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

New speed = $$\frac{6}{7}$$ of usual speed
New time = $$\frac{7}{6}$$ of usual time
$$\therefore \left( {\frac{7}{6}{\text{ of usual time}}} \right) - $$    $$\left( {{\text{usual time}}} \right)$$   $$ = \frac{1}{5}{\text{ hr}}$$
$$\eqalign{ & \Rightarrow \frac{1}{6}{\text{ of usual time = }}\frac{1}{5}{\text{ hr}} \cr & \Rightarrow {\text{usual time = }}\frac{6}{5}{\text{ hr}} \cr & \Rightarrow {\text{usual time = 1 hr 12 min}} \cr} $$

Let the original planned duration of the flight be x hours
Then,
$$\eqalign{ & \Leftrightarrow \frac{{6000}}{x} - \frac{{600}}{{\left( {x + \frac{1}{2}} \right)}} = 400 \cr & \Leftrightarrow \frac{{6000}}{x} - \frac{{12000}}{{\left( {2x + 1} \right)}} = 400 \cr & \Leftrightarrow \frac{{15}}{x} - \frac{{30}}{{\left( {2x + 1} \right)}} = 1 \cr & \Leftrightarrow 2{x^2} + x - 15 = 0 \cr & \Leftrightarrow \left( {x + 3} \right)\left( {2x - 5} \right) = 0 \cr & \Leftrightarrow x = \frac{5}{2} \cr & \Leftrightarrow x = 2\frac{1}{2} \cr} $$

Error = Time taken to cover 10 m at 300 m/sec
    = $$\frac{10}{300}$$ sec
    = $$\frac{1}{30}$$ sec
    = 0.03 sec

Suppose the three trains meet x hours after 9 pm
Let the speed of train C be y km/hr
Distance travelled by train A in (x + 3) hrs = Distance travelled by train B in x hrs
⇒ 60 (x + 3) = 90x
⇒ 30x = 180
⇒ x = 6
⇒ Also, distance travelled by train B in x hrs + distance travelled by train C in x hrs = 1260 km
⇒ 90x + yx = 1260
⇒ 540 + 6y = 1260
⇒ 6y = 720
⇒ y = 120
Hence, speed of train C = 120 km/hr

Time taken = 3 hrs 15 min = $$3\frac{1}{4}$$ hrs = $$\frac{13}{4}$$ hrs
∴ Required distance :
$$\eqalign{ & = \left( {40 \times \frac{{13}}{4}} \right){\text{ km}} \cr & = 130{\text{ km}} \cr} $$

Speed of John = 30 km/h
Speed of Max = 40 km/h
Distance between L and M = 650 km

Distance travelled by john in 3 hours
= 30 x 3
= 90 km
Time taken by Max and John to travel remaining
= 650 - 90
= 560 km
$$\eqalign{ & = \frac{{560}}{{40 + 30}} \cr & = \frac{{560}}{{70}} \cr & = 8\,{\text{hours}} \cr} $$
Distance travel by John to reach point
P = 8 × 30 = 240 km
Distance between P and M
= 240 + 90
= 330 km

$$\eqalign{ & {\text{Required distance :}} \cr & AD = \sqrt {{{\left( {AE} \right)}^2} + {{\left( {DE} \right)}^2}} \cr & = \sqrt {{{\left( {30} \right)}^2} + {{\left( {30} \right)}^2}} \cr & = \sqrt {900 + 900} \cr & = \sqrt {1800} \cr & = 30\sqrt 2 {\text{ miles}} \cr} $$

Time taken by the express train to cover 600 km :
$$ = \left( {\frac{{600}}{{100}}} \right){\text{ hrs}} = 6{\text{ hrs}}$$
Number of stoppages = (600 ÷ 75) - 1 = 7
Duration of stoppage = (3 × 7) min = 21 min
Total time taken = 6 hrs 21 min
Total time taken by local train to cover 50 km (with stoppages)
= 1 hr 2 min
So, the local train covers (50 × 6) = 300 km in 6 hr 12 min
In remaining 9 min, it covers $$ = \left( {\frac{{50}}{{60}} \times 9} \right){\text{ km}} = 7.5{\text{ km}}$$
∴ Required distance = (300 + 7.5) km = 307.5 km

Average speed :
$$\eqalign{ & {\text{ = }}\left( {\frac{{2 \times 500 \times 700}}{{500 + 700}}} \right){\text{ km/hr}} \cr & {\text{ = }}\left( {\frac{{1750}}{3}} \right){\text{ km/hr}} \cr & {\text{ = }}583\frac{1}{3}{\text{ km/hr}} \cr} $$

Let the normal speed of the train be x km/hr
Then, new speed
$$\eqalign{ & = \left( {112\frac{1}{2}\% {\text{ of }}x} \right){\text{ km/hr}} \cr & = \left( {\frac{{225}}{2} \times \frac{1}{{100}} \times x} \right){\text{ km/hr}} \cr & = \left( {\frac{9}{8}x} \right){\text{ km/hr}} \cr} $$
Let the distance covered be d km
Then,
$$\eqalign{ & \Rightarrow \frac{d}{x} - \frac{d}{{\left( {\frac{{9x}}{8}} \right)}} = \frac{{20}}{{60}} \cr & \Rightarrow \frac{d}{x} - \frac{d}{{\left( {\frac{{9x}}{8}} \right)}} = \frac{1}{3} \cr & \Rightarrow \frac{d}{x} - \frac{{8d}}{{9x}} = \frac{1}{3} \cr & \Rightarrow \frac{d}{{9x}} = \frac{1}{3} \cr & \Rightarrow d = 3x \cr} $$
∴ Actual time taken:
$$\frac{d}{x} = \frac{{3x}}{x} = 3{\text{ hours = 180 minutes}}$$