Rate/Distance/Time question

Question

Hey,

I found this question through Khan Academy, it goes as follows:

"A jogger and a walker set out at 9am from the same point, headed in the same direction. The average speed of the jogger is 1 mph slower than twice the speed of the walker. In two hours, the jogger is 3 miles ahead of the walker. Find the rate of the jogger"

Can someone please explain how I'd go about setting this up, preferably in a "Rate * Time = Distance" chart.

Thanks in advance

EMPOWERgmatRichC · Answer

Hi erikvm,

This question relates the speed of the walker to that of the jogger, so we can use 1 variable to refer to both.

Speed of Walker: X miles/hour
Speed of Jogger: (2X-1) miles/hour

Now we can use the distance formula (we were told that the length of time is 2 hours for each person):

(Rate)(Time) = Distance

Walker: (X miles/hour)(2 hours) = 2X miles
Jogger: (2X-1 miles/hour)(2 hours) = (4X - 2) miles

We're also told that, after those 2 hours, the jogger is 3 miles ahead of the walker, so the difference in distance can be represented as.....

(4X - 2) - 2X = 3 miles

Now we simplify....

2X - 2 = 3
2X = 5
X = 2.5

This means that the speed of the Walker is 2.5 miles/hour and the speed of the Jogger is 2(2.5) - 1 = 4 miles/hour

With these numbers, we can actually 'double check' the math....

Walker: (2.5)(2) = 5 miles
Jogger: (4)(2) = 8 miles
Difference = 3 miles, which is a match for what we were told.

GMAT assassins aren't born, they're made,
Rich

chetan2u · Answer

hi  erikvm ,
as u want the answer in the format "Rate * Time = Distance"..
lets first look at the rate..
let walker(W) speed be x, then jogger(J) speed will be 2x-1.....
now lets see the time , we are talking here about dist in 2 hrs.. so time =2 hrs and distance = 3 miles..
the equation here becomes   *2=3....
2x-2=3... x=5/2...
therefore speed of W=5/2mph and J=4mph({2*5/2}-1)

MayankSatkar · Answer

Useful tricks in TSD ...

Apti-Tricks

Walking at 3/4 of his usual speeds, a man reaches his office 10 minutes late. Find his usual time to cover the distance between home and office.
Solution: Usual Time = (Late Time)/(1/(3/4) - 1) = 10/(4/3 - 1)

Running at 4/3 of his usual speeds, a man improves his timing by 10 minutes. Find his usual time to cover the distance.
Solution: Usual Time = (Improved Time)/(1 - 1/(4/3)) = 10/(1- 3/4)

Find more useful tips and tricks here - https://aptidude.com/trickbook

EgmatQuantExpert · Answer

Dear Erik In Distance-Speed problems, like in most Word Problems, if you're not so sure about how to go setting up the solution, a good first step is Visualization. Make the given information come alive in a diagram, and often, you'll be able to see (literally!) what your next step should be. :-D

Like, here is a visual representation of the information in this question: https://s17.postimg.org/qsp9pjvgv/TSD_Visualization.png Let's assume the Jogger's speed is J mph and the walker's speed is W mph. They both start from the same point at the same time. So, we show this common starting point with the dotted gray line. Now, 2 hours later, the jogger is 3 miles ahead of the walker. We've shown the snapshot of 2 hours later with the dotted black line. Now, what does the diagram tell you? (Distance traveled by Jogger in 2 hours) = (Distance traveled by Walker in 2 hours) + 3 Distance = Speed*Time So, the above equation can be written as: J *2 = W *2 + 3 . . . (1) This is equation 1 between the 2 unknowns. To find a unique value of j, we need another equation. And that comes from the fact that So, the key takeaway of our discussion: Representing the given information visually is a great first step in Word Problems Hope this helped!

- Japinder

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Rate/Distance/Time question

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