Question

If a man had walked 2 km /hrs faster. He would have taken 40 minutes less to walk 5 km. Find the speed of walking

Answer 1

Distance Walked = 5km
Original Speed = x km/hr
Time Taken (T1) = Distance/Speed = 5/x

Distance Walked = 5km
New Speed = (x+2) km/hr
Time Taken (T2) = Distance/Speed = 5/(x+2)

Difference between times = 40mins = 2/3 hrs

T1 - T2 = 2/3 hrs

5/(x) - 5/(x+2) = 2/3 hrs
5(x+2) - 5(x) = 2(x)(x+2)/3           (taking LCM)
10 = (2x$x^{2}$+ 4x)/3
30 = (2x$x^{2}$ + 4x)
0 = 2x$x^{2}$ + 4x - 30
0 = 2x(x+5) - 6(x+5)
x= -5 , +3

Since speed can't be negative so, original speed x = 3 km/hr

Answer 2

Step-by-step explanation:

Distance Walked = 5km

Original Speed = x km/hr

Time Taken (T1) = Distance/Speed = 5/x

Distance Walked = 5km

New Speed = (x+2) km/hr

Time Taken (T2) = Distance/Speed = 5/(x+2)

Difference between times = 40mins = 2/3 hrs

T1 - T2 = 2/3 hrs

5/(x) - 5/(x+2) = 2/3 hrs

5(x+2) - 5(x) = 2(x)(x+2)/3 (taking LCM)

10 = (2xx^{2}x

2

+ 4x)/3

30 = (2xx^{2}x

2

+ 4x)

0 = 2xx^{2}x

2

+ 4x - 30

0 = 2x(x+5) - 6(x+5)

x= -5 , +3

Since speed can't be negative so, original speed x = 3 km/hr