Another Challenge Distance Problem

[Abdulla]

Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour and the remainder at 50 miles per hour. If Martha's average speed for the entire trip is represented as a fraction in its reduced form, in terms of x, which of the following is the numerator?

(A) 110
(B) 300
(C) 1,100
(D) 3,000
(E) 30,000

OA is E

[raleigh]

This is a problem where it helps to plug in a value. Since the first part of the trip is x percent of the distance, choosing the distance of the trip to be 100 miles means that the distance traveled at the first rate will be x.

So let D = 100 miles.

Then she traveled x miles at 50 miles per hour. D = RT gives us that the time of this portion of the trip is x/50 hours.

The second part of the trip was 100-x miles at 60 miles per hour. D = RT gives us that this portion of the trip took (100-x)/60 hours.

Average rate is distance/time. The distance is 100 miles, and the time is x/50 + (100-x)/60.

Let's work out the denominator first. x/50 + (100-x)/60 = [6x + 5(100-x)]/300 = (500-x)/300.

So the average rate is 100/[(500-x)/300] = 100*300/(500-x) = 30000/(500-x)

So the numerator of the average rate is 30,000. The trick is to plug in the appropriate distance. We choose 100 because they mention x percent of the total distance or (x/100)*distance.

[Abdulla]

This is a problem where it helps to plug in a value. Since the first part of the trip is x percent of the distance, choosing the distance of the trip to be 100 miles means that the distance traveled at the first rate will be x.

So let D = 100 miles.

Then she traveled x miles at 50 miles per hour. D = RT gives us that the time of this portion of the trip is x/50 hours.

The second part of the trip was 100-x miles at 60 miles per hour. D = RT gives us that this portion of the trip took (100-x)/60 hours.

Average rate is distance/time. The distance is 100 miles, and the time is x/50 + (100-x)/60.

Let's work out the denominator first. x/50 + (100-x)/60 = [6x + 5(100-x)]/300 = (500-x)/300. you missed one zero

So the average rate is 100/[(500-x)/300] = 100*300/(500-x) = 30000/(500-x)

So the numerator of the average rate is 30,000. The trick is to plug in the appropriate distance. We choose 100 because they mention x percent of the total distance or (x/100)*distance.

Thanks raleigh, but I think you made some mistakes in the process, do it again.

Distance= 100
Time 1 = x/60 this is the first difference btw me and your approach.
Time 2= 100-x / 50
Ave = total distance / total time
Ave = 100/ x/60 + 100-x/50 = 100/ 50x + 6000 - 60x / 3000 this is the second difference.

Ave = 100/ 6000 - 10x / 3000 = 100 * 3000 / 6000 - 10x
= 300,000/ 6000 - 10x ###

so my answer will be 300,000 which is not correct .. why ?

[raleigh]

Ave = total distance / total time
Ave = 100/ x/60 + 100-x/50 = 100/ 50x + 6000 - 60x / 3000 this is the second difference.

Your problem that you did not use the definition of average rate correctly. You added the average rate for each segment of the trip.

The average rate is total distance / total time as you said. The total distance is x + (100-x) = 100.

The total time is the sum times for each segment: x/60 + (100-x)/50. This quantity is the denominator.

The ratio should be 100/[x/60 + (100-x)/50]

[Abdulla]

Ave = total distance / total time
Ave = 100/ x/60 + 100-x/50 = 100/ 50x + 6000 - 60x / 3000 this is the second difference.

Your problem that you did not use the definition of average rate correctly. You added the average rate for each segment of the trip.

The average rate is total distance / total time as you said. The total distance is x + (100-x) = 100.

The total time is the sum times for each segment: x/60 + (100-x)/50. This quantity is the denominator.

The ratio should be 100/[x/60 + (100-x)/50]

It's similar to mine bro, the only thing that I forgot to put is the parentheses.

What about now :

Ave = 100/ [x/60 + (100-x)/50] = 100/ [50x + 6000 - 60x] / 3000

Ave = 100/ [6000 - 10x] / 3000 = 100 * 3000 /[6000 - 10x]
= 300,000/ [6000 - 10x] ###

[raleigh]

Factor a ten out of the denominator and it will cancel leaving you with 30,000 in the numerator. The problem asks for the ratio in reduced form. Otherwise, you could multiply by 10/10 as many times as possible to inflate the numerator.

[Abdulla]

Got it bro.. Thank you

[raleigh]

A big problem of mine isn't carefully reading the whole problem and picking out all the information. We're pressed for time that we try to get to our calculations as fast as possible. I'm trying to slow down when I first look at the problem so that way I don't miss anything. They always seem to have incorrect answers corresponding to those small bits that are missed. I'm surprised that 300,000 isn't an answer.