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

Let the speeds of the train and the car be x km/hr and y km/hr respectively.
Then,
$$\eqalign{ & \Rightarrow \frac{{160}}{x} + \frac{{600}}{y} = 8 \cr & \Rightarrow \frac{{20}}{x} + \frac{{75}}{y} = 1.....(i) \cr} $$
And,
$$\eqalign{ & \Rightarrow \frac{{240}}{x} + \frac{{520}}{y} = 8\frac{1}{5} \cr & \Rightarrow \frac{{240}}{x} + \frac{{520}}{y} = \frac{{41}}{5}.....(ii) \cr} $$
Multiplying (i) by 12 and subtracting (ii) from it, we get :
$$\eqalign{ & \Rightarrow \frac{{380}}{y} = 12 - \frac{{41}}{5} \cr & \Rightarrow \frac{{380}}{y} = \frac{{19}}{5} \cr & \Rightarrow y = \left( {380 \times \frac{5}{{19}}} \right) \cr & \Rightarrow y = 100 \cr} $$
Putting y = 100 in equation (i), we get :
$$\eqalign{ & \Rightarrow \frac{{20}}{x} + \frac{3}{4} = 1 \cr & \Rightarrow \frac{{20}}{x} = \frac{1}{4} \cr & \Rightarrow x = 80 \cr} $$
∴ 100 km/hr, 80 km/hr

Suppose the trains meet after x hrs
Then,
Distance covered by 1^st train in x hrs + Distance covered by 2^nd train in $$\left( {x - \frac{3}{4}} \right)$$  hrs = 232 km
$$\eqalign{ & \Rightarrow 48x + 50\left( {x - \frac{3}{4}} \right) = 232 \cr & \Rightarrow 98x = 232 + \frac{{75}}{2} \cr & \Rightarrow 98x = \frac{{539}}{2} \cr & \Rightarrow x = \frac{{539}}{{196}}{\text{ hrs}} \cr} $$
Required distance = Distance travelled by 1^st train in $$\left( {\frac{{539}}{{196}}} \right)$$  hrs
$$\eqalign{ & = \left( {48 \times \frac{{539}}{{196}}} \right){\text{km}} \cr & = 132{\text{ km}} \cr} $$

Speed of Karan = 60 kmph
Time = 9 hrs
Distance = Speed × Time
Distance = 60 × 9
Distance = 540 km
∴ Time taken to cover 540 km at 90 km/ph
= $$\frac{540}{90}$$ hours
= 6 hours

Let the distance covered by man be x km
Journey covered by rail = $$\frac{2x}{15}$$
Journey covered by bus = $$\frac{9x}{20}$$
Remaining covered by cycle = 10 km
$$\eqalign{ & \therefore x\left( {1 - \frac{2}{{15}} - \frac{9}{{20}}} \right) = 10 \cr & \Rightarrow x\left( {\frac{{60 - 8 - 27}}{{60}}} \right) = 10 \cr & \Rightarrow x\left( {\frac{{25}}{{60}}} \right) = 10 \cr & \Rightarrow x = \frac{{10 \times 60}}{{25}} \cr & \Rightarrow x = 24{\text{ km}} \cr} $$

Actual time taken for the journey :
= 4 hrs 15 min - (10 + 2 × 5 + 3) min
= 4 hrs 15 min - 23 min
= 3 hrs 52 min
= $$3\frac{26}{30}$$ hrs
= $$\frac{116}{30}$$ hrs
∴ Average speed :
$$\eqalign{ & = \left( {\frac{{580}}{3} \times \frac{{30}}{{116}}} \right){\text{ km/hr}} \cr & = 50{\text{ km/hr}} \cr} $$

Let the speed of steam engine train be x
Then, speed of electric train = 125% of x = $$\frac{5x}{4}$$
Time taken by steam engine :
= 4 hrs 25 min
= $$4\frac{25}{60}$$ hrs
= $$4\frac{5}{12}$$ hrs
Let the time taken by electric train be $$t$$ hours
Then,
$$\eqalign{ & \Rightarrow x:\frac{{5x}}{4}::t:4\frac{5}{{12}} \cr & \Rightarrow 1:\frac{5}{4}::t:\frac{{53}}{{12}} \cr & \Rightarrow \frac{5}{4}t = \frac{{53}}{{12}} \cr & \Rightarrow t = \left( {\frac{{53}}{{12}} \times \frac{4}{5}} \right) \cr & \Rightarrow t = \frac{{53}}{{15}} \cr & \Rightarrow t = 3\frac{8}{{15}}{\text{ hrs}} \cr} $$

Time taken :
= 5 hrs 25 min
= $$\frac{65}{12}$$ hrs
Let the required distance be x km
Then,
$$\eqalign{ & \Leftrightarrow \frac{x}{{10}} + \frac{x}{1} = \frac{{65}}{{12}} \cr & \Leftrightarrow 11x = \frac{{650}}{{12}} \cr & \Leftrightarrow x = \frac{{325}}{{66}} \cr & \Leftrightarrow x = 4\frac{{61}}{{66}}{\text{ km}} \cr} $$

Let distance = x km and usual rate = y kmph
$$\eqalign{ & \Rightarrow \frac{x}{y} - \frac{x}{{y + 3}} = \frac{{40}}{{60}} \cr & \Rightarrow 2y\left( {y + 3} \right) = 9x.....(i) \cr} $$
And,
$$\eqalign{ & \Rightarrow \frac{x}{{y - 2}} - \frac{x}{y} = \frac{{40}}{{60}} \cr & \Rightarrow y\left( {y - 2} \right) = 3x.....(ii) \cr} $$
On dividing (i) by (ii), we get :
$$\eqalign{ & \Rightarrow 2\left( {y + 3} \right) = 3\left( {y - 2} \right) \cr & \Rightarrow y = 12 \cr} $$
∴ Distance :
$$\eqalign{ & = x{\text{ km}} \cr & = \left( {\frac{{2y\left( {y + 3} \right)}}{9}} \right){\text{ km}} \cr & = \left( {\frac{{2 \times 12 \times 15}}{9}} \right){\text{ km}} \cr & = 40{\text{ km}} \cr} $$

Let C's speed = x km/hr
Then, B's speed = 3x km/hr
And A's speed = 6x km/hr
∴ Ratio of speeds of A, B, C
= 6x : 3x : x
= 6 : 3 : 1
Ratio of times taken
= $$\frac{1}{6}$$ : $$\frac{1}{3}$$ : 1
= 1 : 2 : 6
If C takes 6 minutes, then A takes 1 minute
If C takes 72 minutes, then A takes $$\left( {\frac{1}{6} \times 72} \right)$$  min = 12 min

Relative speed = (2 + 3) = 5 rounds per hour
So, they cross each other 5 times in an hour and 2 times in half an hour.
Hence, they cross each other 7 times before 9.30 am.