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BangOn
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?
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Another way - less than 2 mins
Take Total distance = 100
1) y% of 100 at 50 mi/hr = distance is y miles -- So its y miles at 50 mi/hr
2) remaining (100-y) miles at 70 mi/hr

Avg spd = Total Dist/Total Time = 100/(y/50 + (100-y)/70) -> leads to D.(in case there are more choices with y in deno)
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HiteshPunjabi
Another way - less than 2 mins
Take Total distance = 100
1) y% of 100 at 50 mi/hr = distance is y miles -- So its y miles at 50 mi/hr
2) remaining (100-y) miles at 70 mi/hr

Avg spd = Total Dist/Total Time = 100/(y/50 + (100-y)/70) -> leads to D.(in case there are more choices with y in deno)

Check here: on-his-trip-to-alaska-joe-traveled-y-percent-of-the-total-141661.html#p1140444
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i just plugged in some random nos-

say Total distance=100
y=30%
Then ball parked to find that D is the ans.
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rajathpanta
i just plugged in some random nos-

say Total distance=100
y=30%
Then ball parked to find that D is the ans.

If you plug y=0 or y=100, then you won't need to plug anything for the distance. Check here: on-his-trip-to-alaska-joe-traveled-y-percent-of-the-total-141661.html#p1138511
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rajathpanta
i just plugged in some random nos-

say Total distance=100
y=30%
Then ball parked to find that D is the ans.

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)
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I recently encountered one such questions and couldn't come up with the logic to start the solution.
After solving a bunch of these, the quickest method (at least for me) is to calculate the individual times and then divide the total distance (Which I assume 100 in most cases) with the total time. A little bit of calculation but it works better than plugging in numbers for me.
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