Walk The Talk II !

Let the boy speed be = x. Let the speed of moving walkway be = y. Distance(d) = speed *time.

Both walking in same directions, d = 162(x+y)

Moving in opposite direction, d = 810(x-y)

Multiply first equation by 5 and add the two equation.

6d = x(1620)

d/x(time) = 1620/6 = 270 seconds

let my speed be s. The speed of the walkway be w.

1)Walking in the correct direction d = (s + w)(162 seconds)

2) Walking in the wrong direction d = (s – w)(810 seconds)

Multiply first equation by 5 and then add the two equations.

6d = (s) (1620 seconds)

d/s = 1620/6 = 270 seconds.

For those who like this.......

I tried this like this :

Let the distance be 'd'. Speed of walkway be 'y' m/s. Speed of the man be 'x' m/s.

So while going with the walkway total speed is (x+y) m/s. TIME taken = 162 sec. (given) SO distance(1) = 162*(x+y) m/s. ..............eq(1)

And while going against total speed is (x-y) m/s. NOTE : as speed of the man 'ld be more than speed of the walkway. TIME taken = 5 162 (given five times more) hence distance(2) = d 162*5(x-y) m/s.

As both distance(1) = distance(2) so, 162 (x+y) m/s = d 162 5(x-y) m/s x+y = 5(x-y) x+y = 5x - 5y 4x = 6y y = 2/3x. Substituting y = 2/3x in eq(1) , we get distance = 162 (x + 2/3x) distance = 162*(5/3x) distance = 270 m.

Let M=my speed, W=walkway speed

(M+W)=5(M-W)

M+W=5M-5W

6W=4M

W=4M/6

W=2/3M _ (equation 1)

Now let D=distance,

D=(W+M)162

Substitute equation 1 in to W:

D=(2/3M+M)162

$D=(5/3)M\times162$

$\boxed{D=270M}$

Let x be distance, v1 human speed, v2 belt speed x=(v1+v2) 162=(v1-v2) 5 162 => v1=2/3 v2 => x=v2*270

It took 270secs

I kinda worked my way the hardest one, but it kinda solved the first part of the problem too.

Let the guy speed be Sg and the walkway be Sw. We have that the relative total speed of the system, when the guy is walking towards, is S = Sg + Sw, and when the guy is walking in the opposite way, is S = Sq - Sw.

Given that the general formula for speed is S = D / T where S is speed, D is distance and T is time, walking towards gives us the formula Sg + Sw = D / 162, and walking backwards gives us the formula Sg - Sw = D / 810.

Working with the walkway speed only, we have that (D/162) - Sw = (D/810) + Sw. Solving this equation, we get that our constant D = 405Sw.

Switching the speed formula to isolate the distance, we get D = T * S, which is exactly what we have there, D = 405 * Sw, so, the time the walkway takes to move that distance D is 405s. (Part 1 solved.)

Replacing it in the speed formula, we have that Sg + Sw = (405 * Sw) / 162. Solving, we reach Sg = 1,5 * Sw.

Ta-da! Now we get to the point when we know the ratio between the speeds and we already know that Tw = 405, so, we divide: Tg = 405 / 1,5 = 270. (Part 2 solved.)