There were some dragonfruits in 3 baskets, R, S and T. 25% of the number of dragonfruits in Basket R was equal to 10% of the number of dragonfruits in Basket S. The number of dragonfruits in Basket T was 60% of the total number of dragonfruits. After Seth removed 20% of the dragonfruits in Basket T, there were 136 more dragonfruits in Basket T than in Basket S. In the end, how many dragonfruits should be transferred from Basket S to Basket T so that the number of dragonfruits in Basket R would be the same as Basket S?

Basket R	Basket S	Basket T
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Basket R	Basket S	Basket T
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 R = ¹₁₀ Basket S

Basket R : Basket S
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of dragonfruits in Basket R and Basket S is the combined repeated identity.  Make the total number of dragonfruits in Box R and Basket S the same.  LCM of 7 and 2 = 14

Number of dragonfruits removed from Basket T
= ²⁰₁₀₀ x 21 u
= 4.2 u

Number of dragonfruits left in Basket T
= 21 u - 4.2 u
= 16.8 u

Number of more dragonfruits in Basket T than Basket S
= 16.8 u - 10 u
= 6.8 u

6.8 u = 136

1 u = 136 ÷ 6.8 = 20

	Basket R	Basket S	Basket T
Before	4 u	10 u	16.8 u
Change		- 6 u	+ 6 u
After	4 u	4 u	22.8 u

Number of dragonfruits to be transferred from Basket S to Basket T so that the number of dragonfruits in Basket R would be the same as Basket S.
= 10 u - 4 u
= 6 u
= 6 x 20
= 120

Answer(s): 120