a train meets with an accident after travelling 30 Kms ,after which it moves 4/5th of the it's original speed and arrives the destination 45 mnts late .had the accident happened 18 Kms further on ,it would have 9 mnts before .find the distance of journey and original speed of train
Answers
Let the original speed be x and distance be y
Case I.
Time taken by train to travel
30 km =
30
x
Time taken by train after accident =
y − 30
4/5x
Total time taken =
30
+
y − 30
x 4/5x
Case II :
Time taken by train to travel
48 km =
48
x
Time taken by train after accident =
y − 48
4/5x
Total time taken =
48
+
y − 48
x 4/5x
30
+
y − 30
−
48
+
y − 48
x 4/5x x 4/5x
=
9
[∵ Difference between time is 9 minutes]
60
y − 30
−
y − 48
+
30
−
48
4/5x 4/5x x x
=
9
60
y − y − 30 + 48
+
(− 18)
=
9
4/5x x 60
5(18)
−
18
=
9
4x 4x 60
⇒
90 − 72
=
9
4x 60
x =
18 × 60
= 30
4 × 9
Hence, original speed = 30 kmph
Also,
30
+
y − 30
=
y
+
45
x 4/5x x 60
[Original time + 45 minute = New time]
⇒ 3x – y = –30
⇒ 3(30) – y = –30
⇒ i.e. Distance = 120 km
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