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# A Passenger Train Takes 3 Hours Less for a Journey of 360 Km, If Its Speed is Increased by 10 Km/Hr from Its Usual Speed. What is the Usual Speed? - Mathematics

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#### Question

A passenger train takes 3 hours less for a journey of 360 km, if its speed is increased by 10 km/hr from its usual speed. What is the usual speed?

#### Solution

Let the usual speed of the train be x km/hr

then the increased speed of the train = (x + 10)km/hr

We know that

"Time"="Distance"/"speed"

Time taken by the train under usual speed to cover 360 km = 360/x hr

Time taken by the train under increased speed to cover 3690km = 360/(x+10)hr

Then according to the question

rArr360/x-360/(x+10)=3

rArr(360(x+10)-360x)/(x(x+10))=3

rArr(360x+3600-360x)/(x(x+10))=3

rArr3600/(x(x+10))=3

⇒ 3x(x + 10) = 3600

⇒ 3x2 + 30x = 3600

⇒ 3x2 + 30x - 3600 = 0

⇒ 3(x2 + 10 - 1200) = 0

⇒ x2 + 10 - 1200 = 0

⇒ x2 - 30x + 40x - 1200 = 0

⇒ x(x - 30) + 40(x - 30) = 0

⇒ (x + 40)(x - 30) = 0

⇒ x + 40 = 0

⇒ x = -40

Or

⇒ x - 30 = 0

⇒ x = 30

∵ The speed of the train cannot be negative Therefore. the speed of the train is 30km/hr.

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#### APPEARS IN

RD Sharma Solution for Class 10 Maths (2018 (Latest))