Question

train covered a certain distance at a uniform speed. If the train would

have been 6 km/h faster, it would have taken 4 hours less than the scheduled

time. And, if the train were slower by 6 km/h, it would have taken 6 hours

more than the scheduled time. Find the length of the journey.​

Answer 1

In the above Question , the following information is given -

A train covered a certain distance at a uniform speed.

If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time.

And, if the train were slower by 6 km/h, it would have taken 6 hours more than the scheduled time.

To find -

Find the length of the Journey .

Solution -

A train covered a certain distance at a uniform speed.

Let this distance be D.

Let the scheduled time to cover this distance be t .

Now ,

Speed = [ Distance / Time ] = [ D / t ] kmph

If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time.

New speed = [ D / t ] + 6

New time = t - 4

So ,

[ D / t ] + 6 = [ D ] / [ t - 4 ]

=> [ D + 6t ] / [ t] = [ D ] / [ t - 4 ] ............ { 1 }

And, if the train were slower by 6 km/h, it would have taken 6 hours more than the scheduled time.

New speed = [ D / t ] - 6

New time = t + 6

So ,

[ D / t ] - 6 = [ D ] / [ t + 6 ]

=> [ D - 6t ] / [ t] = [ D ] / [ t + 6 ] ............. { 2 ]

Now , divide the two equations and solve to obtain a third Equation .

Substitute this value into either of the equations to get the value of D .

Finally , we get the value of D as 720 km .

This is the required answer ....

___________________

Answer 2

$\huge\sf\pink{Answer}$

☞ Distance = 720 Km

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ If the train was 6 km faster, it would take 4 hours less

✭ If the train was 6 km slower, it would have taken 6 hours more

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ The Length of the Journey?

$\rule{110}1$

$\huge\sf\purple{Steps}$

So the usual time will be,

➝ $\sf T =\dfrac{D}{S}$

➝ $\sf D = TS$

$\bullet\underline{\textsf{\: As Per the Question}}$

Case 1

➳ $\sf \dfrac{D}{S+6} = T-4$

➳ $\sf \dfrac{TS}{S+6} = T-4$

➳ $\sf TS = TS+6T-4S-24$

➳ $\sf 6T - 4S -24 = 0 \qquad -eq(1)$

Case 2

➢ $\sf \dfrac{D}{S-6} = T+6$

➢ $\sf TS = TS - 6T+6T-36$

➢ $\sf -6T+6S-36 = 0\qquad -eq(1)$

On solving eq(1) and eq(2)

»» S = 30km/h

»» T = 24 hrs

»» D = S × T

»» D = 30 × 24

»» $\sf\orange{D = 720 \ Km}$

$\rule{170}3$