There were some kiwis in 3 bags, P, Q and R. 20% of the number of kiwis in Bag P was equal to 10% of the number of kiwis in Bag Q. The number of kiwis in Bag R was 60% of the total number of kiwis. After Reggie removed 20% of the kiwis in Bag R, there were 160 more kiwis in Bag R than in Bag Q. In the end, how many kiwis should be transferred from Bag Q to Bag R so that the number of kiwis in Bag P would be the same as Bag Q?

Bag P	Bag Q	Bag R
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Bag P	Bag Q	Bag R
Before	2 u	4 u	9 u
Change			- 1.8 u
After	2 u	4 u	7.2 u

20% = ²⁰₁₀₀ = ¹₅

10% = ¹⁰₁₀₀ = ¹₁₀

¹₅ Bag P = ¹₁₀ Bag Q

Bag P : Bag Q
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of kiwis in Bag P and Bag Q is the combined repeated identity.  Make the total number of kiwis in Box P and Bag Q the same.  LCM of 3 and 2 = 6

Number of kiwis removed from Bag R
= ²⁰₁₀₀ x 9 u
= 1.8 u

Number of kiwis left in Bag R
= 9 u - 1.8 u
= 7.2 u

Number of more kiwis in Bag R than Bag Q
= 7.2 u - 4 u
= 3.2 u

3.2 u = 160

1 u = 160 ÷ 3.2 = 50

	Bag P	Bag Q	Bag R
Before	2 u	4 u	7.2 u
Change		- 2 u	+ 2 u
After	2 u	2 u	9.2 u

Number of kiwis to be transferred from Bag Q to Bag R so that the number of kiwis in Bag P would be the same as Bag Q.
= 4 u - 2 u
= 2 u
= 2 x 50
= 100

Answer(s): 100