1st leg: t1 = 113/50
2nd leg: t2 = 113/60

Total time = t1 + t2 = 113/50 + 113/60

Average speed for the round trip = Total distance/Total time = (113 + 113)/(113/50 + 113/60) = 54.5 mph < 55 mph.

So Ans: B

correct: B
route = 113 miles (50 miles / hour) 113/50 hours
return_route = 113 miles (60 miles /hour) 113 / 60 hours
his average speed is: total miles passed / total time passed = (113 + 113)/ (113/50 + 113/60) = 226 * 300/ 1243 = 54.4
average_speed = 54.4 miles per hour
so B is bigger than A

Quick Solution.

Since Tavi spent more time driving at 50 mph than he did at 60 mph. His average speed must be closer to 50 than to 60. His average speed cannot be 55 since the amount of time he spent driving at each speed is different so. if he had spent the same amount of time at each speed then his average for the entire trip would have been 55 mph. So, here Qty B is greater than A.

113/50=2.26 not 2.6.

If the distance is same and speeds are different then you can use this easy formula to compute avg speed:
Avg Speed = 2ab/a+b
where a and b are speeds.
2(50)(60)/50+60
Answer is 54.5 so Quantity B is greater.

We essentially need average time to get average speed

113/50 = 2.26 (t1)
113/60 = 1.8 (t2)

=1.8 + 2.26 = 4.14
Aveg time = 4.14/2 = 2.07

Aveg Speed = 113/2.07 = 54.5
Thus its lesser than 55, ans B

Kudos please!

Is the OA correct?
113/50= 2.26
113/60= 1.88

Total time is 4.14
226 divided by 4.14 is 54.58 which is less than 55

But if I take only 2.2 and 1.8 as the times the total time is 4
which when dividing 226 gives 56.5 which is more than 55

The OA is correct

His speed is between 50 and 60. If you average 50+60/2=55 and the answer would be C. But we cannot simply average the two because the return is faster than the first stretch, which is at 50 mph.

we can use the weighted average concept.

Being the speed between 50 and 60 but because the first stretch is slower we do know that his speed average is NOT 55 but smaller. It does not matter if it is 54.9 or 22.3.

It MUST be < 55.

B is the answer

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