Peter walks from point P to Q and back again, a total of 2 miles. If he ave. 4mph frm P to Q and 5mph on retur

Question

Peter walks from point P to Q and back again, a total of 2 miles. If he averages 4mph frm P to Q and 5mph on return trip, what was his average walking speed for the entire trip?

A) 2 2/9
B)4
C)4 4/9
D)5

The book says its C and I don't know why?

Answer 1

From P to Q it's 1 mile since the total is 2 miles.

S = D / T
D = S x T
T = D / S

The time from P to Q:
T = 1 / 4 hour.

The time from Q to P:
T = 1 / 5

So the total time is 1/4 + 1/5 = 9 / 20 hour

Now the total average speed from P to P again is
S = D / T
S = 2 / (9 / 20) = 2 x 20 / 9 = 40 / 9 = 4 and 4/9

Hope that helps!

Answer 2

I hope you know how to figure area, if not forget I exist. If so you know area of a rectangle is base times height. Distance is a lot like area. Height is a lot like speed and base (or width) is a lot like time. The first rectangle is 4 mph high and 1/2 hour long.

how do I know base*height=area
base*4miles/hours=2miles
don't be confused when I write miles per hour as miles divided by hour. multibly by hours/4miles on both sides 1/2 hour.

The return trip area is 2miles. How do I know you told me 2 miles up 2 miles back the height is 5mph u told me, the question is how wide is the rectangle measured in time.
b*h=a or time*5mph=2m or time=2/5 hours. The trip took 1/2+2/5 hours=5/10+4/10 hours=9/10 hour and covered 4 miles. time*speed=distance
(9/10) *speed = 4 total miles same as speed =40miles/9hours reduces to 4 4/9 mph

Answer 3

Remember the formulas

rate = distance/time

r = average rate
d = one way distance
2d = round trip distance
t = total time

time going = d/4
time returning = d/5
t = d/4 + d/5

r = 2d/[d/4 + d/5] = 2/[1/4 + 1/5] = 2*20/(5 + 4) = 40/9 = 4 4/9

Answer 4

60/27*2=2 2/9