There were some peaches in 3 boxes, S, T and U. 25% of the number of peaches in Box S was equal to 10% of the number of peaches in Box T. The number of peaches in Box U was 60% of the total number of peaches. After Charlie removed 10% of the peaches in Box U, there were 89 more peaches in Box U than in Box T. In the end, how many peaches should be transferred from Box T to Box U so that the number of peaches in Box S would be the same as Box T?

Box S	Box T	Box U
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Box S	Box T	Box U
Before	4 u	10 u	21 u
Change			- 2.1 u
After	4 u	10 u	18.9 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Box S = ¹₁₀ Box T

Box S : Box T
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of peaches in Box S and Box T is the combined repeated identity.  Make the total number of peaches in Box S and Box T the same.  LCM of 7 and 2 = 14

Number of peaches removed from Box U
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of peaches left in Box U
= 21 u - 2.1 u
= 18.9 u

Number of more peaches in Box U than Box T
= 18.9 u - 10 u
= 8.9 u

8.9 u = 89

1 u = 89 ÷ 8.9 = 10

	Box S	Box T	Box U
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

Number of peaches to be transferred from Box T to Box U so that the number of peaches in Box S would be the same as Box T.
= 10 u - 4 u
= 6 u
= 6 x 10
= 60

Answer(s): 60