Jacob drove from Town A to Town B at an average rate of

M7MBA 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? $$\left(A\right)\ \ \frac{x+y+z}{3}$$ $$\left(B\right)\ \ \frac{3xyz}{xy+yz+zx}$$ $$\left(C\right)\ \ \frac{xyz}{x+y+z}$$ $$\left(D\right)\ \ \frac{xy+yz+zx}{x+y+z}$$ $$\left(E\right)\ \ \frac{3\left(x+y+z\right)}{xyz}$$ The Oa is the option B.

To solve, we can use the formula for average rate:

average rate = total distance/total time

We can let the 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)]

Canceling d, we are left with:

3xyz/(yz + xz + xy)

Answer: B

M7MBA wrote: ↑

Sun May 06, 2018 12:06 am

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? $$\left(A\right)\ \ \frac{x+y+z}{3}$$ $$\left(B\right)\ \ \frac{3xyz}{xy+yz+zx}$$ $$\left(C\right)\ \ \frac{xyz}{x+y+z}$$ $$\left(D\right)\ \ \frac{xy+yz+zx}{x+y+z}$$ $$\left(E\right)\ \ \frac{3\left(x+y+z\right)}{xyz}$$ The Oa is the option B.

May someone helps me, please? I don't know how to solve this PS question. I'd really appreciate any help.

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 B
So, total distance traveled = 3d

Time = distance/speed
time of 1st journey = d/x
time of 2nd journey = d/y
time of 3rd journey = d/z

Total time = d/x + d/y + dz
To simplify, rewrite with common denominator: dyz/xyz + dxz/xyz + dxy/xyz
So, 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