Question

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.​

Answer 1

Answer:

Let us consider

Time is taken = t

Speed = x

Distance = d

Time = Distance/Speed

or

Speed = Distance/time

x = d/t

d = xt………………………………….(1)

Case 1:

d = (x + 10)(t – 2)

d = xt – 2x +10t -20

d = d – 2x +10t -20 [from (1) xt = d]

10t – 2x = 20…………………………………..(2)

Case2:

d = (x – 10)(t + 3)

d = xt + 3x -10t – 30

d = d + 3x -10t – 30 [from (1) xt = d]

3x -10t = 30…………………………………..(3)

Adding equation (2) and (3)

(10t – 2x + 3x -10t) = 20 + 30

x = 50

Substituting the value of x in 2

10t – 2 × 50 = 20

10t = 100 + 20

10t = 120

t = 12 hours

Now substitute the value of t and in x in equation (1)

Distance d = xt

d = 50 × 12

d = 600 km

Therefore distance covered by the train is 600km.

Step-by-step explanation:

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Answer 2

Answer:

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Let us take

Time taken to be = t sec

Speed be = x km/h

Distance be d km

we know that

Time = Distance/Speed

or it can be as

Speed = Distance/time

x = d/t

d = xt………………………………….(1)

Let us take it as (1) case

d = (x + 10)(t – 2)

d = xt – 2x +10t -20

d = d – 2x +10t -20 [from (1) xt = d]

10t – 2x = 20…………………………………..(2)

Let us take it as (2) case

d = (x – 10)(t + 3)

d = xt + 3x -10t – 30

d = d + 3x -10t – 30 [from (1) xt = d]

3x -10t = 30…………………………………..(3)

Adding equation (2) and (3)

(10t – 2x + 3x -10t) = 20 + 30

x = 50 km/h

put the value of x in eq(2)

10t – 2 × 50 = 20

10t = 100 + 20

10t = 120

time= 12 hours

Now

put the value of t at x in eq (1)

Distance d = xt

d = 50 × 12

d = 600 km

and

distance covered by the train is 600km