There were some grapefruits in 3 baskets, X, Y and Z. 20% of the number of grapefruits in Basket X was equal to 10% of the number of grapefruits in Basket Y. The number of grapefruits in Basket Z was 60% of the total number of grapefruits. After Carl removed 10% of the grapefruits in Basket Z, there were 205 more grapefruits in Basket Z than in Basket Y. In the end, how many grapefruits should be transferred from Basket Y to Basket Z so that the number of grapefruits in Basket X would be the same as Basket Y?

Basket X	Basket Y	Basket Z
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Basket X	Basket Y	Basket Z
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 X = ¹₁₀ Basket Y

Basket X : Basket Y
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of grapefruits in Basket X and Basket Y is the combined repeated identity.  Make the total number of grapefruits in Box X and Basket Y the same.  LCM of 3 and 2 = 6

Number of grapefruits removed from Basket Z
= ¹⁰₁₀₀ x 9 u
= 0.9 u

Number of grapefruits left in Basket Z
= 9 u - 0.9 u
= 8.1 u

Number of more grapefruits in Basket Z than Basket Y
= 8.1 u - 4 u
= 4.1 u

4.1 u = 205

1 u = 205 ÷ 4.1 = 50

	Basket X	Basket Y	Basket Z
Before	2 u	4 u	8.1 u
Change		- 2 u	+ 2 u
After	2 u	2 u	10.1 u

Number of grapefruits to be transferred from Basket Y to Basket Z so that the number of grapefruits in Basket X would be the same as Basket Y.
= 4 u - 2 u
= 2 u
= 2 x 50
= 100

Answer(s): 100