There were some peaches in 3 baskets, B, C and D. 25% of the number of peaches in Basket B was equal to 10% of the number of peaches in Basket C. The number of peaches in Basket D was 60% of the total number of peaches. After Billy removed 10% of the peaches in Basket D, there were 445 more peaches in Basket D than in Basket C. In the end, how many peaches should be transferred from Basket C to Basket D so that the number of peaches in Basket B would be the same as Basket C?

Basket B	Basket C	Basket D
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Basket B	Basket C	Basket D
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 Basket B = ¹₁₀ Basket C

Basket B : Basket C
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of peaches in Basket B and Basket C is the combined repeated identity.  Make the total number of peaches in Box B and Basket C the same.  LCM of 7 and 2 = 14

Number of peaches removed from Basket D
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of peaches left in Basket D
= 21 u - 2.1 u
= 18.9 u

Number of more peaches in Basket D than Basket C
= 18.9 u - 10 u
= 8.9 u

8.9 u = 445

1 u = 445 ÷ 8.9 = 50

	Basket B	Basket C	Basket D
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

Number of peaches to be transferred from Basket C to Basket D so that the number of peaches in Basket B would be the same as Basket C.
= 10 u - 4 u
= 6 u
= 6 x 50
= 300

Answer(s): 300