Neave, Luke and Howard have some coins. The ratios of their coins are as follows:
Neave : (Neave + Luke + Howard) = 1 : 5
Luke : (Neave + Howard) = 1 : 3
Howard has 60 coins more than Luke. In the end, Howard gives away all he has to Luke and Neave so that both Luke and Neave end up with the same number of coins, how many coins has Luke received from Howard?
Luke |
Howard |
Neave |
Total |
4x4 |
1x4 |
5x4 |
1x5 |
3x5 |
4x5 |
5 u |
11 u |
4 u |
20 u |
The total number of coins is repeated. Make the total number of coins the same. LCM of 5 and 4 is 20.
Number of coins that Howard has more than Luke
= 11 u - 5 u
= 6 u
6 u = 60
1 u = 60 ÷ 6 = 10
|
Luke |
Howard |
Neave |
Before |
5 u |
11 u |
4 u |
Change |
+ 5 u |
- 11 u |
+ 6 u |
After |
10 u |
0 u |
10 u |
Number of coins that Luke has received from Howard
= 10 u - 5 u
= 5 u
= 5 x 10
= 50
Answer(s): 50