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?
Answers
SOLUTION :
Given: Total distance of a journey = 150 km
Let the ongoing speed of the person be x km/h and the returning speed of the person is (x + 10) km/h.
Time taken by the person while going to cover 150 km= 150/x hrs
[ Time = Distance/speed]
Time taken by the train while returning to cover 150 km= 150/(x + 10) hrs
A.T.Q
150/ x - 150/(x + 10) = 5/2
[Time = 2.5 h = 25/10 = 5/2h]
[150(x + 10) - 150x] /(x(x + 10) = 5/2
[By taking LCM]
150x + 1500 - 150x /(x² + 10x) = 5/2
1500 / x² + 10x = 5/2
5(x² + 10x ) = 2 × 1500
[By cross multiplication ]
5x² + 50x = 3000
5x² + 50x - 3000 = 0
5(x² + 10x - 600) = 0
x² + 10x - 600 = 0
x² - 20x + 30x - 600 = 0
[By middle term splitting]
x(x - 20) + 30 ( x - 20) = 0
(x - 20) (x + 30) = 0
x = 20 or x = - 30
Since, speed can't be negative, so x ≠ - 30
Therefore, ongoing speed of the person be = x = 20 km/h
And returning speed of the person is (x + 10) km/h = 20 + 10 = 30 km/h
Hence the ongoing speed of the person is 25 km/h & returning speed of the person is 30 km/h
HOPE THIS ANSWER WILL HELP YOU...
Let ,
Forward speed = x
(If he returned at a speed of 10 km/hr more than the speed of going)
then, return speed = x + 10
Distance = same 150 miles 1
original time – time with increased speed
= 2.5 hour
s=d/t
s= speed
d = distance
t = time
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
[ Ignore negative i.e -30 because speed cannot be negative ]