Carcass wrote:
Jacob drove from Town A to Town B at an average rate of x miles per hour, then returned along the same route at y miles per hour. If he then drove back to Town B at z miles per hour along the same route, what was Jacob’s average rate of speed for the entire trip, in miles per hour?
A. \(\frac{(x+y+z)}{3}\)
B. \(\frac{3xyz}{(xy+yz+zx)}\)
C. \(\frac{xyz}{(x+y+z)}\)
D. \(\frac{(xy+yz+zx)}{(x+y+z)}\)
E. \(\frac{3(x+y+z)}{xyz}\)
Average speed = (total distance traveled)/(total travel time)
= (total distance)/(
time of 1st journey +
time of 2nd journey +
time of 3rd journey)
Let d = the distance between Town A and Town BSo, total distance traveled =
3dTime = distance/speed
time of 1st journey = d/x time of 2nd journey = d/ytime of 3rd journey = d/zTotal time =
d/x +
d/y +
dzTo simplify, rewrite with common denominator:
dyz/xyz +
dxz/xyz +
dxy/xyzSo, total time = (dyz + dxz + dxy)/xyz
Average speed = (total distance)/(total time)
=
3d/[(dyz + dxz + dxy)/xyz]
= (3dxyz)/(dyz + dxz + dxy)
Divide top and bottom by d to get: (3xyz)/(yz + xz + xy)
Answer: B
Cheers,
Brent