Question

Denise has 6 hours to spend for an upcoming race. She completes her training by running full speed the distance of the race and walking back the same distance to cool down. If she runs at a speed of 7mph and walks back at a speed of 3 mph, how long should she plan to spend walking back?

Answer 1

To solve this problem,you need to use the formula.
d = rd (distance = rates x time)
She runs at a speed of 7 mph and walks at a speed of 3 mph.
Her distance running is d = 7tr where tr is the time she spends running Her distance walking isd = 3twwhere tw is the time she spends walking The distances are the same so7tr = 3tw We also know that the total time is 4 hourstr + tw = 4tr = 4-tw Substitute this value of tr in the first equation7tr = 3tw7(4-tw) = 3tw28-7tw = 3tw28 = 10tw2.8 = tw Denise will spend 2.8 hours (2 hours, 48 minutes) walking back and 1.2 hours (1 hour, 12 minutes running)i hope this helps:)

Answer 2

She walks back for 4.2 hours

Explanation:

Notice that we don't know the distance of the race (so let's name it "d").

She clearly would spend less time running (t1) the distance d at 7 mph, than when coming back walking (t2) at 3 mph.

But we know that the addition of both times must be 6 hours which is what she has to spend:

t1 + t2 = 6

Considering the equation that relates distance (D), time(t) , and speed (v) as:

v = D/t

we can solve for the times t1 and t2 in each trip (running vs walking){

t1 = d/7

t2 = d/3

then when we add these two equations term by term, we get:

t1 + t2 = d/7 + d/3

extract d common factor;

t1 + t2 = d (1/7 + 1/3)

replace t1 + t2 with the 6 hours:

6 = d (10/21)

multiply both sides by 21, and divide both sides by 10 to isolate d:

21 * 6 / 10 = d

d = 12.6 miles

Now knowing the distance she run and walked, we can find the time she spent walking back (t2)

t2 = 12.6 /3 = 4.2 hours