Ahmad and Rael had some coins in the ratio of 4 : 3. Rael and Gabriel had some coins in the ratio of 8 : 7. Ahmad sold 120 coins to Gabriel and ended up with half the number of coins Rael had. How many coins did Gabriel have in the end?
| Ahmad |
Rael |
Gabriel |
| 4x8 |
3x8 |
|
| |
8x3 |
7x3 |
| 32 u |
24 u |
21 u |
The number of coins that Rael had at first is repeated. Make the number of coins that Rael had at first the same. LCM of 3 and 8 is 24.
| |
Ahmad |
Rael |
Gabriel |
| Before |
32 u |
24 u |
21 u |
| Change |
- 120 |
|
+ 120 |
| After |
12 u |
24 u |
21 u + 120 |
Number of coins that Ahmad had in the end
= 24 u ÷ 2
= 12 u
Number of coins that Ahmad sold to Gabriel
= 32 u - 12 u
= 20 u
20 u = 120
1 u = 120 ÷ 20 = 6
Number of coins that Gabriel had in the end
= 21 u + 120
= 21 x 6 + 120
= 126 + 120
= 246
Answer(s): 246
