Anthony covers a certain distance on a bike. Had he moved 3mph faster-

(s + 3)*(t - 2/3) = st = Certain Distance
3t - (2/3)s = 2 .........(I)

(s - 2)*(t + 2/3) = st = Certain Distance
(2/3)s -2t = 4/3 .........(II)

Add (I) and (II) to get
t = 10/3
s = 12
Distance = 12*10/3 = 40 miles

hi abhimahna

seems something wrong with the above equations...
Please correct me if my understanding is wrong...

thanks
rohit

T1, T2 and T3 are the times, I mentioned.

Note that when someone is faster, time taken is less and when someone is slower time taken is more.

So, as per the question stem, I believe 1st statement is correct and I wrote the 2nd equation in reverse order. It is a typo. Edited the original solution.

Please let me know if you find any other flaw.

We can let the distance = d and his rate = r. Thus, we can create the equations (notice that 40 minutes = 2/3 hour):

d/(r + 3) = d/r - 2/3

and

d/(r - 2) = d/r + 2/3

Subtracting the first equation from the second, we have:

d/(r - 2) - d/(r + 3) = 4/3

Multiplying by 3(r - 2)(r + 3), we have:

3d(r + 3) - 3d(r - 2) = 4(r - 2)(r + 3)

3dr + 9d - 3dr + 6d = 4(r - 2)(r + 3)

15d = 4(r - 2)(r + 3)

d = 4(r - 2)(r + 3)/15

Substituting this back into the first equation, we have:

[4(r - 2)(r + 3)/15](r + 3) = [4(r - 2)(r + 3)/15]/r - 2/3

4(r - 2)/15 = 4(r - 2)(r + 3)/(15r) - 2/3

Multiplying by 15r, we have:

4r(r - 2) = 4(r - 2)(r + 3) - 10r

4r^2 - 8r = 4(r^2 + r - 6) - 10r

4r^2 - 8r = 4r^2 + 4r - 24 - 10r

-8r = -6r - 24

-2r = -24

r = 12

So d = 4(12 - 2)(12 + 3)/15= 4(10)(15)/15 = 40 miles.

Answer: E

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