There were some pears in 3 boxes, M, N and P. 20% of the number of pears in Box M was equal to 10% of the number of pears in Box N. The number of pears in Box P was 60% of the total number of pears. After Daniel removed 20% of the pears in Box P, there were 128 more pears in Box P than in Box N. In the end, how many pears should be transferred from Box N to Box P so that the number of pears in Box M would be the same as Box N?

Box M	Box N	Box P
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Box M	Box N	Box P
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 Box M = ¹₁₀ Box N

Box M : Box N
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of pears in Box M and Box N is the combined repeated identity.  Make the total number of pears in Box M and Box N the same.  LCM of 3 and 2 = 6

Number of pears removed from Box P
= ²⁰₁₀₀ x 9 u
= 1.8 u

Number of pears left in Box P
= 9 u - 1.8 u
= 7.2 u

Number of more pears in Box P than Box N
= 7.2 u - 4 u
= 3.2 u

3.2 u = 128

1 u = 128 ÷ 3.2 = 40

	Box M	Box N	Box P
Before	2 u	4 u	7.2 u
Change		- 2 u	+ 2 u
After	2 u	2 u	9.2 u

Number of pears to be transferred from Box N to Box P so that the number of pears in Box M would be the same as Box N.
= 4 u - 2 u
= 2 u
= 2 x 40
= 80

Answer(s): 80