Dana had 10 more coins than Min. After Min gave away
47 of her coins, Dana had thrice as many coins as Min. How many coins did Min have at first?
| |
Dana |
Min |
Comparing Dana and
Min at first |
10 more |
|
| Before |
9 u |
7 u |
| Change |
|
- 4 u |
| After |
9 u |
3 u |
Comparing Dana and
Min in the end |
3x3 =
9 u |
1x3 =
3 u |
Number of coins that Min had in the end is repeated. Make the number of coins that Min had in the end the same. LCM of 1 and 3 is 3.
Number of coins that Dana had at first = 9 u
Number of coins that Dana had more than Min at first
= 9 u - 7 u
= 2 u
2 u = 10
1 u = 10 ÷ 2 = 5
Number of coins that Min had at first
= 7 u
= 7 x 5
= 35
Answer(s): 35
