There were some mangoes in 3 bags, P, Q and R. 25% of the number of mangoes in Bag P was equal to 10% of the number of mangoes in Bag Q. The number of mangoes in Bag R was 60% of the total number of mangoes. After Riordan removed 20% of the mangoes in Bag R, there were 68 more mangoes in Bag R than in Bag Q. In the end, how many mangoes should be transferred from Bag Q to Bag R so that the number of mangoes in Bag P would be the same as Bag Q?

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

	Bag P	Bag Q	Bag R
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 Bag P = ¹₁₀ Bag Q

Bag P : Bag Q
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

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

Number of mangoes removed from Bag R
= ²⁰₁₀₀ x 21 u
= 4.2 u

Number of mangoes left in Bag R
= 21 u - 4.2 u
= 16.8 u

Number of more mangoes in Bag R than Bag Q
= 16.8 u - 10 u
= 6.8 u

6.8 u = 68

1 u = 68 ÷ 6.8 = 10

	Bag P	Bag Q	Bag R
Before	4 u	10 u	16.8 u
Change		- 6 u	+ 6 u
After	4 u	4 u	22.8 u

Number of mangoes to be transferred from Bag Q to Bag R so that the number of mangoes in Bag P would be the same as Bag Q.
= 10 u - 4 u
= 6 u
= 6 x 10
= 60

Answer(s): 60