There were some persimmons in 3 bags, U, V and W. 25% of the number of persimmons in Bag U was equal to 10% of the number of persimmons in Bag V. The number of persimmons in Bag W was 60% of the total number of persimmons. After Justin removed 20% of the persimmons in Bag W, there were 340 more persimmons in Bag W than in Bag V. In the end, how many persimmons should be transferred from Bag V to Bag W so that the number of persimmons 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			- 4.2 u
After	4 u	10 u	16.8 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 persimmons in Bag U and Bag V is the combined repeated identity.  Make the total number of persimmons in Box U and Bag V the same.  LCM of 7 and 2 = 14

Number of persimmons removed from Bag W
= ²⁰₁₀₀ x 21 u
= 4.2 u

Number of persimmons left in Bag W
= 21 u - 4.2 u
= 16.8 u

Number of more persimmons in Bag W than Bag V
= 16.8 u - 10 u
= 6.8 u

6.8 u = 340

1 u = 340 ÷ 6.8 = 50

	Bag U	Bag V	Bag W
Before	4 u	10 u	16.8 u
Change		- 6 u	+ 6 u
After	4 u	4 u	22.8 u

Number of persimmons to be transferred from Bag V to Bag W so that the number of persimmons in Bag U would be the same as Bag V.
= 10 u - 4 u
= 6 u
= 6 x 50
= 300

Answer(s): 300