There were some grapefruits in 3 bags, L, M and N. 25% of the number of grapefruits in Bag L was equal to 10% of the number of grapefruits in Bag M. The number of grapefruits in Bag N was 60% of the total number of grapefruits. After Liam removed 25% of the grapefruits in Bag N, there were 115 more grapefruits in Bag N than in Bag M. In the end, how many grapefruits should be transferred from Bag M to Bag N so that the number of grapefruits in Bag L would be the same as Bag M?

Bag L	Bag M	Bag N
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Bag L	Bag M	Bag N
Before	4 u	10 u	21 u
Change			- 5.25 u
After	4 u	10 u	15.75 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag L = ¹₁₀ Bag M

Bag L : Bag M
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of grapefruits in Bag L and Bag M is the combined repeated identity.  Make the total number of grapefruits in Box L and Bag M the same.  LCM of 7 and 2 = 14

Number of grapefruits removed from Bag N
= ²⁵₁₀₀ x 21 u
= 5.25 u

Number of grapefruits left in Bag N
= 21 u - 5.25 u
= 15.75 u

Number of more grapefruits in Bag N than Bag M
= 15.75 u - 10 u
= 5.75 u

5.75 u = 115

1 u = 115 ÷ 5.75 = 20

	Bag L	Bag M	Bag N
Before	4 u	10 u	15.75 u
Change		- 6 u	+ 6 u
After	4 u	4 u	21.75 u

Number of grapefruits to be transferred from Bag M to Bag N so that the number of grapefruits in Bag L would be the same as Bag M.
= 10 u - 4 u
= 6 u
= 6 x 20
= 120

Answer(s): 120