There were some mangoes in 3 boxes, G, H and J. 25% of the number of mangoes in Box G was equal to 10% of the number of mangoes in Box H. The number of mangoes in Box J was 60% of the total number of mangoes. After Paul removed 10% of the mangoes in Box J, there were 445 more mangoes in Box J than in Box H. In the end, how many mangoes should be transferred from Box H to Box J so that the number of mangoes in Box G would be the same as Box H?

Box G	Box H	Box J
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Box G	Box H	Box J
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 Box G = ¹₁₀ Box H

Box G : Box H
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

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

Number of mangoes removed from Box J
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of mangoes left in Box J
= 21 u - 2.1 u
= 18.9 u

Number of more mangoes in Box J than Box H
= 18.9 u - 10 u
= 8.9 u

8.9 u = 445

1 u = 445 ÷ 8.9 = 50

	Box G	Box H	Box J
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

Number of mangoes to be transferred from Box H to Box J so that the number of mangoes in Box G would be the same as Box H.
= 10 u - 4 u
= 6 u
= 6 x 50
= 300

Answer(s): 300