There were some kiwis in 3 bags, M, N and P. 25% of the number of kiwis in Bag M was equal to 10% of the number of kiwis in Bag N. The number of kiwis in Bag P was 60% of the total number of kiwis. After Ryan removed 10% of the kiwis in Bag P, there were 178 more kiwis in Bag P than in Bag N. In the end, how many kiwis should be transferred from Bag N to Bag P so that the number of kiwis in Bag M would be the same as Bag N?

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

	Bag M	Bag N	Bag P
Before	4 u	10 u	21 u
Change			- 2.1 u
After	4 u	10 u	18.9 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag M = ¹₁₀ Bag N

Bag M : Bag N
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

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

Number of kiwis removed from Bag P
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of kiwis left in Bag P
= 21 u - 2.1 u
= 18.9 u

Number of more kiwis in Bag P than Bag N
= 18.9 u - 10 u
= 8.9 u

8.9 u = 178

1 u = 178 ÷ 8.9 = 20

	Bag M	Bag N	Bag P
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

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

Answer(s): 120