lalania1 wrote:

this question comes from Prep4Gmat app

Frederique traveled x/8 of the total distance of a trip at an average speed of 40 miles per hour, where 1=<x=<7. She traveled the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Frederique's average speed for the entire trip?

A) (180-x)/2

B) (x+60)/4

C) (300-x)/8

D) 800/(120+x)

E) 960/(x+16)

Dear

lalania1,

I'm happy to respond.

Here's a blog you might find germane:

GMAT Distance and Work: Rate FormulaHere's my solution.

There are two legs in the trip. Call the total distance D. In the first leg, she covers a distance of D*(x/8) at a speed of 40 mph, and in the second leg, she covers a distance of D*((8 - x)/8) at a speed of 60 mph.

We need to calculate the time in each leg. Since D = RT, we know that T = D/R.

\(T_1 = \frac{x}{8}*D \div 40 = \frac{xD}{8*40}\)

\(T_2 = \frac{(8-x)}{8}*D \div 60 = \frac{(8-x)D}{8*60}\)

Leave the denominators unmultiplied for the moment.

\(T_{total} = T_1 + T_2 = \frac{xD}{8*40} + \frac{(8-x)D}{8*60}\)

\(T_{total} = \frac{3xD}{8*120} + \frac{2(8-x)D}{8*120}\)

\(T_{total} = \frac{(16 + x)D}{8*120} = \frac{(16 + x)D}{960}\)

Now, to find the average velocity, divide the total distance D by the total time.

\(v_{average} = \frac{D}{T_{total}} = D \times \frac{960}{(16 + x)D} = \frac{960}{(16 + x)}\)

Answer =

(E)Does all this make sense?

Mike

_________________

Mike McGarry

Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)