Ekland
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?
\(\frac{(x+y+z)}{3}\)
\(\frac{3xyz}{(xy+yz+zx)}\)
\(\frac{xyz}{(x+y+z)}\)
\(\frac{(xy+yz+zx)}{(x+y+z)}\)
\(\frac{3(x+y+z)}{xyz}\)
To solve, we can use the formula for average rate:
average rate = total distance/total time
We can let distance from Town A to Town B (or vice versa) = d, and since he went from A to B, then from B to A, and then from A to B, his total distance traveled was 3d. Recall that time = distance/rate, so the time to get from Town A to Town B = d/x, the time it takes to get from Town B back to Town A = d/y, and the time it takes to go back to Town B = d/z. Thus:
average rate = 3d/(d/x + d/y + d/z)
To combine the three fractions in the denominator, we use the common denominator xyz:
average rate = 3d/(yzd/xyz + xzd/xyx + xyd/xyz)
average rate = 3d/[(yzd + xzd + xyd)/xyz]
average rate = 3d * xyz/(yzd + xzd + xyd)
average rate = 3d * xyz/[d(yz + xz + xy)]
The ds cancel and we are left with:
3xyz/(yz + xz + xy)
Answer: B