some coins in the ratio of 2 : 1. Albert and Peter had some coins in the ratio of 6 : 5. Mark sold 63 coins to Peter and ended up with half the number of coins Albert had. How many coins did Peter have in the end?
| Mark |
Albert |
Peter |
| 2x6 |
1x6 |
|
| |
6x1 |
5x1 |
| 12 u |
6 u |
5 u |
The number of coins that Albert had at first is repeated. Make the number of coins that Albert had at first the same. LCM of 1 and 6 is 6.
| |
Mark |
Albert |
Peter |
| Before |
12 u |
6 u |
5 u |
| Change |
- 63 |
|
+ 63 |
| After |
3 u |
6 u |
5 u + 63 |
Number of coins that Mark had in the end
= 6 u ÷ 2
= 3 u
Number of coins that Mark sold to Peter
= 12 u - 3 u
= 9 u
9 u = 63
1 u = 63 ÷ 9 = 7
Number of coins that Peter had in the end
= 5 u + 63
= 5 x 7 + 63
= 35 + 63
= 98
Answer(s): 98
