There were some mangoes in 3 baskets, Q, R and S. 20% of the number of mangoes in Basket Q was equal to 10% of the number of mangoes in Basket R. The number of mangoes in Basket S was 60% of the total number of mangoes. After Flynn removed 20% of the mangoes in Basket S, there were 32 more mangoes in Basket S than in Basket R. In the end, how many mangoes should be transferred from Basket R to Basket S so that the number of mangoes in Basket Q would be the same as Basket R?

Basket Q	Basket R	Basket S
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Basket Q	Basket R	Basket S
Before	2 u	4 u	9 u
Change			- 1.8 u
After	2 u	4 u	7.2 u

20% = ²⁰₁₀₀ = ¹₅

10% = ¹⁰₁₀₀ = ¹₁₀

¹₅ Basket Q = ¹₁₀ Basket R

Basket Q : Basket R
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of mangoes in Basket Q and Basket R is the combined repeated identity.  Make the total number of mangoes in Box Q and Basket R the same.  LCM of 3 and 2 = 6

Number of mangoes removed from Basket S
= ²⁰₁₀₀ x 9 u
= 1.8 u

Number of mangoes left in Basket S
= 9 u - 1.8 u
= 7.2 u

Number of more mangoes in Basket S than Basket R
= 7.2 u - 4 u
= 3.2 u

3.2 u = 32

1 u = 32 ÷ 3.2 = 10

	Basket Q	Basket R	Basket S
Before	2 u	4 u	7.2 u
Change		- 2 u	+ 2 u
After	2 u	2 u	9.2 u

Number of mangoes to be transferred from Basket R to Basket S so that the number of mangoes in Basket Q would be the same as Basket R.
= 4 u - 2 u
= 2 u
= 2 x 10
= 20

Answer(s): 20