Question

a man travels from home to town and back in a motor cycle.he travels to home from town at a speed which is 20km/hour more than his journey to the town from home .the average speed of his total journey was 48 km/hour

Answer 1

$let \: speed \: of \: journey \: town \: from \: home \: be \: x \\ speed \: home \: from \: town \: = x + 20 \\ d \: be \: distence \: from \: home \: to \: town \\ average \: speed = \frac{total \: distance}{total \: time} \\ 48 = \frac{d + d}{ \frac{d}{x} + \frac{d}{x + 4} } \\ 48 = \frac{2x(x + 4)}{x + 4 + x} \\ 48 (2x + 4) = 2x(x + 4) \\ 96x + 192 = 2 {x}^{2} + 8 \\ 2 {x}^{2} - 96x - 184 = 0 \\ {x}^{2} - 48x - 92 = 0 \\ (x - 46)(x + 2) = 0 \\ x = 46km \: per \: hour$

Answer 2

Answer: 40km/hr and 60km/hr

Step-by-step explanation:

Let the speed from town to home be x

Then, speed from home to town = x+20

Total distance covered = 5+5 = 10km

Total time taken = 5/x + 5/x+20 = 10x+100/x2+20x

Average speed= Total distance/ Total time

or, 48 = 10/10x+100/x2+20x

or, 48 = 10(x2+20x) / 10(x+10)

or, 48x+480 = x2+20x

or, x2-28x-480=0

or,( x+12) (x-40) =0

Therefore x= 40kmph and x+20 = 60kmph