Simon travels between two cities

Let the total distance between cities be 2K miles.
Average speed of the trip = T miles/hour
Average speed for the first half of the trip = X miles/hour
Let, Average speed for the second half of the trip = Y miles/hour
Time taken to cover the first half of the trip = K/X
Time taken to cover the first half of the trip = K/Y

Average speed of the trip = Total distance/Total time taken = T = 2K/((K/Y)+(K/X))
T = 2K/((K/Y)+(K/X))
T = 2K/(K*((X+Y)/XY))
2XY = TX + TY
Y = TX/(2X-T)
Y = X /((2X/T)-1) (Dividing numerator and denominator by T)

IMO E

chieftang wrote:Simon traveled between two cities. For the first half of the trip, he drove at a constant rate of X miles per hour. Over the entire trip he averaged a rate of T miles per hour. What was his rate of travel for the second half of his trip in terms of X and T, assuming it was constant?

(A) T - (X/2)
(B) 2T - (X/2)
(C) 2TX / (X-T)
(D) TX / (2X-2T)
(E) X / ((2X/T)-1)


Source: Original 700 level question

Let the distance = 24 miles.
Let X = 4 miles per hour.
Time for the first half of the trip = d/r = 12/4 = 3 hours.
Let the rate for the second half of the trip = 12 miles per hour.
Time for the second half of the trip = d/r = 12/12 = 1 hour.
T = average speed for the whole trip = total distance/total time = 24/(3+1) = 6 miles per hour.

Since the question asks for the rate during the second half of the trip, our target is 12.
Now we plug X=4 and T=6 into the answers to which yields our target of 12.

Only answer choice E works:
X/((2X/T) - 1) = 4/((2*4)/6 - 1) = 4/(1/3) = 12.

The correct answer is E.

My solution:
This is a harmonic mean question because distances are equal.

Harmonic mean:
2XY/(X+Y) = T

In this question, we're asked to solve for Y, so...
2XY = TX+TY
Y(2X-T) = TX
Y = TX/(2X-T)
Y = X/((2X/T)-1)

My OA: E

(E) QED.