Eva has 4 times as many shortbreads as Kimberly. The total number of Eva's and Kimberly's shortbreads is 3 times of Dana's. Eva and Kimberly have 50 shortbreads more than Dana's. How many more shortbreads does Eva have than Kimberly?
| Eva |
Kimberly |
Dana |
| 4x3 |
1x3 |
|
| 3x5 |
1x5 |
| 12 u |
3 u |
5 u |
The total number of shortbreads that Eva and Kimberly have is repeated. Make the total number of shortbreads that Eva and Kimberly have the same. LCM of 5 and 3 is 15.
Total number of shortbreads that Eva and Kimberly have
= 12 u + 3 u
= 15 u
Number of more shortbreads that Eva and Kimberly have than Dana
= 15 u - 5 u
= 10 u
10 u = 50
1 u = 50 ÷ 10 = 5
Number of more shortbreads that Eva has than Kimberly
= 12 u - 3 u
= 9 u
= 9 x 5
= 45
Answer(s): 45
