If a car covers a certain distance at x kmph and an equal distance at y kmph,
the average speed of the whole journey = $\dfrac{2xy}{x+y}$kmph
Hence, average speed = $\dfrac{2 \times 3 \times 2}{2 + 3} = \dfrac{12}{5}$km/hr
Total time taken = 5hours
$\Rightarrow$ Distance travelled = $\dfrac{12}{5} \times 5 $= 12km
$\Rightarrow$ Distance between his house and office =$\dfrac{12}{2}$= 6km
Let the speed of the train be $ x $ km/hr and that of the car be $ y $ km/hr.
| Then,$ \dfrac{120}{x} $+$ \dfrac{480}{y} $= 8 $\Rightarrow \dfrac{1}{x} $+$ \dfrac{4}{y} $=$ \dfrac{1}{15} $....(i) |
| And,$ \dfrac{200}{x} $+$ \dfrac{400}{y} $=$ \dfrac{25}{3} $ $\Rightarrow \dfrac{1}{x} $+$ \dfrac{2}{y} $=$ \dfrac{1}{24} $....(ii) |
Solving (i) and (ii), we get: $ x $ = 60 and $ y $ = 80.
$\therefore$ Ratio of speed = 60 : 80 = 3 : 4.
| $ \dfrac{(1/2)x}{21} $+$ \dfrac{(1/2)x}{24} $= 10 |
| $\Rightarrow \dfrac{x}{21} $+$ \dfrac{x}{24} $= 20 |
$\Rightarrow$ 15$ x $ = 168 x 20
| $\Rightarrow x $ =$ \left(\dfrac{168 \times 20}{15} \right) $= 224 km. |
Let the distance travelled on foot be $ x $ km.
Then, distance travelled on bicycle = $\left(61 -x\right)$ km.
| So,$ \dfrac{x}{4} $+$ \dfrac{(61 -x)}{9} $= 9 |
$\Rightarrow$ 9$ x $ + 4$\left(61 -x\right)$ = 9 x 36
$\Rightarrow$ 5$ x $ = 80
$\Rightarrow x $ = 16 km.
Speed = 80 km/hr = $80 \times \dfrac{5}{18} m/s = 40 \times \dfrac{5}{9}m/s = \dfrac{200}{9}m/s = 22\dfrac{2}{9}$m/s
Let the actual distance travelled be $ x $ km.
| Then,$ \dfrac{x}{10} $=$ \dfrac{x + 20}{14} $ |
$\Rightarrow$ 14$ x $ = 10$ x $ + 200
$\Rightarrow$ 4$ x $ = 200
$\Rightarrow x $ = 50 km.
Let Abhays speed be $ x $ km/hr.
| Then,$ \dfrac{30}{x} $-$ \dfrac{30}{2x} $= 3 |
$\Rightarrow$ 6$ x $ = 30
$\Rightarrow x $ = 5 km/hr.