Howard, Tom and Gabriel had 504 coins. After Howard used 12 coins and Tom used
18 of her coins, Howard had thrice as many coins left as Tom. Gabriel bought another 21 coins, he had four times the number of coins Tom had left. How many coins did Gabriel have at first?
| |
Howard |
Tom |
Gabriel |
Total |
| Before |
21 u + 12 |
8 u |
28 u - 21 |
504 |
| Change |
- 12 |
- 1 u |
+ 21 |
|
| After |
21 u |
7 u |
28 u |
|
Comparing Howard
and Tom in the end |
3 |
1 |
|
|
Comparing Gabriel
and Tom in the end |
|
1 |
4 |
|
Number of coins that Howard had in the end
= 3 x 7 u
= 21 u
Number of coins that Gabriel had in the end
= 4 x 7 u
= 28 u
Total number of coins at first
= 21 u + 12 + 8 u + 28 u - 21
= 57 u - 9
57 u - 9 = 504
57 u = 504 + 9
57 u = 513
1 u = 513 ÷ 57 = 9
Number of coins that Gabriel had at first
= 28 u - 21
= 28 x 9 - 21
= 252 - 21
= 231
Answer(s): 231
