Question

If a train runs at 60 km/hr it reaches its destination late by 15 minutes. But, if it runs at 85 kmph it is late by only 4 minutes. Find the distance to be covered by the train.

Answer 1

Let the distance to be covered by train be ‘d’

Using the formula, time take (t) = `"Distance"/"Speed"`

Case 1:

If speed = 60km/h

The time taken is 15 minutes more than usual `("t" + 15/60)`

Let the usual time taken be ‘t’ hrs.

Caution: Since speed is given in km/hr, we should take care to maintain all units such as time should be in hour and distance should be in min.

Given that in case 1, it takes 15 min. more

15 m = `15/60` hr

= `1/4` hr

 ∴ Substituting in formula,

∴ `"d"/60 = "t" + 1/4`

Since usually it takes ‘t’ hr, but when running at 60 k, it kes 15 min `(1/4 "hr")` extra

Multiplying by 60 on both sides

d = `60 xx "t" + 60 xx 1/4`

60t + 15  ...(1)

Case 2:

Speed is given as 85 km/h

Time taken is only 4 min `(4/60 "hr")` more than usual time

∴ time taken = `("t" + 1/35)  "hr". (4/60 = 1/15)`

Using the formula,

`"Distance"/"Speed"` = time

Multiplying by 85 on both sides

`"d"/85 xx 85 = 85 xx "t" + 85 xx 1/15`

∴ d = `85"t" + 17/3`  ...(2)

From (1) and (2), we will solve for ‘r’

Equating and eliminating ‘d’ we get

∴ By transposing, we get

`15 - 17/3` = 85t − 60t

`(45 - 17)/3` = 25t

∴ 25t = `28/3`

∴ t = `28/(3 xx 25)`

= `28/75` hr   ...`(28/75 xx 60 = 22.4  "min")`

Substituting this value of ‘t’ in eqn. (1), we get

d = 60t + 15

= `60 xx 28/75 + 15`

= `1680/75 + 15`

= 22.4 + 15

= 37.4 km