On his trip to Alaska, Joe traveled y percent of the total distance at an average speed of 50 miles per hour and the rest of the distance at an average speed of 70 miles per hour. What was Joe’s average speed in miles per hour for the whole trip in terms of y?


(A) 70(50 – y)

(B) (y + 60)/4

(C) 10,000 – 199y

(D) 17,500/(y + 250)

(E) y/2 + 35

If the first part is at 50 mile/h and the second part is 70 miles/h the overall speed is the toral distance / total time.

Pick a number for distance we can use 350. so the time for first part is 7 and for the rest is 5.

in the end we will have 26. something, but testing answer choices nothing comes up (

where I'm wrong

PS: I could use a bit of logic and come to the answer easier, but .............
_________________

As you need a quick logical answer let me try
Lety distance be X
so Time 1 = XY/100*50
Time 2 = X(100-Y)/100*70

Avg Speed = Total Distance / Total Time
= X/ Time 1 + Time 2

Ans D
reason is simple...X cancels and Y reamins in denominator
only 1 option has the same
_________________

I deducted the above within 30 secs but took another 3 mins to solve the whole mess to get to the ans choice.
As you rightly said if we notice Y remains in deno for only one choice, life becomes easier.

I tend to go the conventional way every time not the GMAT way

Any other tricks like this which could help save TIME?

Check here: on-his-trip-to-alaska-joe-traveled-y-percent-of-the-total-141661.html#p1140444
_________________

No need to assume the value of y. Assume total distance = 100. Therefore total time taken = (y/50) + (100-y)/70 = (y + 250)/175

Average speed = 100/( (y + 250)/175) = 175000/ (y + 250)