Vroom! Vroom! Vroom!

Let Y initially travel a distance $a$ on the bike with X, Then equating the time X takes to drop Y and reach till Z with the time Z takes to reach that spot, we get

$\frac x7=\frac a{56} +\frac{a-x}{56} \tag{1}$ $\therefore a=\frac{9x}2$ Then equating the total time Y takes to reach B and Z takes to reach B, we get $\frac{2.8}8 + \frac a{56} = \frac x7 + \frac{a+2.8-x}{56}$

This gives $x=2.4$ after substituting $a$ from equation $1$

If we were to write their travelled distance as an equation, where each had their distance respresented by their letter, it would look like this: $X = 56 \times T_{1} - 56 \times T_{2} + 56 \times T_{3}$ $Y = 56 \times T_{1} + 8 \times T_{2} + 8 \times T_{3}$ $Z = 7 \times T_{1} + 7 \times T_{2} + 56 \times T_{3}$ Where T 1 is the time when X is giving a lift to Y, T 2 is the time when X is travelling back to Z, hence explaining the negative 56 before it in the formula for X, and T 3 is the time when X is giving a lift to Z. Given that at the end of T 3 they are all at the same place, that means all equations equal eachother. Solving for the T's gives: $T_{1} = \frac{9}{7} \times T_{2}$ $T_{2} = \frac{3}{4} \times T_{3}$ We also know that the last bit Y travels is equal to 2.8, so: $8 \times (T_{2} + T{3}) = 2.8$ Solving for T_2 gives $T_{2} = 0,15 hr$ We want to know the first part of Z's equation, so: $x = 7 \times (T_{1} + T_{2}) = 7 \times \frac{18}{7} \times T_{2} = 2.4$ So: $10x = \boxed{24}$

>Well, Y walks 2.8km and with this $\frac{8km}{1 h}=\frac{2,8km}{x h}$ , we know that Y takes $\frac{7}{20}h$ to arrive in B.

> So, in $\frac{7}{20}h$ the X meets Z and arrives at B. X travels a distance "w" until meets with Z, now returning, travels "w" once more and, finally, to get to B, 2.8 miles more.

>Thus, X travels $2w + 2,8$ in the time $\frac{7}{20}h$ . Solving this: $\frac{56km}{1 h}=\frac{xkm}{ \frac{7}{20}h }$ we got that X travels $\frac{98}{5}km$ in the time $\frac{7}{20}h$ .

> So, solving $2w + 2,8 = \frac{98}{5}$ , $w = 8,4km$

>The last step is write the equations based on: $time = \frac{distance}{speed}$ at moment that X meets Z. The distance which Z walked is $x$ , and X travels $x + 16,8$ .

> For X: $t = \frac{x + 16,8}{56}$ and for Z: $t = \frac{x}{7}$ , both times are equals, so: $\frac{x}{7} = \frac{x + 16,8}{56}$ , $\boxed{x = \frac{12}{5}}$ .

>>>The exercise asks $10 \times x$ , so $10 \times \frac{12}{5} = \boxed{24}$

First of all let us assume that the total distance from A to B is 'd'. Let's consider the time taken by Y, First Y is given a lift for d-2.8/56 hr. Then Y walks to B at a constant rate of 8 km/hr. Hence the total time taken by Y to reach B is d-2.8/56 + 2.8/8 = d-2.8/56 + 0.35 Similarly for Z, we have d-x/56 + x/7 as total time. But Z and Y reach B at same time so d-2.8/56 + 0.35 = d-x/56 + x/7 Which on further simplification gives us the ans: $\boxed{24}$

This is how I did it!

Let us say initially $X$ gave a lift to $Y$ and traveled a distance of $d$ km. After which $Y$ travels the remaining distance, i.e. $2.8$ km by foot. So, the total distance between $A$ and $B$ is $d + 2.8$ km.

Let's calculate $d$ :

As $Y$ 's initial velocity for $d$ km is $56$ km/hr and then for $2.8$ km, his velocity is $8$ km/hr, we can write the time taken by $Y$ to travel from $A$ to $B$ as,

$t = \frac{d}{56} + \frac{2.8}{8}$

$t = \frac{d + 19.6}{56}$ .......... (i)

Now, consider $X$ 's trajectory. $X$ first travels for $d$ km to drop $Y$ , then travels back $d$ km to pick up $Z$ and finally travels the entire distance from $A$ to $B$ . The velocity is the same for the whole trajectory of $X$ . Since $X$ reaches at the same time to $B$ as $Y$ , the time at which $A$ does so is,

$t = \frac{d + d + length(AB)}{56}$

$t = \frac{d + d + (d + 2.8)}{56}$

$t = \frac{3d + 2.8}{56}$ .......... (ii)

Equating (i) and (ii), we have

$\frac{d + 19.6}{56} = \frac{3d + 2.8}{56}$

$\frac{d + 19.6}{56} = \frac{3d + 2.8}{56}$

$d + 19.6 = 3d + 2.8$

$2d = 16.8$

$d = 8.4$ km

So, the distance between $A$ and $B$ is $d + 2.8 = 8.4 + 2.8 = 11.2$ km.

And the time they took to travel from $A$ to $B$ is

$t = \frac{d + 19.6}{56} = \frac{8.4 + 19.6}{56} = \frac{1}{2}$ hour.

Now, consider $Z$ 's path. It is given that he walked $x$ km. So, the distance after which $X$ dropped him off is $11.2 - x$ km.

Hence,

$t = \frac{11.2 - x}{56} + \frac{x}{7}$

$\frac{1}{2} = \frac{11.2 - x}{56} + \frac{x}{7}$

$\frac{1}{2} = \frac{11.2 - x}{56} + \frac{8x}{56}$

$\frac{1}{2} = \frac{11.2 - x + 8x}{56}$

$\frac{1}{2} = \frac{11.2 + 7x}{56}$

$1 = \frac{11.2 + 7x}{28}$

$28 = 11.2 + 7x$

$16.8 = 7x$

$x = 2.4$

Hence, $10x = 10 \times 2.4 = 24$

That's the answer!

Let total distance beteen A and B be L.....walking distance of Z =x...walking distance of y=2.8... Now...if before walking y travelled a distance of (L-2.8) with X on his motor cycle.therefore total time =(L-2.8)/56 + 2.8/8...here 56 is the velocity of motor cycle and 8 is the walking velocity of y....thus for z total time=(L-x)/56 + x/7....then for X, he comes travels (L-2.8)km and comes again therefore 2(L-2.8) then then comes and takes Z at a distance (L-x) from A....and then travels remaining x distance .....therefore [2(L-2.8)+(L-x)+x]/56=> time taken by X ....now they reach at same time such equate this three equation and solve for x....

Let total distance = d kms Since Y has walked 2.8 kms , distance of lift of Y = ( d-2.8 ) kms Therefore time of travel of Y = time taken to walk distance of 2.8 kms + time taken during the lift with X Therefore time of travel of Y = 2.8/8+(d-2.8)/56 Time taken by Z = time taken to walk x kms + time taken with X on motor cycle to travel remaining (d-x) kms = x/7 + (d-x)/56 Sicne both the times are same ,. 3.5+(d-2.8)/56 = x/7 + (d-x)/56 Solving we get x = 2.4 Therefore 10x = 24

Since all three people arrived at the same time, we know that the time $T$ they took to reach B are equal. Let us consider Y's journey. At first, X gave him a ride at a speed of 56 km/h for a certain unknown distance $D$ . After that, Y walked 2.8 km at 8 km/h. With this information, we can construct an equation in T and D.

$T = \frac{D}{56} + \frac{2.8}{8}$

Let us now consider Z's motion. Z travelled for some distance $x$ while X and Y were busy cruising ahead. This distance $x$ was covered with a speed of 7 km/h.

Then, X, being the nice guy that he is, returned for Z on his motorbike. After picking up Z, X rode his motorbike all the way to B. This distance covered on the bike is given by $D + 2.8 - x)$ . (If you can't see why, I suggest you draw a diagram showing the distances covered). And of course, this distance is covered with a speed of 56 km/h. This allows us to construct an equation in $T$ , $D$ and $x$ .

$T = \frac{x}{7} + \frac{D + 2.8 - x}{56}$

At first glance, these equations don't seem sufficient to solve for $x$ (Three variables and only two equations). Yet, upon equating the right-hand sides of both equations:

$\frac{D + 19.6}{56} = \frac{8x + D + 2.8 - x}{56}$

This simplifies to $7x = 16.8$ , yielding $x = 2.4$ or $10x = 24$ , as the question asks.

first we find the time taken by Y to reach B, but in terms of x and the total distance,AB. Y walks 2.8 km, and rides (AB-2.8)km on the motorcycle. we know that time=distance/speed. then, it took Y (2.8/8) h to complete walking, and ((AB-2.8)/56) h on the motorcycle. hence, by adding the two duration and simplifying, the time taken by Y to reach B will be: t1= ((16.8+AB)/56) h . now , we do the same thing with Z. Z walks xkm and rides the motorcycle (AB-x)km. so , Z spends (x/7) h walking and ((AB-x)/56)h riding to reach B. so , by adding the two duration and simplifying, the time taken by Z to reach B will be: t2 = ((AB+7x)/56) h. now, since we know that the three persons reached B at the same time, we deduce that t1=t2. then by solving the equation we get the answer.

We let $k$ to be the total distance from $A$ to $B$ , and write the time traveled from $A$ to $B$ by $Y$ , in terms of $k$ . So we have

$\frac{k-2.8}{56} hrs + \frac{2.8}{8} hrs = \frac{2(k-2.8)+k}{56} hrs$

$\frac{k-2.8}{56} hrs + 0.35 hrs = \frac{3k-56}{56} hrs$

$\frac{2k - 2.8}{56} hrs = 0.35 hrs$

$(2k-2.8) hrs = 19.6 hrs$

$k = 11.2 km$

Then, we substitute $k=11.2$ into the RHS, we know $Y$ uses 30 minutes to go from $A$ to $B$ . After that, we find the length of journey $Z$ has walked.

$\frac{11.2-x}{56} hrs + \frac{x}{7} hrs = 30$

$\frac{11.2 +7x}{56} hrs = 30$

$\frac{1.6 +x}{8} hrs = 30$

$60(1.6 + x) = 240$

$1.6 + x = 4$

$x = 2.4 km$

$10x=\boxed{24}$