There were some starfruits and chikoos in Container D and Container E. In Container D, the ratio of the starfruits to the number of chikoos was 5 : 1. In Container E, the ratio of the number of starfruits to the number of chikoos was 3 : 1. There were 3 times as many fruits in Container D as in Container E. After another 50 chikoos were put into Container E, the ratio of the number of starfruits to the number of chikoos in Container E became 1 : 2. How many fruits were there in Container E in the end?

Container D		Container E
3x4 = 12 u		1x4 = 4 u
Starfruits	Chikoos	Starfruits	Chikoos
5x2	1x2	3	1
10 u	2 u	3 u	1 u

The total number of fruits in Container D is repeated.  Make the total number of fruits in Container D the same.  LCM of 3 and 6 is 12.

The total number of fruits in Container E at first is repeated.  Make the total number of fruits in Container E the same.  LCM of 1 and 4 is 4.

	Container D		Container E
	Starfruits	Chikoos	Starfruits	Chikoos
Before	10 u	2 u	3 u	1 u
Change				+ 50
After	10 u	2 u	1x3 = 3 u	2x3 = 6 u

Number of starfruits in Container E remains unchanged.  Make the number of starfruits in Container E the same.  LCM of 3 and 1 is 3.

Number of chikoos put into Container E
= 6 u - 1 u
= 5 u

5 u = 50

1 u = 50 ÷ 5 = 10

Number of fruits in Container E in the end
= 3 u + 6 u
= 9 u
= 9 x 10
= 90

Answer(s): 90