AAPL wrote:Peter traveled from home to school and returned along the same route in the evening at a speed of 6 mph. What was Peters average speed for the two way trip?

1) Peters average speed is 0.8 mph greater than his speed from home to school.
2) Peter traveled from home to school at 4 mph.

The OA is D.

I'm not sure about this question, I tried to solve it as follows

Avg. speed = total distance/ total time
Let speed from home to school = v, and distance from home to school = x
total distance = x + x = 2x
and total time = x/v + x/6
thus avg. speed = 2x/(x/v + x/6)
= 2/ ( 1/v + 1/6)
= 12v/6+v (1)

statement 1,
avg. speed = v+0.8
Now substitute the value of avg. speed = v+0.8, we have
v+0.8 = 12v/v+6
Solving this we have;
v^2 - 5.2v + 4.8 =0
(v-1.2)(v-4) =0
v= 1.2 or v = 4.
Since here two values of v are possible, (so correspondingly we will have two different values of average speed 2 and 4.8) hence st.1 alone is not sufficient

statement 2,
Here v = 4, we will simply substitute this value in equation 1 and will get avg. speed = 4.8 as our answer.

Hence answer should be B but the OA is different.

Please, can someone help? Thanks!