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

Box G	Box H	Box J
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Box G	Box H	Box J
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 G = ¹₁₀ Box H

Box G : Box H
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

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

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

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

Number of more pears in Box J than Box H
= 7.2 u - 4 u
= 3.2 u

3.2 u = 128

1 u = 128 ÷ 3.2 = 40

	Box G	Box H	Box J
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 H to Box J so that the number of pears in Box G would be the same as Box H.
= 4 u - 2 u
= 2 u
= 2 x 40
= 80

Answer(s): 80