Barry walks from one end to the other of a 30-meter long

Say Barry's speed is \(b\) meter per second and walkaway speed is \(w\) meter per second, then as \(Speed=\frac{Distance}{Time}\) we'll have that:

\(b+w=\frac{30}{30}=1\);
\(b-w=\frac{30}{120}=\frac{1}{4}\);

Sum these two equations; \(2b=\frac{5}{4}\) --> \(b=\frac{5}{8}\).

\(Time=\frac{Distance}{Speed}=\frac{30}{(\frac{5}{8})}=48\).

Answer: A.

Hope it helps.
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If it is w-b, it means that speed of walkway is more than than his speed, then how will he reach the other end of the walkway.

Let Barry's speed be x m/s and walkway's speed by y m/s. We need to find 30/x.

If Barry walks in the direction of the moving walkway, the total speed is x + y. Time taken is 30/(x+y)
Thus, 30/(x+y) = 30

=> x + y = 1 ... (1)

If Barry walks against the moving walkway's direction, total speed is x-y. Time take is 30(x-y)
Thus, 30/(x-y) = 120

=> x - y = 1/4 ... (2)

Add (1) and (2),
=> 2x = 5/4
=> x = 5/8

therefore, 30/x = 48

Right answer is A.

We can let the rate of the walkway = w and Barry’s rate = r.

Since he walks from one end to the other of a 30-meter moving walkway at a constant rate in 30 seconds, assisted by the walkway:

w + r = 30/30

w + r = 1

He reverses direction and continues walking with the same speed, but this time it takes him 120 seconds because he is traveling against the direction of the moving walkway:

r - w = 30/120

r - w = 1/4

Adding the two equations together, we have:

2r = 1¼

2r = 5/4

r = (5/4)/2 = ⅝

Thus, if the walkway were not moving, it would take Barry 30/(5/8) = 240/5 = 48 seconds to walk its length.

Answer: A
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Yes, the speed of the person/boat etc will be higher than the speed of walkway/stream etc to ensure that the person reaches the other side even while moving against the motion. These are not cases of relative speed. They are cases of "effective speed" i.e. two speeds act on the same object and we need to find the effective speed of the object. Boats & streams, airplanes and wind, people and escalators/travellators etc are all cases of effective speed. The speed of one object acts on the other.

Relative speed involves two objects maintaining their own different speeds and we find the relative speed between them. One object's speed does not "act" on the other.

Check this video for when to use relative speed: https://youtu.be/wrYxeZ2WsEM

Let's try something else JFF ('Just for Fun')

B = Barry's Rate
W= Walkway's

So we have 30 (b+w) = 120 (b-w)
5w = 3b

D=30 (8/5b) = 48b

Time takes for Barry

t = D/R = 48b / b = 48 seconds

Answer: A

Hi Bunuel,
May I understand the thought process when you formed these 2 equations?
\(b+w=\frac{30}{30}=1\);
\(b-w=\frac{30}{120}=\frac{1}{4}\);

Thanks!

This was quite helpful. Thank you so much

How do we know that it is b-w? Why cant it be w-b. The question never says anything about the speeds of b or w individually.

Make sense, thanks.

We have 30 (a+b) = 30
a+b = 1

We also get that 120 (a-b) = 30
So a-b = 1

So we have that a = 5/8

So 30*8/5 = 48

Answer is A

Hope it helps!
Cheers

J

Do we actually need to know that the distance is 30 meters?

IF we learn that w=3b/5 we can find distance in terms of 'b' and then divide by rate 'b' to figure out that it will take Ben 48 seconds to reach the end of the walkway

Please advice
Cheers!
J

No we don't need this value of 30 meters since the distance remains the same while travelling up and down and thus, speeds are inversely proportional to the time taken. The information is superfluous.


Hope it helps!!!

s1 = 30/30 = 1 metre/sec
s2 = 30/120 = 0.25 metres/sec
now
x+y=1
x-y=0.25
or,
2x=1.25 ==> x = 0.625 metres/sec
and y = 0.375 metres/sec
time taken by barry = 30000/625=48 sec

So basically since barry is moving at a rate and its given that walkway is also moving,thus speed of the walkway must be considered.Right?

let b=B's speed
w=walkway speed
b+w=30 meters/30 sec=1 mps
b-w=30 meters/120 sec=1/4 mps
adding the two equations,
b=5/8 mps
30 meters/(5/8) mps=48 sec
A

As this is a Rates & Work problem, we should use a chart to keep track of the different legs of the journey. We will need a row for the journey walking with the walkway and one for the journey against.
The distance traveled for each leg of the journey was 30 meters. We also have the time each trip took. The only unknown is the rates for each leg of the trip. There are really two unknowns: The rate at which Barry walked, and the rate at which the walkway moved. Label the rate Barry walked b, and the rate at which the walkway moved w:


Distance = Rate × Time
With walkway 30 = b + w × 30
Against walkway 30 = b – w × 120

This allows us to create two equations:

30 = 30(b + w) ⇒ 30 = 30b + 30w
30 = 120(b – w) ⇒ 30 = 120b – 120w

We now have enough information to solve for b and w. The question asks for the time it would take Barry to walk the distance by himself, so we want to solve for b. We can solve for b with Combination. Multiply the first equation by 4 and add the equations together:

120 = 120b + 120w
+ 30 = 120b – 120w
150 = 240b

Now we can solve for b:


Now we have one final equation to solve. How long did it take Barry to walk 30 meters?

KarishmaB Bunuel

While reading the second half of the question we write b - w = 1/4. Aren't we assuming b > w? But after thinking for a while I realized that b > w is something we can infer since berry ultimately reaches the other end of the walkway. If w > b then he would never have completed the distance. Is this logic correct?

I understand that in this situation while completing the journey of 30m in 30sec the speed of the walkway and that of berry is added i.e. b + w (something similar to a boat traveling upstream) However, isn't it true that one object is moving at b m/s while the other at w m/s? In other words, don't we have two objects moving in the same direction and hence relative speed is b - w? OR is this thinking wrong since the "walkway" is not to be considered as an "object" moving in a direction BUT rather some additional "force" that is acting on the object and thus altering (in this case increasing) the speed of the original object?

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