There were some grapefruits in 3 bags, R, S and T. 25% of the number of grapefruits in Bag R was equal to 10% of the number of grapefruits in Bag S. The number of grapefruits in Bag T was 70% of the total number of grapefruits. After Ian removed 20% of the grapefruits in Bag T, there were 726 more grapefruits in Bag T than in Bag S. In the end, how many grapefruits should be transferred from Bag S to Bag T so that the number of grapefruits in Bag R would be the same as Bag S?

Bag R	Bag S	Bag T
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Bag R	Bag S	Bag T
Before	6 u	15 u	49 u
Change			- 9.8 u
After	6 u	15 u	39.2 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag R = ¹₁₀ Bag S

Bag R : Bag S
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of grapefruits in Bag R and Bag S is the combined repeated identity.  Make the total number of grapefruits in Box R and Bag S the same.  LCM of 7 and 3 = 21

Number of grapefruits removed from Bag T
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of grapefruits left in Bag T
= 49 u - 9.8 u
= 39.2 u

Number of more grapefruits in Bag T than Bag S
= 39.2 u - 15 u
= 24.2 u

24.2 u = 726

1 u = 726 ÷ 24.2 = 30

	Bag R	Bag S	Bag T
Before	6 u	15 u	39.2 u
Change		- 9 u	+ 9 u
After	6 u	6 u	48.2 u

Number of grapefruits to be transferred from Bag S to Bag T so that the number of grapefruits in Bag R would be the same as Bag S.
= 15 u - 6 u
= 9 u
= 9 x 30
= 270

Answer(s): 270