Problem Solving

VJesus12 wrote:Anthony covers a certain distance on a bike. Had he moved 3mph faster, he would have taken 40 Minutes less. Had he moved 2mph slower, he would have taken 40 minutes more. The distance (in miles) is -

A. 30
B. 35
C. 36
D. 37.5
E. 40

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