There were some starfruits in 3 baskets, B, C and D. 25% of the number of starfruits in Basket B was equal to 10% of the number of starfruits in Basket C. The number of starfruits in Basket D was 60% of the total number of starfruits. After Lee removed 20% of the starfruits in Basket D, there were 136 more starfruits in Basket D than in Basket C. In the end, how many starfruits should be transferred from Basket C to Basket D so that the number of starfruits 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			- 4.2 u
After	4 u	10 u	16.8 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 starfruits in Basket B and Basket C is the combined repeated identity.  Make the total number of starfruits in Box B and Basket C the same.  LCM of 7 and 2 = 14

Number of starfruits removed from Basket D
= ²⁰₁₀₀ x 21 u
= 4.2 u

Number of starfruits left in Basket D
= 21 u - 4.2 u
= 16.8 u

Number of more starfruits in Basket D than Basket C
= 16.8 u - 10 u
= 6.8 u

6.8 u = 136

1 u = 136 ÷ 6.8 = 20

	Basket B	Basket C	Basket D
Before	4 u	10 u	16.8 u
Change		- 6 u	+ 6 u
After	4 u	4 u	22.8 u

Number of starfruits to be transferred from Basket C to Basket D so that the number of starfruits in Basket B would be the same as Basket C.
= 10 u - 4 u
= 6 u
= 6 x 20
= 120

Answer(s): 120