There were some pears in 3 bags, T, U and V. 25% of the number of pears in Bag T was equal to 10% of the number of pears in Bag U. The number of pears in Bag V was 70% of the total number of pears. After Oliver removed 10% of the pears in Bag V, there were 1164 more pears in Bag V than in Bag U. In the end, how many pears should be transferred from Bag U to Bag V so that the number of pears in Bag T would be the same as Bag U?

Bag T	Bag U	Bag V
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Bag T	Bag U	Bag V
Before	6 u	15 u	49 u
Change			- 4.9 u
After	6 u	15 u	44.1 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag T = ¹₁₀ Bag U

Bag T : Bag U
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of pears in Bag T and Bag U is the combined repeated identity.  Make the total number of pears in Box T and Bag U the same.  LCM of 7 and 3 = 21

Number of pears removed from Bag V
= ¹⁰₁₀₀ x 49 u
= 4.9 u

Number of pears left in Bag V
= 49 u - 4.9 u
= 44.1 u

Number of more pears in Bag V than Bag U
= 44.1 u - 15 u
= 29.1 u

29.1 u = 1164

1 u = 1164 ÷ 29.1 = 40

	Bag T	Bag U	Bag V
Before	6 u	15 u	44.1 u
Change		- 9 u	+ 9 u
After	6 u	6 u	53.1 u

Number of pears to be transferred from Bag U to Bag V so that the number of pears in Bag T would be the same as Bag U.
= 15 u - 6 u
= 9 u
= 9 x 40
= 360

Answer(s): 360