1)over a distance of 120km, the average speed of a train is 40km/hr faster than that of a car.If the train covers the distance in 30 min less time,find it's average speed.
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Average car speed = C km/h
Average train speed = C+40 km/h
Car time = Distance / (Car speed) = 120 / C
Train time = Distance / (Train speed) = 120 / (C+40)
But Car time = Train time + 1/2 h
So 120/C = 120/(C+40) + 1/2
Multiply both sides by C.(C+40)
120(C+40) = 120C + 1/2.C(C+40)
120C + 4800 = 120C + 1/2.C^2 + 20C
Subtract 120C from both sides, and rearrange terms
1/2.C^2 + 20C - 4800 = 0
Multiply both sides by 2
C^2 + 40C - 9600 = 0
We must find two values which sum to 40 and whose product is -9600. By trial and error, these are 120 and -80.
(C+120)(C-80) = 0
So C is either -120 km/h or 80 km/h.
But since C is positive, C is 80 km/h.
Therefore the train's speed is C = 40 = 120 km/h.
Average train speed = C+40 km/h
Car time = Distance / (Car speed) = 120 / C
Train time = Distance / (Train speed) = 120 / (C+40)
But Car time = Train time + 1/2 h
So 120/C = 120/(C+40) + 1/2
Multiply both sides by C.(C+40)
120(C+40) = 120C + 1/2.C(C+40)
120C + 4800 = 120C + 1/2.C^2 + 20C
Subtract 120C from both sides, and rearrange terms
1/2.C^2 + 20C - 4800 = 0
Multiply both sides by 2
C^2 + 40C - 9600 = 0
We must find two values which sum to 40 and whose product is -9600. By trial and error, these are 120 and -80.
(C+120)(C-80) = 0
So C is either -120 km/h or 80 km/h.
But since C is positive, C is 80 km/h.
Therefore the train's speed is C = 40 = 120 km/h.
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Balance by the time,
120/r - 120/(r+40) = 1/2
=> 120 * 40 = (1/2)r(r+40)
r^2 + 40r - 80*120 = 0
(r+120)(r-80) = 0
r = 80
r+40 = 120 km/hr <== Answer
120/r - 120/(r+40) = 1/2
=> 120 * 40 = (1/2)r(r+40)
r^2 + 40r - 80*120 = 0
(r+120)(r-80) = 0
r = 80
r+40 = 120 km/hr <== Answer