There were some pomegranates in 3 boxes, D, E and F. 25% of the number of pomegranates in Box D was equal to 10% of the number of pomegranates in Box E. The number of pomegranates in Box F was 60% of the total number of pomegranates. After Daniel removed 20% of the pomegranates in Box F, there were 136 more pomegranates in Box F than in Box E. In the end, how many pomegranates should be transferred from Box E to Box F so that the number of pomegranates in Box D would be the same as Box E?

Box D	Box E	Box F
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Box D	Box E	Box F
Before	4 u	10 u	21 u
Change			- 4.2 u
After	4 u	10 u	16.8 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Box D = ¹₁₀ Box E

Box D : Box E
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of pomegranates in Box D and Box E is the combined repeated identity.  Make the total number of pomegranates in Box D and Box E the same.  LCM of 7 and 2 = 14

Number of pomegranates removed from Box F
= ²⁰₁₀₀ x 21 u
= 4.2 u

Number of pomegranates left in Box F
= 21 u - 4.2 u
= 16.8 u

Number of more pomegranates in Box F than Box E
= 16.8 u - 10 u
= 6.8 u

6.8 u = 136

1 u = 136 ÷ 6.8 = 20

	Box D	Box E	Box F
Before	4 u	10 u	16.8 u
Change		- 6 u	+ 6 u
After	4 u	4 u	22.8 u

Number of pomegranates to be transferred from Box E to Box F so that the number of pomegranates in Box D would be the same as Box E.
= 10 u - 4 u
= 6 u
= 6 x 20
= 120

Answer(s): 120