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# The Time Taken by a Person to Cover 150 Km Was 2.5 Hrs More than the Time Taken in the Return Journey. If He Returned at a Speed of 10 Km/Hr More than the Speed of Going, What Was the Speed per Hour in Each Direction? - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Factorization

#### Question

The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction?

#### Solution

Let the ongoing speed of person be x km/hr. Then,

Returning speed of the person is = (x + 10)km/hr.

Time taken by the person in going direction to cover 150km = 150/xhr

Time taken by the person in returning direction to cover 150km = 150/(x+10)hr

Therefore,

150/x-150/(x+10)=5/2

(150(x+10)-150x)/(x(x+10))=5/2

(150x+1500-150x)/(x^2+10x)=5/2

1500/(x^2+10)=5/2

1500(2)=5(x2+10x)

3000 = 5x2 + 50x

5x2 + 50x - 3000 = 0

5(x2 + 10x - 600) = 0

x2 + 10x - 600 = 0

x2 - 20x + 30x - 600 = 0

x(x - 20) + 30(x - 20) = 0

(x - 20)(x + 30) = 0

So, either

x - 20 = 0

x = 20

Or

x + 30 = 0

x = -30

But, the speed of the train can never be negative.

Thus, when x = 20 then

= x + 10

= 20 + 10

= 30

Hence, ongoing speed of person is x = 20 km/hr

and returning speed of the person is x = 30 km/hr respectively.

Is there an error in this question or solution?

#### APPEARS IN

Solution The Time Taken by a Person to Cover 150 Km Was 2.5 Hrs More than the Time Taken in the Return Journey. If He Returned at a Speed of 10 Km/Hr More than the Speed of Going, What Was the Speed per Hour in Each Direction? Concept: Solutions of Quadratic Equations by Factorization.
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