Journey on a train!

Algebra Level pending

A train covered a distance at a uniform speed.

If the train would have been 10 k m h 10\frac{km}{h} faster, it would have taken 1 1 hour less than the schedule time.

And if the train were slower by 6 k m h 6\frac{km}{h} , it would have taken 1 1 hour more than the schedule time.

Find the length of the journey in kilometres.


The answer is 120.

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2 solutions

Let the required distance be d d km. and the original speed of the train be u u km/hr. Then

d ( 1 u 1 u + 10 ) = 1 d = u 2 + 10 u 10 d\left (\dfrac{1}{u}-\dfrac{1}{u+10}\right ) =1\implies d=\dfrac{u^2+10u}{10}

d ( 1 u 6 1 u ) = 1 d = u 2 6 u 6 d\left (\dfrac{1}{u-6}-\dfrac{1}{u}\right ) =1\implies d=\dfrac{u^2-6u}{6} .

So, u 2 + 10 u 10 = u 2 6 u 6 \dfrac{u^2+10u}{10}=\dfrac{u^2-6u}{6}

3 u 2 + 30 u = 5 u 2 30 u u = 30 \implies 3u^2+30u=5u^2-30u\implies u=30 km/hr. (Since u 0 u\neq 0 ).

Hence, d = 900 + 300 10 = 120 d=\dfrac{900+300}{10}=\boxed {120} km.

Marvin Kalngan
Jun 7, 2020

d = v t d=vt [ 1 ] [1]

d = ( v + 10 ) ( t 1 ) = v t v + 10 t 10 d=(v+10)(t-1)=vt-v+10t-10 [ 2 ] [2]

d = ( v 6 ) ( t + 1 ) = v t + v 6 t 6 d=(v-6)(t+1)=vt+v-6t-6 [ 3 ] [3]

Subtracting [ 3 ] [3] from [ 2 ] [2] , we get

2 v + 4 = 16 t 2v+4=16t [ 4 ] [4]

Subtracting [ 1 ] [1] from [ 2 ] [2] , we get

v = 10 t 10 v=10t-10 [ 5 ] [5]

Substituting [ 5 ] [5] into [ 4 ] [4] , we get

t = 4 t=4

It follows that

v = 10 t 10 = 10 ( 4 ) 10 = 30 v=10t-10=10(4) - 10=30

Finally,

d = v t = ( 30 ) ( 4 ) = 120 k m d=vt=(30)(4)=\text{\boxed{120 km}}

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