Question

A train travels at a certain average speed for a distance of 120 km and then travels a distance of 130 km at an average speed of 5 km/hr more than its original speed. If it takes 4 hours to complete the total journey, then average speed of train during 130km journey and if original speed is decreaced by 10 km/hr and is constant in whole journey, then the. time taken. by the train to complete the journey is?​

Answer 1

Answer:

The original average speed is 42 km/hr

Step-by-step explanation:

Let, the original average speed be x km/hr.

Time taken at a speed of x km/hr (t

1

)=

x

63

hrs

Time taken at a speed of x+6 km/hr (t

2

)=

x+6

72

hrs

t

1

+t

2

=3

x

63

+

x+6

72

=3

⟹

x(x+6)

63((x+6)+72x

=3

⟹63x+378+72x=3x(x+6)

⟹378+135x=3x

2

+18x

⟹3x

2

+18x−135x−378=0

⟹3x

2

−117x−378=0

⟹3(x

2

−39x−126)=0

⟹x

2

−39x−126=0

⟹x

2

−42x+3x−126=0

⟹x(x−42)+3(x−42)=0

⟹(x−42)(x+3)=0

Either x=42 or x=−3

∵ speed cannot be negative. x=42 is considered.

Answer 2

Solution:

Step 1: Let the original average speed of the train be x km/hr.

As we know,

Speed = Distance / Time taken

Time taken = Distance / Speed            ------------------   (i)

For distance 120 km, using equation (i), we get

Step 2: Time taken at a speed of x km/hr, $t_1$ = $\frac{120}{x}$  hr   [∵ At x km/hr speed, it covered a distance of 120 km]

Step 3: Now average speed is increased by 5 km/hr

∴ Time taken at a speed of (x+5) km/hr, $t_2 = \frac{130}{x+5}$  hr     [∵ At (x+5) km/hr speed, it covered a distance of 130 km]

Step 4: Total time taken to complete the journey, t = 4 hours

                 $t_1+t_2 = 4$

                 $\frac{120}{x} + \frac{130}{x+5} = 4$

              $\frac{120(x+5)+130x}{x(x+5)}$ = 4

             $\frac{120x+600+130x}{x^2 +5x} = 4$

          $250x+600 = 4(x^2+5x)$

          $250x+600= 4x^2+20x$

           $4x^2+20x-250x-600 = 0$

         $4x^2-230x-600=0$

        $4x^2+10x-240x-600 = 0$

         $2x(2x+5)-120(2x+5) =0$

          $(2x-120)(2x+5) = 0$                                      

      Solution 1:-                 $2x-120=0\\2x=120\\x=\frac{120}{2} = 60$      

                                         

     Solution 2:-                  $2x+5=0\\2x=-5\\x=\frac{-5}{2}$

We get,

x = 60, (-5/2)

X is the original average speed that can't be negative. So, we neglect (-5/2).

Hence, the original average speed of the train = x = 60 km/hr

Step 5: Now, the original speed is decreased by 10 km/hr and is constant in whole journey. And we have to find the time taken by the train to complete the journey.

New average speed, $x_{new}$ = (60-10) km/hr = 50 km/hr

Total distance, d = 120+130 = 250 km

Let the time taken be T.

Time taken = Distance/Speed

$T = \frac{d}{x_{new}}$

$T = \frac{250}{50} hr$

T = 5 hours

Hence, the time taken by the train to complete the journey is 5 hours.

#SPJ3