A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
Answers
SOLUTION :
Given: Total distance of a journey = 150 km
Let the usual speed of the train be x km/h and the increased speed of the train is (x + 5) km)h.
Time taken by the train with usual speed to cover 150 km= 150/x hrs
Time taken by the train with increased speed to cover 150 km= 150/(x + 5) hrs
[ Time = Distance/speed]
150/(x + 5) = 150/ x - 1
150/ x - 150/(x + 5) = 1
[150(x + 5) - 150x] /(x(x + 5) = 1
[By taking LCM]
150x + 750 - 150x /(x² + 5x) = 1
750 / x² + 5x = 1
(x² + 5x ) = 750
x² + 5x - 750 = 0
x² - 25x + 30x - 750 = 0
[By middle term splitting]
x(x - 25) + 30 ( x - 25) = 0
(x - 25) (x + 30) = 0
x = 25 or x = - 30
Since, speed can't be negative, so x = - 30
Therefore, usual speed of the train be = x = 25 km/h
Hence the usual speed of the train is 25 km/h .
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Answer :
25 km/hr
Step-by-step explanation :
Given that ;
Total distance of the journey = 150 km.
Let the usual speed of the train be x km/hr. And, the increased speed of the train be (x + 5) km/hr.
We know that,
Time = Distance / Speed
Time taken by the train with usual speed to cover 150 km = ( 150 / x ) hrs.
Time taken by the train with increased speed to cover 150 km = 150 / ( x + 5 ) hrs.
Now, we have ;
150 /(x + 5) = ( 150 / x ) - 1
⇒ 150 / x - 150 /(x + 5) = 1
⇒ [ 150(x + 5) - 150x ] / x ( x + 5 ) = 1
⇒ 150x + 750 - 150x / x² + 5x = 1
⇒ 750 / x² + 5x = 1
⇒ x² + 5x = 750
⇒ x² + 5x - 750 = 0
⇒ x² + 30x - 25x - 750 = 0
⇒ x ( x + 30 ) - 25 ( x + 30 ) = 0
⇒ ( x + 30 ) ( x - 25 ) = 0
⇒ x + 30 = 0 or x - 25 = 0
⇒ x = - 30 or x = 25
Since, speed cannot be negative, so the value of x = 25.
Hence, the usual speed of train is 25km/hr.