Speed Time And Distance - Aptitude MCQ Questions and Solutions with Explanations

Distance between Home & School $$ = \frac{{\frac{7}{2} \times \frac{9}{2}}}{{\frac{9}{2} - \frac{7}{2}}} \times \frac{{16}}{{60}} = 4.2{\text{ km}}$$

$$\eqalign{ & {\text{Time}} = \frac{{450 + 300 + 300}}{{\left( {162 + 108} \right) \times \frac{5}{{18}}}}\,\sec \cr & = \frac{{1050 \times 18}}{{270 \times 5}}\,\sec \cr & = 14\,\sec \cr} $$

If train crosses the platform i.e. it covers the distance equal to the length of train and platform. In the question train crosses the man who stands on the platform in 10 seconds and crosses the man + platform in 25 seconds i.e. train crosses the platform whose length is 300 metres in 25 - 10 = 15 seconds, here train's length is not added.
So speed of the train $$ = \frac{{300}}{5} \Rightarrow 20\,{\text{m/sec}}$$
Length of the train = 10 × 20 = 200 metres (If train crosses the only man in 10 seconds)
Time taken by the train to cross a platform 200 m long
$$\eqalign{ & = \frac{{{\text{Length of train}} + {\text{Platform}}}}{{{\text{Speed}}}} \cr & = \frac{{200 + 200}}{{20}} \cr & = \frac{{400}}{{20}} \cr & = 20 \cr} $$
Time taken by train = 20 seconds

Time = $$\frac{{1800}}{{44 + 46}}$$
Time = 20 hours
20 × 44 = 880 km from station A

15 min. rest
Total time = 1 hr 10 min
Distance same
Speed ⟶ 32 : 24
              4 : 3
Time ⟶ 3 : 4
7 unit ⟶ 70 min
1 unit ⟶ 10 min
Total distance = $$32 \times \frac{1}{2} + \frac{{24}}{{60}} \times 40$$
= 16 + 16
= 32 km

Initial speed = $$\frac{{12 \times 60}}{{12}}$$  = 60 km/h
New speed = 60 - 5 = 55 km/h
Time taken to cover 22 km = $$\frac{{22 \times 60}}{{55}}$$  = 24 min

$$\eqalign{ & \frac{{300}}{{x + 10}} - \frac{{300}}{x} = 5 \cr & 60\left[ {\frac{{10}}{{x\left( {x + 10} \right)}}} \right] = 1 \cr & 600 = x\left( {x + 10} \right) \cr & {\text{Put }}x = 20 \cr & {\text{Average speed}} = \frac{{2 \times 20 \times 30}}{{20 + 30}} \cr & = \frac{{2 \times 600}}{{50}} \cr & = 24{\text{ km/hr}} \cr} $$

Let length of engine = $$x$$ m
According to question
$$\eqalign{ & \frac{{x + 400}}{{100}} = \frac{{x + 235}}{{60}} \cr & \Rightarrow 6x + 2400 = 10x + 2350 \cr & \Rightarrow 4x = 50 \cr & \Rightarrow 2x = 25{\text{ m}} \cr} $$

Let the speed of outward journey be $$x$$ km/hr
Hence, the speed of return journey will be ($$x$$ + 10) km/hr
So, time taken in return journey = $$\frac{{300}}{{x + 10}}{\text{hr}}$$
Time taken in outward journey = $$\frac{{300}}{x}{\text{hr}}$$
According to the question,
$$\eqalign{ & \Rightarrow \frac{{300}}{x} - \frac{{300}}{{x + 10}} = 5 \cr & \Rightarrow \frac{{300\left( {x + 10} \right) - 300x}}{{x\left( {x + 10} \right)}} = 5 \cr & \Rightarrow 300x + 3000 - 300x = 5x\left( {x + 10} \right) \cr & \Rightarrow 3000 = 5{x^2} + 50x \cr & \Rightarrow 5{x^2} + 50x - 3000 = 0 \cr & \Rightarrow {x^2} + 30x - 20x - 600 = 0 \cr & \Rightarrow x\left( {x + 30} \right) - 20\left( {x + 30} \right) = 0 \cr & \Rightarrow \left( {x - 20} \right)\left( {x + 30} \right) = 0 \cr & \Rightarrow x = 20,\, - 30 \cr} $$
⇒ x = 20 km/hr as negative speed is not possible.
∴ The speed for outward journey is 20 km/hr

$$\eqalign{ & {T_L} = \left( {48 + 60} \right) \times \frac{5}{{18}} \times 12 \cr & = 108 \times \frac{5}{3} \times 2 \cr & = 360 \cr} $$