There were some lemons in 3 boxes, C, D and E. 25% of the number of lemons in Box C was equal to 10% of the number of lemons in Box D. The number of lemons in Box E was 70% of the total number of lemons. After Pierre removed 10% of the lemons in Box E, there were 1164 more lemons in Box E than in Box D. In the end, how many lemons should be transferred from Box D to Box E so that the number of lemons in Box C would be the same as Box D?

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

	Box C	Box D	Box E
Before	6 u	15 u	49 u
Change			- 4.9 u
After	6 u	15 u	44.1 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Box C = ¹₁₀ Box D

Box C : Box D
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

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

Number of lemons removed from Box E
= ¹⁰₁₀₀ x 49 u
= 4.9 u

Number of lemons left in Box E
= 49 u - 4.9 u
= 44.1 u

Number of more lemons in Box E than Box D
= 44.1 u - 15 u
= 29.1 u

29.1 u = 1164

1 u = 1164 ÷ 29.1 = 40

	Box C	Box D	Box E
Before	6 u	15 u	44.1 u
Change		- 9 u	+ 9 u
After	6 u	6 u	53.1 u

Number of lemons to be transferred from Box D to Box E so that the number of lemons in Box C would be the same as Box D.
= 15 u - 6 u
= 9 u
= 9 x 40
= 360

Answer(s): 360