There were some peaches in 3 baskets, W, X and Y. 25% of the number of peaches in Basket W was equal to 10% of the number of peaches in Basket X. The number of peaches in Basket Y was 70% of the total number of peaches. After Michael removed 10% of the peaches in Basket Y, there were 1455 more peaches in Basket Y than in Basket X. In the end, how many peaches should be transferred from Basket X to Basket Y so that the number of peaches in Basket W would be the same as Basket X?

Basket W	Basket X	Basket Y
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Basket W	Basket X	Basket Y
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 Basket W = ¹₁₀ Basket X

Basket W : Basket X
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of peaches in Basket W and Basket X is the combined repeated identity.  Make the total number of peaches in Box W and Basket X the same.  LCM of 7 and 3 = 21

Number of peaches removed from Basket Y
= ¹⁰₁₀₀ x 49 u
= 4.9 u

Number of peaches left in Basket Y
= 49 u - 4.9 u
= 44.1 u

Number of more peaches in Basket Y than Basket X
= 44.1 u - 15 u
= 29.1 u

29.1 u = 1455

1 u = 1455 ÷ 29.1 = 50

	Basket W	Basket X	Basket Y
Before	6 u	15 u	44.1 u
Change		- 9 u	+ 9 u
After	6 u	6 u	53.1 u

Number of peaches to be transferred from Basket X to Basket Y so that the number of peaches in Basket W would be the same as Basket X.
= 15 u - 6 u
= 9 u
= 9 x 50
= 450

Answer(s): 450