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

Speed = 10/sec
Speed = 10 × $$\frac{18}{5}$$ km/hr
Speed = 36 km/hr

Total distance = 120 km
Total time = 15 hours
He covers half of the journey in $$\frac{3}{5}$$th of the time
= 15 × $$\frac{3}{5}$$ hours
= 9 hours
Now, remaining distance :
= (120 - 60) km
= 60 km
And, remaining time :
= (15 - 9) hours
= 6 hours
Average speed to cover a distance of 60 km will be :
$$\eqalign{ & = \frac{{60\,\,{\text{km}}}}{{6\,\,{\text{hour}}}} = 10\,\,{\text{km/hr}} \cr & \left\{ {{\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right\} \cr} $$

Given :
Speed of running train = 60 km/hr
Length of running train = 110 metres
⇒ Length of standing train = 170 metres
⇒ Speed of the standing train = 0 km/hr
⇒ Time taken by running train to cross the standing train :
$$\eqalign{ & \Rightarrow {\text{Time}} = \frac{{\left( {100 + 170} \right){\text{ metres}}}}{{60{\text{ km/hr}}}} \cr & \Rightarrow {\text{Time }} = \frac{{280 \times 18}}{{60 \times 5}} \cr & \Rightarrow {\text{Time = 16}}{\text{.8 seconds}} \cr} $$

Time = $$\frac{270}{60 + 75}$$
Time = 2 hours
So, time at which they meet = 11.00 am

Alternate
Assume that they meet x hours after 8 am
Then,
Train 1, starting from A, travels x hours till the trains meet
Distance travelled by train 1 in x hours = 60x km

Train 2, starting from B, travels (x - 1) hours till the trains meet
Distance travelled by train 2 in (x - 1)hours = 75(x - 1) km

Total distance travelled
= Distance travelled by train 1 + Distance travelled by train 2
⇒ 330 = 60x + 75(x -1)
⇒ 12x + 15(x - 1) = 66
⇒ 12x + 15x - 15 = 66
⇒ 27x = 66 + 15 = 81
⇒ 3x = 9
⇒ x = 3
Hence, the trains meet 3 hours after 8 am, i.e. at 11 am

Total distance covered by man in (1.30 pm - 10.00 am)
= $$3\frac{1}{2}$$ hours at a speed of 12 km/hr
= 12 × $$3\frac{1}{2}$$
= 42 km (Total distance)
Time taken by his elder brother to catch him :
= $$3\frac{1}{2}$$ - 1 hour 15 min
∴ Brother's time
= 3 hr 30 min - 1 hr 15 min
= 2 hr 15 min
= $$2\frac{15}{60}$$
= $$2\frac{1}{4}$$
= $$\frac{9}{4}$$ hours
⇒ Brother's speed :
= $$\frac{{42}}{{\frac{9}{4}}}\,\,\left( {{\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right)$$
= $$18\frac{2}{3}{\text{ km/hr}}$$

$$\because $$ They meet after 6 hours if they walk towards each other i.e., their speed will be added.
So, their relative speed in opposite direction
$$ = \frac{{{\text{Distance }}}}{{{\text{Time }}}} = \frac{{60}}{6}$$
Relative speed in opposite direction :
$$\left( \rightleftharpoons \right) = 10{\text{ km/h}}.....{\text{(i)}}$$
According to the question,
$$\eqalign{ & \Rightarrow \frac{2}{3}A + 2B = \frac{{60}}{5} \cr & \Rightarrow \frac{2}{3}A + 2B = 12 \cr & \Rightarrow A + 3B = 18 \cr & \Rightarrow B's{\text{ Speed = }}\frac{{18 - A}}{3} \cr & \Rightarrow A + B = 10 \cr & \Rightarrow A + \frac{{18 - A}}{3} = 10 \cr & \Rightarrow 3A + 18 - A = 30 \cr & \Rightarrow 2A = 12 \cr & \Rightarrow A{\text{'s speed = 6 km/h}} \cr} $$

$$\eqalign{ & \,\,\,\,\,{\text{A}}\,\,\,\,\,\,\,\,\,\,\,\,{\text{B}}\,\,\,\,\,\,\,\,\,\,\,{\text{C}} \cr & {\text{1000}}\,\,\,\,\,\,{\text{960}}\,\,\,\,\,\,{\text{930}} \cr} $$
⇒ B can give C a start of 30 metres in a 960 metres race
⇒ 960 units → 30
⇒     1 unit → $$\frac{1}{32}$$
⇒ 1000 units → $$\frac{1000}{32}$$
  = $$31\frac{1}{4}$$ metres

$$\eqalign{ & {\text{Average speed }} = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & {\text{Average speed}} = \frac{{50 + 40 + 90}}{{2 + 2 + 6}} \cr & {\text{Average speed}} = \frac{{180}}{{10}} \cr & {\text{Average speed}} = 18{\text{ km/hr}} \cr} $$

According to the question,
⇒ Using Pythagoras theorem
⇒ (ML)² = (MO)² + (LO)²
⇒ (ML)² = (12)² + (5)²
⇒ ML = 13 km

Their relative speed in same direction :
= 36 - 9
= 27 km/hr
Time taken by train to cross the man = $$\frac{150}{27}$$
$$\left( {{\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}}} \right)$$
$$\eqalign{ & {\text{Time}} = \frac{{150}}{{27 \times \frac{5}{{18}}}} \cr & \left[ {1{\text{ km/hr}} = \frac{5}{{18}}{\text{m/s}}} \right] \cr & {\text{Time = }}\frac{{150 \times 18}}{{27 \times 5}} \cr & {\text{Time = }}\frac{{30 \times 2}}{3} \cr & \therefore {\text{ Time = 20 seconds}} \cr} $$