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

Speed of bus :
= 72 km/hr
= $$\left( {\frac{{72 \times 5}}{{18}}} \right)$$  m/sec
= 20 m/sec
Let distance covered by bus in 5 sec be x
∴ Distance = Speed × Time
⇒ x = 20 × 5
⇒ x = 100 metres

Deepa's original speed = 30 km/hr
Deepa's new speed :
= (30 + 10) km/hr = 40 km/hr
Distance = (30 × 6) km = 180 km
Hema's original speed :
= $$\frac{180}{4}$$ km/hr = 45 km/hr
Hema's new speed :
= (45 + 5) km/hr = 50 km/hr
Difference in time :
$$\eqalign{ & = \left( {\frac{{180}}{{40}} - \frac{{180}}{{50}}} \right){\text{ hrs}} \cr & = \frac{9}{{10}}{\text{ hrs}} \cr & = \left( {\frac{9}{{10}} \times 60} \right){\text{ min}} \cr & = 54{\text{ min}} \cr} $$

Suppose B takes x hours to walk d km
Then, A takes (x + 2) hours to walk d km
A's speed = $$\left( {\frac{d}{{x + 2}}} \right)$$  km/hr and
B's speed = $$\left( {\frac{d}{{x}}} \right)$$  km/hr
A's new speed = $$\left( {\frac{2d}{{x + 2}}} \right)$$  km/hr
$$\eqalign{ & \therefore \frac{d}{{\left( {\frac{d}{x}} \right)}} - \frac{d}{{\left( {\frac{{2d}}{{x + 2}}} \right)}} = 1 \cr & \Leftrightarrow x - \left( {\frac{{x + 2}}{2}} \right) = 1 \cr & \Leftrightarrow x - 2 = 2 \cr & \Leftrightarrow x = 4 \text{ hours} \cr} $$

Total distance travelled :
= (39 + 25) km
= 64 km
Total time taken
= (45 + 35) min
= 80 min
= $$\frac{4}{3}$$ hr
∴ Average speed :
= $$\left( {64 \times \frac{3}{4}} \right)$$  km/hr
= 48 km/hr

Let distance = x km and usual speed = y kmph
$$\eqalign{ & \Rightarrow \frac{x}{y} - \frac{x}{{y + 6}} = 4 \cr & \Rightarrow 6x = 4y\left( {y + 6} \right).....(i) \cr} $$
And,
$$\eqalign{ & \Rightarrow \frac{x}{{y - 6}} - \frac{x}{y} = 6 \cr & \Rightarrow 6x = 6y\left( {y - 6} \right).....(ii) \cr} $$
From (i) and (ii), we get :
$$\eqalign{ & \Rightarrow 4y\left( {y + 6} \right) = 6y\left( {y - 6} \right) \cr & \Rightarrow 2\left( {y + 6} \right) = 3\left( {y - 6} \right) \cr & \Rightarrow y = 30 \cr} $$
∴ Length of the journey :
$$\eqalign{ & = x{\text{ km}} \cr & = \left( {\frac{{4y\left( {y + 6} \right)}}{6}} \right){\text{ km}} \cr & = \left( {\frac{{4 \times 30 \times 36}}{6}} \right){\text{ km}} \cr & = 720{\text{ km}} \cr} $$

Quantity of water let in by the leak in 1 minute :
$$\eqalign{ & = \left( {\frac{{3\frac{3}{4}}}{{12}}} \right){\text{tonnes}} \cr & = \left( {\frac{{15}}{4} \times \frac{1}{{12}}} \right){\text{tonnes}} \cr & = \frac{{15}}{{18}}{\text{tonnes}} \cr} $$
Quantity of water thrown out by the pumps in 1 minute :
$$\eqalign{ & = \left( {\frac{{12}}{{60}}} \right){\text{tonnes}} \cr & = \frac{1}{5}{\text{ tonnes}} \cr} $$
Net quantity of water filled in the ship in 1 min :
$$\eqalign{ & = \left( {\frac{{15}}{{48}} - \frac{1}{5}} \right){\text{tonnes}} \cr & = \frac{{27}}{{240}}{\text{ tonnes}} \cr} $$
$$\frac{{27}}{{240}}$$ tonnes water is filled in 1 minute
60 tonnes water is filled in :
$$\eqalign{ & = \left( {\frac{{240}}{{27}} \times 60} \right){\text{min}} \cr & {\text{ = }}\frac{{1600}}{3}{\text{min}} \cr & = \frac{{80}}{9}{\text{hrs}} \cr} $$
Hence, required speed :
$$\eqalign{ & = \left( {\frac{{40}}{{\frac{{80}}{9}}}} \right){\text{km/hr}} \cr & = \left( {40 \times \frac{9}{{80}}} \right){\text{km/hr}} \cr & = \frac{9}{2}{\text{km/hr}} \cr & = 4\frac{1}{2}{\text{km/hr}} \cr} $$

Time taken by the two cyclists to cover one round of the track is $$\frac{300}{7}$$ sec and $$\frac{300}{8}$$ sec respectively.
∴ Required time :
= L.C.M. of $$\frac{300}{7}$$ and $$\frac{300}{8}$$
= 300 sec

Let ther distance covered be 2x km
$${{\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}}}$$
Time taken to covers the first half and second half of the journey in t₁ hours and t₂ hours
$$\eqalign{ & \Rightarrow \frac{a}{{60}} = {t_1}.....(i) \cr & \Rightarrow \frac{a}{{45}} = {t_2}.....(ii) \cr} $$
Adding (i) and (ii) we get
$$\eqalign{ & \Leftrightarrow \frac{a}{{60}} + \frac{a}{{45}} = {t_1} + {t_2} \cr & \Leftrightarrow \frac{a}{{60}} + \frac{a}{{45}} = 5\frac{{15}}{{60}} \cr & \Leftrightarrow \frac{a}{{60}} + \frac{a}{{45}} = 5\frac{1}{4} \cr & \Leftrightarrow \frac{{3a + 4a}}{{180}} = \frac{{21}}{4} \cr & \Leftrightarrow 7a = \frac{{21}}{4} \times 180 \cr & \Rightarrow a = \frac{{21 \times 180}}{{4 \times 7}} \cr & \Rightarrow a = 135{\text{ km}} \cr} $$
∴ Length of total journey :
= (2 × 135) km
= 270 km

Time taken to cover 60 miles = 80 min = $$\frac{4}{3}$$ hrs
∴ Speed of the train = $$\left( {60 \times \frac{3}{4}} \right)$$  mph = 45 mph
Speed of the aeroplane = (2 × 45) mph = 90 mph
Distance covered by the aeroplane in 60 min = 90 miles
Distance covered by the aeroplane in 20 min :
= $$\left( {\frac{90}{60} \times 20} \right)$$  miles
= 30 miles

Distance covered in 2 hrs 15 min, i.e.,
= $$2\frac{1}{4}$$ hrs = $$\left( {80 \times \frac{9}{4}} \right)$$  hrs = 180 hrs
Time taken to cover remaining distance :
$$\eqalign{ & = \left( {\frac{{350 - 180}}{{60}}} \right){\text{ hrs}} \cr & = \frac{{17}}{6}{\text{ hrs}} \cr & = 2\frac{5}{6}{\text{ hrs}} \cr & = 2{\text{ hrs }}50\min \cr} $$
Total time taken :
(2 hrs 15 min + 2 hrs 50 min) = 5 hrs 5 min
So, Anna reached city A at 10.25 am