Simon, Riordan and Warren had a total of 76 coins. The ratio of the number of coins that Riordan had to the number of coins that Warren had was 8 : 3. After Simon and Riordan each gave away half of their coins, the 3 children had 47 coins left. How many coins did Riordan and Warren have in the end?
| |
Simon |
Riordan |
Warren |
Total |
| Before |
2 p |
8 u |
3 u |
76 |
| Change |
- 1 p |
- 4 u |
|
- 29 |
| After |
1 p |
4 u |
3 u |
47 |
Total number of coins that Simon and Riordan gave away
= 76 - 47
= 29
1 p + 4 u = 29 --- (1)
1 p + 4 u + 3 u = 47
1 p + 7 u = 47 --- (2)
(2) - (1)
(1 p + 7 u) - (1 p + 4 u) = 47 - 29
7 u - 4 u = 18
3 u = 18
1 u = 18 ÷ 3 = 6
Total number of coins that Riordan and Warren had in the end
= 4 u + 3 u
= 7 u
= 7 x 6
= 42
Answer(s): 42
