Question

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

Answer 1

hii mate!!

Forward speed x
return speed x + 10

↪Distance = same 150 miles 1
original time – time with increased speed = 2.5 hour
t=d/t

150 / x - 150 / ( x + 10 ) = 2.50
LCD= x ( x + 10 )
multiply by LCD
150 ( x + 10 ) - 150 x = 2.5 x ( x + 10 )
150 x + 1500 - 150 x = 2.5 X^2 + 25 x
1500 = 2.5 X^2 + 25 x
2.5 X^2 + 25 x - -1500 = 0
↪Find roots of the quadratic equation
a= 2.5 b= 25 c= -1500

x1= ( -25 + sqrt( 625 + 15000 )) / 5
x1=( -25 + 125 )/ 5
x1= 20

x2= ( -7 - sqrt( 49 - 20 ) / 2
x2=( -25 - 125 )/ 5
x2= -30
⭐Ignore negative 
forward speed = 20 kmph
return speed = 30 kmph 

$hope \: it \: helps$

$be \: brainly$

Answer 2

Forward speed x

return speed x + 10

⇝Distance = same 150 miles 1

original time – time with increased speed = 2.5 hour

t=d/t

150 / x - 150 / ( x + 10 ) = 2.50

LCD= x ( x + 10 )

multiply by LCD

150 ( x + 10 ) - 150 x = 2.5 x ( x + 10 )

150 x + 1500 - 150 x = 2.5 X^2 + 25 x

1500 = 2.5 X^2 + 25 x

2.5 X^2 + 25 x - -1500 = 0

⇝Find roots of the quadratic equation

a= 2.5 b= 25 c= -1500

x1= ( -25 + sqrt( 625 + 15000 )) / 5

x1=( -25 + 125 )/ 5

x1= 20

x2= ( -7 - sqrt( 49 - 20 ) / 2

x2=( -25 - 125 )/ 5

x2= -30

forward speed = 20 kmph

return speed = 30 kmph

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