There were some starfruits in 3 boxes, X, Y and Z. 25% of the number of starfruits in Box X was equal to 10% of the number of starfruits in Box Y. The number of starfruits in Box Z was 70% of the total number of starfruits. After Lee removed 20% of the starfruits in Box Z, there were 242 more starfruits in Box Z than in Box Y. In the end, how many starfruits should be transferred from Box Y to Box Z so that the number of starfruits in Box X would be the same as Box Y?

Box X	Box Y	Box Z
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Box X	Box Y	Box Z
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 X = ¹₁₀ Box Y

Box X : Box Y
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of starfruits in Box X and Box Y is the combined repeated identity.  Make the total number of starfruits in Box X and Box Y the same.  LCM of 7 and 3 = 21

Number of starfruits removed from Box Z
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of starfruits left in Box Z
= 49 u - 9.8 u
= 39.2 u

Number of more starfruits in Box Z than Box Y
= 39.2 u - 15 u
= 24.2 u

24.2 u = 242

1 u = 242 ÷ 24.2 = 10

	Box X	Box Y	Box Z
Before	6 u	15 u	39.2 u
Change		- 9 u	+ 9 u
After	6 u	6 u	48.2 u

Number of starfruits to be transferred from Box Y to Box Z so that the number of starfruits in Box X would be the same as Box Y.
= 15 u - 6 u
= 9 u
= 9 x 10
= 90

Answer(s): 90