David and Michael shared some coins. David took
511 of the coins. After David lost 20 coins, Michael had twice as many coins as David. How many coins did Michael have in the end?
|
David |
Michael |
Before |
5 u |
6 u |
Change |
- 20 |
|
After |
1x3 = 3 u |
2x3 = 6 u |
The number of coins that Michael had remains the same. Make the number of coins that Michael had the same. LCM of 6 and 2 is 6.
Number of coins that David lost
= 5 u - 3 u
= 2 u
2 u = 20
1 u = 20 ÷ 2 = 10
Number of coins that Michael had in the end
= 6 u
= 6 x 10
= 60
Answer(s): 60