A passenger train takes 3 hours less for a journey of 360 km, if its speed is increased
by 10 km per hour. What is the usual speed?
that the lengths of the tangents drawn in a circle from an external point are
Answers
Answer:
The usual speed of train is 30 km/hr.
Step-by-step explanation:
Let the usual speed be s.
time=\frac{distance}{speed}time=speeddistance
It is given that train takes 3 hours less for a journey of 360 km if its speed is increased by 10 km/hr from the usual speed.
\frac{360}{s}-\frac{360}{s+10}=3s360−s+10360=3
\frac{360s+3600-360s}{s(s+10)}=3s(s+10)360s+3600−360s=3
3600=3s(s+10)3600=3s(s+10)
Divide both sides by 3.
1200=s^2+10s1200=s2+10s
s^2+10s-1200=0s2+10s−1200=0
s^2+40s-30s-1200=0s2+40s−30s−1200=0
s(s+40)-30(s+40)=0s(s+40)−30(s+40)=0
(s+40)(s-30)=0(s+40)(s−30)=0
s=-40,30s=−40,30
Step-by-step explanation:
The usual speed of train is 30 km/hr.
Step-by-step explanation:
Let the usual speed be s.
time=\frac{distance}{speed}time=speeddistance
It is given that train takes 3 hours less for a journey of 360 km if its speed is increased by 10 km/hr from the usual speed.
\frac{360}{s}-\frac{360}{s+10}=3s360−s+10360=3
\frac{360s+3600-360s}{s(s+10)}=3s(s+10)360s+3600−360s=3
3600=3s(s+10)3600=3s(s+10)
Divide both sides by 3.
1200=s^2+10s1200=s2+10s
s^2+10s-1200=0s2+10s−1200=0
s^2+40s-30s-1200=0s2+40s−30s−1200=0
s(s+40)-30(s+40)=0s(s+40)−30(s+40)=0
(s+40)(s-30)=0(s+40)(s−30)=0
s=-40,30s=−40,30
Let the usual speed of a train = x km/hr
According to the condition, we get
T2 - T1 = 3 .... (Time = Distance/Speed)
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