Question

The time taken by a person to cover 150 km was 2hrs. 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 his speed in each direction in
2​

Answer 1

Step-by-step explanation:

⇒ Distance given =150km

⇒ Let the forward speed be x and the return speed will be x+10

We know that, Time=

speed

Distance

∴

x

150

=

x+10

150

+2.5

⇒ 150(x+10)−150x=2.5×x(x+10)

⇒ 150x+1500−150x=2.5x

2

+25x

⇒ 2.5x

2

+25x−1500=0

⇒ 25x

2

+250x−15000=0

⇒ x

2

+10x−600=0

⇒ x

2

+30x−20x−600=0

⇒ x(x+30)−20(x+30)=0

⇒ (x+30)(x−20)=0

The value of cannot be negative.

∴ Forward speed will be 20km/hr

⇒ Return speed =20+10=30km/hr

⇒ The required product =20×30=600km/hr

Answer 2

ANSWER:

⇒  Distance given =150km

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)⇒  150x+1500−150x=2.5x2+25x

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)⇒  150x+1500−150x=2.5x2+25x⇒  2.5x2+25x−1500=0

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)⇒  150x+1500−150x=2.5x2+25x⇒  2.5x2+25x−1500=0⇒  25x2+250x−15000=0

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)⇒  150x+1500−150x=2.5x2+25x⇒  2.5x2+25x−1500=0⇒  25x2+250x−15000=0⇒  x2+10x−600=0

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)⇒  150x+1500−150x=2.5x2+25x⇒  2.5x2+25x−1500=0⇒  25x2+250x−15000=0⇒  x2+10x−600=0⇒  x2+30x−20x−600=0

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)⇒  150x+1500−150x=2.5x2+25x⇒  2.5x2+25x−1500=0⇒  25x2+250x−15000=0⇒  x2+10x−600=0⇒  x2+30x−20x−600=0⇒  x(x+30)−20(x+30)=0

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)⇒  150x+1500−150x=2.5x2+25x⇒  2.5x2+25x−1500=0⇒  25x2+250x−15000=0⇒  x2+10x−600=0⇒  x2+30x−20x−600=0⇒  x(x+30)−20(x+30)=0⇒  (x+30)(x−20)=0

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)⇒  150x+1500−150x=2.5x2+25x⇒  2.5x2+25x−1500=0⇒  25x2+250x−15000=0⇒  x2+10x−600=0⇒  x2+30x−20x−600=0⇒  x(x+30)−20(x+30)=0⇒  (x+30)(x−20)=0The value of cannot be negative.

⇒  Distance given =150km⇒  Let the forward speed be x and the return speed will be x+10We know that, Time=speedDistance∴  x150=x+10150+2.5⇒  150(x+10)−150x=2.5×x(x+10)⇒  150x+1500−150x=2.5x2+25x⇒  2.5x2+25x−1500=0⇒  25x2+250x−15000=0⇒  x2+10x−600=0⇒  x2+30x−20x−600=0⇒  x(x+30)−20(x+30)=0⇒  (x+30)(x−20)=0The value of cannot be negative.

∴  Forward speed will be 20km/hr

⇒  Distance given =150km

⇒  Let the forward speed be x and the return speed will be x+10We know that, 

Time=speedDistance

∴  x150=x+10150+2.5

⇒  150(x+10)−150x=2.5×x(x+10)

⇒  150x+1500−150x=2.5x2+25x

⇒  2.5x2+25x−1500=0

⇒  25x2+250x−15000=0

⇒  x2+10x−600=0

⇒  x2+30x−20x−600=0

⇒  x(x+30)−20(x+30)=0

⇒  (x+30)(x−20)=0

The value of cannot be negative.

∴  Forward speed will be 20km/hr

⇒  Return speed =20+10=30km/hr