There were some chikoos in 3 boxes, R, S and T. 20% of the number of chikoos in Box R was equal to 10% of the number of chikoos in Box S. The number of chikoos in Box T was 60% of the total number of chikoos. After Fred removed 10% of the chikoos in Box T, there were 82 more chikoos in Box T than in Box S. In the end, how many chikoos should be transferred from Box S to Box T so that the number of chikoos in Box R would be the same as Box S?

Box R	Box S	Box T
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Box R	Box S	Box T
Before	2 u	4 u	9 u
Change			- 0.9 u
After	2 u	4 u	8.1 u

20% = ²⁰₁₀₀ = ¹₅

10% = ¹⁰₁₀₀ = ¹₁₀

¹₅ Box R = ¹₁₀ Box S

Box R : Box S
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of chikoos in Box R and Box S is the combined repeated identity.  Make the total number of chikoos in Box R and Box S the same.  LCM of 3 and 2 = 6

Number of chikoos removed from Box T
= ¹⁰₁₀₀ x 9 u
= 0.9 u

Number of chikoos left in Box T
= 9 u - 0.9 u
= 8.1 u

Number of more chikoos in Box T than Box S
= 8.1 u - 4 u
= 4.1 u

4.1 u = 82

1 u = 82 ÷ 4.1 = 20

	Box R	Box S	Box T
Before	2 u	4 u	8.1 u
Change		- 2 u	+ 2 u
After	2 u	2 u	10.1 u

Number of chikoos to be transferred from Box S to Box T so that the number of chikoos in Box R would be the same as Box S.
= 4 u - 2 u
= 2 u
= 2 x 20
= 40

Answer(s): 40