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We set up an equation to model the situation:

Using $d=rt$ , we find that:

$d=(r+6)(t-4)=(r-6)(t+6)=rt$ $d=rt+6t-4r-24=rt-6t+6r-36=rt$

Cancelling the $rt$ terms we get:

$6t-4r-24=-6t+6r-36=0$

Which gives us the two equations $6t-4r-24=0$ and $-6t+6r-36=0$ .

Adding the two equations gives us $2r=60$ so $r=30$ . Subsituting this solution gives us $t=24$ .

Now we multiply these two (since $d=rt$ ) to get our answer of $\boxed{720}$ .

Let,

d - length of the journey in km,

S - speed of train in km/hr and

t - scheduled time of journey

Then,

$\frac{d}{(S+6)}$ = t − 4 -------- (1)

and

$\frac{d}{(S-6)}$ = t + 6 --------- (2)

Subtracting the equation (1) from equation(2) we get:

$\frac{d}{(S-6)}$ − $\frac{d}{(S+6)}$ = 10 -------- (iii)

Now , t = $\frac{d}{S}$

Substitute in equation (1) and solve for d and S

We get,

S= $\boxed{30}$ and d= $\boxed{720}$

Hence, the total distance of journey is $\boxed{720km}$

lets say distance covered=x, velocity=v, tome taken =t, therfore x=v t, from the given conditions: x=(v+6) (t-4), x=(v-6) (t+6), on solving we get 6t-4v=24 and 12t-10v=-12. on solving we get v=30 and t=24 tgherfore x=24 30=720

First of all we need to take out all the posible equation

x/y=k---first equation

next the description mention that x/y-4=k+6,x/y+6=k-6

so we solve the equation and we got

x=ky-4k+6y-24and x=ky+6k-6y-36

then we substitute the equation and we got 2k=60,k=30 6y=6(30)-36 y=30-6 y=24

from the first equation x=ky x=24*30=720

Let the initial speed be "x", and the length of the journey be "y". The initial time is y/x. Given: (x+6)(y/x-4)=y. Thus- y-4x+6y/x-24=y ; y=2/3*x^2+4x

Given: (x-6)(y/x+6)=y ; (x-6)((2/3 x^2+4x)/x+6)=2/3 x^2+4x 2/3 x^2+4x+6x-4x-24-36= 2/3 x^2+4x 2x-60=0 2x=60 x=30

y(journey)= 2/3 30^2+4 30= 600+120= 720

Let the uniform speed be s and the time takes(in hours) to be t, Then, Distance traveled will be s t. It has been given that if the speed is 6 more than the regular speed then the time taken is 4 hours less than the regular time. This means that if the speed is s+6 then the time is t-4. Since, Distance=Speed Time, S t=(s+6) (t-4)....[This is our first equation...] Similarly, when speed is s-6, the time taken is t+6. So, s t=(s-6) (t+6).....[This is our second equation] When we equate the two equations, we get s=30km/hr and t=24 hours. Thus distance=s t=24 30=720 km

let us take the original velocity of the train be u km/h and total distance travelled be x km so the original time(scheduled time) be t hours.When speed increased by 6km/h then our new speed is (u+6)km/h then the new time of travel is t1=x/(u+6) hrs this time is 4 hr less than sheduled time so we can form an equation: x/u - x/(u+6)=4............(1) similarly if the train travels 6km/h less than the original speed than the new speed is (u-6)km/h so the new time taken t2=x/(u-6) hrs according to the question it takes 6 hour more to travel the same distance so another equation can be formed: x/(u-6) -x/u=6...........(2) so solving equation (1)and(2) first we find u=30km/h and inserting this value in any one of the equation would give x=720 km. (Rather than just doing calculation i tried to focus on the concept i hope every one understand this) :)

Let the length of the journey be d km and the speed of train be S km/hr.

Then, d (S+6) =t−4 -------- (i) and d (S−6) =t+6 --------- (ii)

Subtracting the 1 equation from another we get: d (S−6) − d (S+6) =10 -------- (iii)

Now , t= d S

Substitute in equation (i) and solve for d and S We get S=30 d= 720 km

Lets say Distance= d, Time =t and Velocity=v d=v x t (1) At speed of 6Km/hr faster, time wud have been 4 hrs less so d = (v+6) x (t-4) (2) At speed of 6Km/hr slower, time wud have been 6 hrs more so d=(v-6) x ( t+6) (3) as distance remains the same so equating (1) and (2) we get 6t-4v=24 (4) and equating (1) and (3) we get v-t=6 (5) solving (4) and (5) we get v=30Km/hr and t=24 hrs So as d=v x t d=30 x 24 = 720 Km

let x be scheduled time, y be speed of train

When the train 6km/hr faster than the normal speed, it will 4 hours less than the scheduled time,so (x-4)hr ; (y+6)km/hr

When the train 6km/hr slower than the normal speed, it will 6 hours more than the scheduled time, so (x+6)hr ; (y-6)km/hr

From the question, if the train moves in uniform speed, x, then it reach to the destination on the scheduled time,y, so the distance is x hr * y km/hr=xy km

Since the distance is same, so (x-4)hr * (y+6)km/hr = (x+6)hr * (y-6)km/hr xy+6x-4y-24 = xy-6x+6y-36 6x-5y+6=0 --(i)

xy=xy+6x-4y-24 6x-4y-24=0 --(ii)

(ii) subtract (i) : y-30=0 y = 30 x = 24 distance = x*y = 24 * 30 = 720

xy = xy-6x+6-36 6x-6y+36=0 -- (iii)

(i) subtract (iii) : y-30 = 0 y = 30 Since y have only one value, so the x also have only one value, therefore only have one solution for the distance.

Schedule: speed = v and time = t Faster then schedule: speed = v + 6 and time = t - 4 Slower then schedule: speed = v - 6 and time = t + 6 Total distance = x = speed . time

so we get: v.t=x and (v+6).(t-4)=x and (v-6).(t+6)=x We solve t = 24 and v = 30 Now x = v.t = 30.24 = 720

After all the math on my paper, I saw that there was multiplechoice...

s=vt (uniform speed) …………………………….(i) s=(v+6)(t-4)(condition 1st)……………………….(ii) s=(v-6)(t+6)(condition 2nd)………………………(iii) Solving (ii) & (iii), we get, vt - 4v + 6t – 24 = vt + 6v – 6t – 36 or, 10v – 12t = 12 or, 5v – 6t = 6 ………………………………………..(iv) Solving (i) & (ii), we get, 4v – 6t = -24 …………………………………………….(v) Subtracting (v) from (iv), we get, 5v – 6t - 4v + 6t = 6 +24 Or, v = 30, putting the value of v = 30 in (iv) we get t = 24 then, s = 30 * 24 = 720

let scheduled velocity is: V scheduled time is: T traveled distance: S so, out come is three equation according to the question

S=(V+6) (T-4)=(V-6) (T+6)=V T so taking (V+6) (T-4)=V T we find , 6 V-6 T=36 and taking (V-6) (T+6)=V T we find , -4 V+6*T=24 solving these final two equations we find V=30 and T=24 so S= 720
that's all :-) :-) :-)

x=normal velocity when the train hadn't been faster or slower y=normal time when the train hadn't been faster or slower z= time taken by the train to finish the journey

3 equations:

1. xy=z

2. (x+6)(y-4)=z

3. (x-6)(y+6)=z

solve these 3 equations to get the answer

Let the distance be 'D', Let the speed be 's', Let the time taken be 't' therefore, we know that, D=st ........(1) D=(s+6)(t-4).........(2) D=(s-6)(t+4)..........(3) solve the three equations simultaneously to find the values of 's' and 't' which come out to be s=30 km/hr and t=24 hrs. and hence, D=30*24=720kms.