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

Distance covered by the first plane till 2 pm, i.e., in 2 hrs :
= (500 × 2) miles
= 1000 miles
Distance covered by the second plane till 2 pm, i.e., in $$1\frac{1}{2}$$ hrs :
= $$\left( {400 \times \frac{3}{2}} \right)$$   miles
= 600 miles
∴ Required distance :
$$\eqalign{ & = AB \cr & = \sqrt {{{\left( {1000} \right)}^2} + {{\left( {600} \right)}^2}} \cr & = \sqrt {1000000 + 360000} \cr & = \sqrt {1360000} {\text{ miles}} \cr & {\text{ = 200}}\sqrt {34} {\text{ miles}} \cr & {\text{ = 1160 miles}} \approx {\text{1200 miles}} \cr} $$

Total distance travelled in 12 hours
= (35 + 37 + 39 +...... upto 12 terms)
This is an A.P. with first term, a = 35,
Number of terms, n = 12,
Common difference, d = 2
∴ Required distance :
$$\eqalign{ & = \frac{{12}}{2}\left[ {2 \times 35 + \left( {12 - 1} \right) \times 2} \right] \cr & = 6\left( {70 + 22} \right) \cr & = 552{\text{ km}} \cr} $$

Average speed :
$$\eqalign{ & = \left( {\frac{{2 \times 56 \times 53}}{{56 + 53}}} \right){\text{km/hr}} \cr & = \left( {\frac{{5936}}{{109}}} \right){\text{km/hr}} \cr & = 54.45{\text{ km/hr}} \cr } $$

Difference between timings = 24 min = $$\frac{24}{60}$$ hr = $$\frac{2}{5}$$ hr
Let the length of the journey be x km
Then,
$$\eqalign{ & \Leftrightarrow \frac{x}{{40}} - \frac{x}{{50}} = \frac{2}{5} \cr & \Leftrightarrow \frac{x}{{200}} = \frac{2}{5} \cr & \Leftrightarrow x = \left( {\frac{2}{5} \times 200} \right) \cr & \Leftrightarrow x = 80{\text{ km}} \cr} $$

Let A's speed = x km/hr
Then, B's speed = (x + 4) km/hr
Clearly, time taken by B to cover (60 + 12) i.e., 72 km
= Time taken by A to cover (60 - 12) i.e., 48 km
$$\eqalign{ & \therefore \frac{{72}}{{x + 4}} = \frac{{48}}{x} \cr & \Rightarrow 72x = 48x + 192 \cr & \Rightarrow 24x = 192 \cr & \Rightarrow x = 8 \cr} $$
Hence, A's speed = 8 km/hr

Let the speed of the man be x m/sec
Then, distance travelled by the man in 5 min 52 sec
= Distance travelled by sound in 8 sec
$$\eqalign{ & \Leftrightarrow x \times 352 = 330 \times 8 \cr & \Leftrightarrow x = \left( {\frac{{330 \times 8}}{{352}}} \right){\text{m/sec}} \cr & \Leftrightarrow x = \left( {\frac{{330 \times 8}}{{352}} \times \frac{{18}}{5}} \right){\text{km/hr}} \cr & \Leftrightarrow x = 27{\text{ km/hr}} \cr} $$

Let the distance be x km
(Time taken to walk x km) + (Time taken to drive x km) = $$\frac{27}{4}$$ hrs
⇒ (Time taken to walk 2x km) + (Time taken to drive 2x km) = $$\frac{27}{2}$$ hrs
But time taken to drive 2x km = $$\frac{19}{4}$$ hrs
∴ Time taken to walk 2x km :
= $$\frac{27}{2}$$ - $$\frac{19}{4}$$
= $$\frac{35}{4}$$ hrs
= 8 hrs 45 min

Let the distance travelled by a car be x km
First $$\frac{x}{3}$$ km distance cover at speed of 10 km/hr
Second $$\frac{x}{3}$$ km distance cover at speed = 20 km/hr
Third $$\frac{x}{3}$$ km distance cover at speed = 60 km/hr
According to the question,
Total time ,
$$\eqalign{ & = \frac{x}{{3 \times 10}} + \frac{x}{{3 \times 20}} + \frac{x}{{3 \times 60}} \cr & = \frac{x}{{30}} + \frac{x}{{60}} + \frac{x}{{180}} \cr & = \frac{{6x + 3x + x}}{{180}} \cr & = \frac{{10x}}{{180}} \cr & = \frac{x}{{18}} \cr} $$
Average speed :
$$\eqalign{ & = \frac{{x \times 18}}{x}{\text{ km/hr}} \cr & = 18{\text{ km/hr}} \cr} $$

Suppose Jane travelled x miles by water,
$$\frac{4x}{7}$$ miles on foot and $$\frac{2x}{5}$$ miles on horseback.
Then,
$$\eqalign{ & \Leftrightarrow x + \frac{{4x}}{7} + \frac{{2x}}{5} = 3036 \cr & \Leftrightarrow \frac{{69x}}{{35}} = 3036 \cr & \Leftrightarrow x = \left( {\frac{{3036 \times 35}}{{69}}} \right) \cr & \Leftrightarrow x = 1540 \cr} $$
∴ Distance travelled on foot :
$$\eqalign{ & = \left( {\frac{4}{7} \times 1540} \right){\text{ miles}} \cr & {\text{ = 880 miles}} \cr} $$

Ratio of speeds = 2 : 3 : 4
∴ Ratio of times taken
= $$\frac{1}{2}$$ : $$\frac{1}{3}$$ : $$\frac{1}{4}$$
= 6 : 4 : 3