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Given: Odyssey runs from the Trojan Horse, which does not move, to the gates of Troy and back to the Trojan Horse by the same path.
Asked: Was his average speed less than 20 kilometres per hour for the entire journey ?

Let the distance between Trojan horse and Troy be D km and time taken from Trojan horse to Troy be t1 and time taken back to Trojan horse from Troy be t2.
Average speed for the entire journey = 2D/(t1+t2)
v1 = D/t1
v2 = D/t2

(1) Odyssey's average speed from the Trojan Horse to the gates of Troy was greater than 40 kilometres per hour.
v1 = D/t1 > 40 kmph
D > 40t1
Average speed for the entire journey > 80t1/(t1+t2)
If t2 = 0; Average speed for the entire journey > 80 kmh
But if t2 is infinite; Average speed for the entire journey = 0 kmh
NOT SUFFICIENT

(2) Odyssey's average speed from the gates of Troy to the Trojan Horse was less than 10 kilometres per hour.
v2 = D/t2 < 10 kmph
D < 10 t2
Average speed for the entire journey < 20t2/(t1+t2)
If t1=0; Average speed for the entire journey < 20 kmph
But if t2 is infinite; Average speed for the entire journey = 0 kmph
In both extreme case, Average speed for the entire journey < 20 kmph
SUFFICIENT

IMO B
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1: does not tell us anything about his speed coming back, insuff. He could run at 40kmph back and it would be, or 0.5kmph back and it wouldn't be.
2: this is sufficient to tell us he cannot run faster than 20kmph. If half the trip is spent running at half the desired speed, one cannot make up the difference in the second leg. For instance, say Troy is 20km away, he would need to complete the round trip in 2 hours to run 20kmph. However, at 10kmph, the first leg takes him 2 hours, meaning no matter how fast he returns, he cannot exceed an average of 20kmph. This is sufficient to tell us he did not average less than 20kmph.

IMO B.
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Average speed formula when distance is constant is 2/(1/x+1/y)

from statement 1 -> y>50
from statement 2 ->x<10

put statement 2 in formula- 20*y/(10+y) or simplify 20/(1+10/y)

which shows statement is sufficient irrespective of y

Hence B
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