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SSC CHSL Model Paper 8
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© examsnet.com
Question : 56
Total: 100
A car after travelling 100 km from point A meets with an accident and then proceeds at
3
4
of its original speed and arrives at the point B 90 minutes late. If the car met the accident 60 km further on, it would have reached 15 minutes sooner. Find the original speed of the car.
60 km/hr
80 km/hr
100 km/hr
120 km/hr
Validate
Solution:
Let the distance between A and B be D km and the real speed of the car be S km/hr.
First time the car takes 90 minutes more and second time the car takes 75 minutes more than the scheduled time.
So,
T
1
−
T
2
=
15
60
Difference in time taken during the two journeys:
15
60
=
(
100
s
+
D
−
100
3
s
∕
4
)
−
(
160
s
+
D
−
160
3
s
∕
4
)
⇒
15
60
=
100
s
+
4
D
3
s
−
400
3
s
−
160
s
−
4
D
3
s
+
640
3
s
⇒
15
60
=
=
240
3
s
−
60
s
⇒
15
60
=
20
s
⇒ s = 80 km/hr
© examsnet.com
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