Bobby gave
29 of his coins to Luke,
27 of his remaining coins to Bryan and
35 of the rest of his coins to Oscar. He then had 18 coins left. How many coins had he at first?
| Luke |
Remaining coins |
Total |
| 2 u |
7 u |
9 u |
| |
Bryan |
Rest of the coins |
|
| |
2 u |
5 u |
|
| |
|
Oscar |
Left |
|
| |
|
3 u |
2 u |
|
| |
|
|
18 |
|
Number of coins that Bobby had before = 9 u
Number of coins that Bobby gave to Luke = 2 u
Number of coins that Bobby gave to Bryan = 2 u
Number of coins that Bobby gave to Oscar = 3 u
Number of coins that Bobby had left = 2 u
2 u = 18
1 u = 18 ÷ 2 = 9
Number of coins that Bobby had at first
= 9 u
= 9 u x 9
= 81
Answer(s): 81
