There were some mangoes in 3 boxes, D, E and F. 25% of the number of mangoes in Box D was equal to 10% of the number of mangoes in Box E. The number of mangoes in Box F was 70% of the total number of mangoes. After Japheth removed 20% of the mangoes in Box F, there were 1210 more mangoes in Box F than in Box E. In the end, how many mangoes should be transferred from Box E to Box F so that the number of mangoes in Box D would be the same as Box E?

Box D	Box E	Box F
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Box D	Box E	Box F
Before	6 u	15 u	49 u
Change			- 9.8 u
After	6 u	15 u	39.2 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Box D = ¹₁₀ Box E

Box D : Box E
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of mangoes in Box D and Box E is the combined repeated identity.  Make the total number of mangoes in Box D and Box E the same.  LCM of 7 and 3 = 21

Number of mangoes removed from Box F
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of mangoes left in Box F
= 49 u - 9.8 u
= 39.2 u

Number of more mangoes in Box F than Box E
= 39.2 u - 15 u
= 24.2 u

24.2 u = 1210

1 u = 1210 ÷ 24.2 = 50

	Box D	Box E	Box F
Before	6 u	15 u	39.2 u
Change		- 9 u	+ 9 u
After	6 u	6 u	48.2 u

Number of mangoes to be transferred from Box E to Box F so that the number of mangoes in Box D would be the same as Box E.
= 15 u - 6 u
= 9 u
= 9 x 50
= 450

Answer(s): 450