There were some guavas in 3 baskets, A, B and C. 20% of the number of guavas in Basket A was equal to 10% of the number of guavas in Basket B. The number of guavas in Basket C was 60% of the total number of guavas. After Owen removed 10% of the guavas in Basket C, there were 41 more guavas in Basket C than in Basket B. In the end, how many guavas should be transferred from Basket B to Basket C so that the number of guavas in Basket A would be the same as Basket B?

Basket A	Basket B	Basket C
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Basket A	Basket B	Basket C
Before	2 u	4 u	9 u
Change			- 0.9 u
After	2 u	4 u	8.1 u

20% = ²⁰₁₀₀ = ¹₅

10% = ¹⁰₁₀₀ = ¹₁₀

¹₅ Basket A = ¹₁₀ Basket B

Basket A : Basket B
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of guavas in Basket A and Basket B is the combined repeated identity.  Make the total number of guavas in Box A and Basket B the same.  LCM of 3 and 2 = 6

Number of guavas removed from Basket C
= ¹⁰₁₀₀ x 9 u
= 0.9 u

Number of guavas left in Basket C
= 9 u - 0.9 u
= 8.1 u

Number of more guavas in Basket C than Basket B
= 8.1 u - 4 u
= 4.1 u

4.1 u = 41

1 u = 41 ÷ 4.1 = 10

	Basket A	Basket B	Basket C
Before	2 u	4 u	8.1 u
Change		- 2 u	+ 2 u
After	2 u	2 u	10.1 u

Number of guavas to be transferred from Basket B to Basket C so that the number of guavas in Basket A would be the same as Basket B.
= 4 u - 2 u
= 2 u
= 2 x 10
= 20

Answer(s): 20