A car starting from airport reaches the bus station in 45 mts with an average speed of 40 km/hr.If the speed of the car is increased by 10 km/hr.How much less time the car take to cover the distance?
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Answered by
0
answer is 36 minutes.
since,v=d÷t
v1=40
t1=45
putted in formula
40=d÷45
d=45×40
=1800. (1)
Now,
v2=v1+10
=40+10
=50
V2=d÷T2
50=1800÷T2. from(1)
T2=1800÷50
=36.
since,v=d÷t
v1=40
t1=45
putted in formula
40=d÷45
d=45×40
=1800. (1)
Now,
v2=v1+10
=40+10
=50
V2=d÷T2
50=1800÷T2. from(1)
T2=1800÷50
=36.
Answered by
0
In 1hr=60 minutes the car coveres 40km
∴, in 1 minute the car coveres =40/60 km
∴, in 45 minutes the car coveres = 40/60×45=2/3×45=30 km
When the speed of the car is increased by 10 km/hr i.e., 50 km/hr then,
the car coveres 50 km in 60 minutes
∴, the car coveres 1 km in =60/50 minutes
∴, the car coveres 30 km in=60/50×30=6/5×30=36 minutes.
∴, the car takes (45-36)=9 mintues less time to cover the distance.
∴, in 1 minute the car coveres =40/60 km
∴, in 45 minutes the car coveres = 40/60×45=2/3×45=30 km
When the speed of the car is increased by 10 km/hr i.e., 50 km/hr then,
the car coveres 50 km in 60 minutes
∴, the car coveres 1 km in =60/50 minutes
∴, the car coveres 30 km in=60/50×30=6/5×30=36 minutes.
∴, the car takes (45-36)=9 mintues less time to cover the distance.
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