Pierre, Simon and Ian had a total of 134 coins. The ratio of Simon's coins to Ian's coins was 4 : 7 at first. Pierre and Simon each gave away
12 of their coins. Given that the three boys had 102 coins left, how many coins did Pierre have in the end?
|
Pierre |
Simon |
Ian |
Total |
Comparing Simon and Ian at first |
|
4 u |
7 u |
|
Before |
2 p |
2x2 = 4 u |
7 u |
134 |
Change |
- 1 p |
-1x2 = - 2 u |
|
- 32 |
After |
1 p |
1x2 = 2 u |
7 u |
102 |
Total number of coins that Pierre and Simon gave away
= 134 - 102
= 32
The number of coins that Simon had at first is repeated. Make the number of coins that Simon had at first the same. LCM of 4 and 2 is 4.
1 p + 2 u = 134 - 102
1 p + 2 u = 32
1 p = 32 - 2 u --- (1)
1 p + 2 u + 7 u = 102
1 p + 9 u = 102
1 p = 102 - 9 u --- (2)
(1) = (2)
32 - 2 u = 102 - 9 u
9 u - 2 u = 102 - 32
9 u - 2 u = 70
7 u = 70
1 u = 70 ÷ 7 = 10
Substitute 1 u = 10 into (1).
1 p = 32 - 2 u
1 p = 32 - 2 x 10
1 p = 32 - 20
1 p = 12
Number of coins that Pierre had in the end
= 1 p
= 12
Answer(s): 12