Bill travels from point A to point B at a constant speed of 120

Let the distance be x
then for returning part we have
if time = A units
A*20/2+ A*60/2 = x (S*T=D)
10A+30A= x
40A=x---- A= x/40-----(total time for return)


for up journey dist.= x (both side same distance) and speed = 120kmph
avg speed = total dist./total time

(x+x)/(x/120+ x/40)==2x* 120/ 4x
thus avg speed =60kmph

Ans B

let 8d=roundtrip distance
average speed=8d/(d/20+3d/60+4d/120)=60 kph
B

Another approach:

Working from the return trip while 'half the time traveling 20 kilometers per hour, and half the time traveling 60 kilometers per hour'

Let half time =1 hr.....Therefore

With 20 km/hr........Bill traveled distance= 20 km

With 60 km/hr........Bill traveled distance= 60 km

Total distance = 80 km

With 20 km/hr........Bill traveled time= 80/120 = 2/3 hr

Total distance = 2 *80= 180 km

Total time = 1+ 1+2/3= 2+2/3= 8/3 hr

Average speed = 2d/total time = 160 / (8/3) = 60 km/hr

Answer: B

Happy Easter, Brent

average speed whilst going to B =40km/hr
average constant speed whilst returning to A =120km/hr
for the same distance, speed inversely related to time t(a->b) : t(b->a)=1:3
average speed = total dis./total time =(40*3x+120x)/4x=60 km/hr -> B

Dear Brent,

I know the following rule but I do if it works, although it provides the answer.

If an object travels an Two equal distances, then the average speed will be slightly less than the mean of two speeds.

I applied same concept here

Mean= (120+20+60) =200/3 = 66.66
so it should less than 66.66

Answer is B

Any trick or problem do you see?

thanks

One foreseeable problem is the word "slightly" (as in "...the average speed will be slightly less than the mean of two speeds"
For example, if I travel from A to B at 60 miles per hour and then return to point A at 1 mile per hour, the MEAN of the two speeds is 30.5 miles per hour. However, the ACTUAL average speed = 120/61 miles per hour (i.e., about 1.97 miles per hour), which is A LOT less than the mean of 30.5 miles per hour.

There will also be issues with fact that you're dealing with 3 different speeds (and your rule applies to 2 speeds).

Cheers,
Brent

Let's assign a nice value to the distance between point A and point B.
Given the 3 speeds (120 kmh, 20 kmh and 60 kmh), it seems that a distance of 120 kilometers will work nicely.

Bill travels from point A to point B at a constant speed of 120 kilometers per hour.
Time = distance/rate
= 120/120 = 1 hour

On his return trip to point A, Bill spends half the time traveling 20 kilometers per hour, and half the time traveling 60 kilometers per hour.
Let x = the distance traveled at 20 kilometers per hour
So, 120 - x = the distance traveled at 60 kilometers per hour

Word equation: TIME traveled at 20 kmh = TIME traveled at 60 kmh
Time = distance/rate
So, we get: x/20 = (120-x)/60
Cross multiply: 60x = 20(120-x)
Expand: 60x = 2400 - 20x
Solve: x = 30
So, Bill drove 30 kilometers at 20 kilometers per hour, which means he drove the other 90 kilometers at 60 kilometers per hour.

TIME traveled at 20 kmh = 30/20 = 1.5 hours
Since the two travel times are EQUAL, we also know that Bill drove 1.5 hours at 60 kilometers per hour

So, the TOTAL DISTANCE = 120 + 120 = 240 kilometers
So, the TOTAL TRAVEL TIME = 1 + 1.5 + 1.5 = 4 hours

Average speed = (TOTAL DISTANCE)/(TOTAL TRAVEL TIME) = 240/4 = 60 kilometers per hour

Answer:
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Algebric approach

Let time taken be T1 and T2-----

Distance covered from A to B = 120 x T1

Distance covered from B to A = (T2/2)x20 + (T2/2)x60= 40 x T2

Since distance covered is same hence---
120 x T1 = 40 x T2
=> (T1/T2) = 1/3

Therefore average velocity = Total distance/ Total time

V= (120 T1 + 40 T2)/(T1 + T2)

Dividing both numerator and denominator by T2 and using (T1/T2)= 1/3

V= 60 kmph

I got A with the following equation:
Ave. Speed = Total distance/total time. Since Bill travels from A to B and then back to A, the total distance should be 2d. The total time is d/120 (first trip) + half the distance with 20 mph + halfe the distance with 60 mph => 2d/(d/120 + 0.5d/20 + 0.5d/60) which is 2d/(d/120 + 3d/120 + d/120) so i end up with 240d/5d, the d's cancel out and i get 42. Can you explain to me how you got to the 4d in the denominator instead of 5d?

Be careful. The question says "On his return trip to point A, Bill spends half the time traveling 20 kilometers per hour"
In your solution, Bill spends half the DISTANCE traveling 20 kilometers per hour"

Cheers,
Brent

[quote="Be careful. The question says "On his return trip to point A, Bill spends half the time traveling 20 kilometers per hour"
In your solution, Bill spends half the DISTANCE traveling 20 kilometers per hour"

Cheers,
Brent[/quote]
Thank you, I got it now!

Given: Bill travels from point A to point B at a constant speed of 120 kilometers per hour. Upon reaching point B, he immediately heads back to point A. On his return trip to point A, Bill spends half the time traveling 20 kilometers per hour, and half the time traveling 60 kilometers per hour.

Asked: What is Bill’s average speed, in kilometers per hour, for the entire roundtrip?

Let us assume distance from point A to point B = 120 kms

From point A to point B: -
Distance travelled = 120 kms
Speed = 120 km/hr
Time taken = 1 hour

From point B to point A: -
Average speed = (20t + 60t)/2t = 40 km/hr
Distance travelled = 120 kms
Time taken = 120 /40 = 3 hours

Average speed = 120 * 2 kms / 4 hours = 60 km/hr

IMO B

My approach

D S T
d 120 d/120
x 20 x/20
d-x 60 d-x/60

Given x/20=d-x/60 which gives d= 4x

Avg speed = Total distance/ Total time

which gives = 2d/ (d/120+x/20+d-x/60)
= 240d/3d+4x
= 240d/4d = 60.... since 4x = d