There were some starfruits in 3 bags, U, V and W. 25% of the number of starfruits in Bag U was equal to 10% of the number of starfruits in Bag V. The number of starfruits in Bag W was 60% of the total number of starfruits. After Jeremy removed 10% of the starfruits in Bag W, there were 178 more starfruits in Bag W than in Bag V. In the end, how many starfruits should be transferred from Bag V to Bag W so that the number of starfruits in Bag U would be the same as Bag V?

Bag U	Bag V	Bag W
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Bag U	Bag V	Bag W
Before	4 u	10 u	21 u
Change			- 2.1 u
After	4 u	10 u	18.9 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag U = ¹₁₀ Bag V

Bag U : Bag V
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of starfruits in Bag U and Bag V is the combined repeated identity.  Make the total number of starfruits in Box U and Bag V the same.  LCM of 7 and 2 = 14

Number of starfruits removed from Bag W
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of starfruits left in Bag W
= 21 u - 2.1 u
= 18.9 u

Number of more starfruits in Bag W than Bag V
= 18.9 u - 10 u
= 8.9 u

8.9 u = 178

1 u = 178 ÷ 8.9 = 20

	Bag U	Bag V	Bag W
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

Number of starfruits to be transferred from Bag V to Bag W so that the number of starfruits in Bag U would be the same as Bag V.
= 10 u - 4 u
= 6 u
= 6 x 20
= 120

Answer(s): 120