Howard gave
29 of his coins to Zane,
27 of his remaining coins to Jeremy and
25 of the rest of his coins to Riordan. He then had 30 coins left. How many coins had he at first?
| Zane |
Remaining coins |
Total |
| 2 u |
7 u |
9 u |
| |
Jeremy |
Rest of the coins |
|
| |
2 u |
5 u |
|
| |
|
Riordan |
Left |
|
| |
|
2 u |
3 u |
|
| |
|
|
30 |
|
Number of coins that Howard had before = 9 u
Number of coins that Howard gave to Zane = 2 u
Number of coins that Howard gave to Jeremy = 2 u
Number of coins that Howard gave to Riordan = 2 u
Number of coins that Howard had left = 3 u
3 u = 30
1 u = 30 ÷ 3 = 10
Number of coins that Howard had at first
= 9 u
= 9 u x 10
= 90
Answer(s): 90
