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

Bag N	Bag P	Bag Q
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Bag N	Bag P	Bag Q
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 N = ¹₁₀ Bag P

Bag N : Bag P
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of starfruits in Bag N and Bag P is the combined repeated identity.  Make the total number of starfruits in Box N and Bag P the same.  LCM of 7 and 2 = 14

Number of starfruits removed from Bag Q
= ²⁵₁₀₀ x 21 u
= 5.25 u

Number of starfruits left in Bag Q
= 21 u - 5.25 u
= 15.75 u

Number of more starfruits in Bag Q than Bag P
= 15.75 u - 10 u
= 5.75 u

5.75 u = 115

1 u = 115 ÷ 5.75 = 20

	Bag N	Bag P	Bag Q
Before	4 u	10 u	15.75 u
Change		- 6 u	+ 6 u
After	4 u	4 u	21.75 u

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

Answer(s): 120