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

Average speed :
$$\eqalign{ & = \frac{{\left( {2 \times \frac{5}{2} \times \frac{{13}}{4}} \right)}}{{\left( {\frac{5}{2} + \frac{{13}}{4}} \right)}}{\text{ km/hr}} \cr & {\text{ = }}\left( {\frac{{65}}{4} \times \frac{4}{{23}}} \right){\text{ km/hr}} \cr & = \left( {\frac{{65}}{{23}}} \right){\text{ km/hr}} \cr} $$
Total time taken :
= 4 hr 36 min
$$= 4\frac{36}{60}\,\, hr$$
$$= 4\frac{3}{5}\,\, hr$$
$$= \frac{23}{5}\,\, hr$$
Total distance covered uphill and downhill :
$$\eqalign{ & = \left( {\frac{{65}}{{23}} \times \frac{{23}}{5}} \right){\text{ km}} \cr & = 13{\text{ km}} \cr} $$
∴ Distance walked uphill :
$$\eqalign{ & = \left( {\frac{{13}}{2}} \right){\text{ km}} \cr & = 6\frac{1}{2}{\text{ km}} \cr} $$

Let the correct time to complete the journey be x min
Distance covered in (x + 11) min at 40 kmph
= Distance covered in (x + 5) min at 50 kmph
$$\eqalign{ & \therefore \frac{{\left( {x + 11} \right)}}{{60}} \times 40 = \frac{{\left( {x + 5} \right)}}{{60}} \times 50 \cr & \Leftrightarrow x = 19{\text{ min}} \cr} $$

Let the cyclist's speed without wind be x km/hr
And the speed of the wind be y km/hr
Then,
$$\eqalign{ & \Rightarrow \frac{1}{{x + y}} = \frac{3}{{60}} \cr & \Rightarrow x + y = 20.....(i) \cr} $$
And
$$\eqalign{ & \Rightarrow \frac{1}{{x - y}} = \frac{4}{{60}} \cr & \Rightarrow x - y = 15.....(ii) \cr} $$
Adding (i) and (ii), we get:
2x = 35 or x = 17.5
Putting x = 17.5 in (i), we get : y = 2.5
Time taken to drive 17.5 km without wind = 1 hr
Time taken to drive 1 km without wind
$$\eqalign{ & = \left( {\frac{1}{{17.5}}} \right){\text{ hr}} \cr & {\text{ = }}\left( {\frac{1}{{17.5}} \times 60} \right){\text{ min}} \cr & {\text{ = 3}}\frac{3}{7}{\text{ min}} \cr} $$

To be (18 + 12) km apart, they take 1 hour
To be 47.5 km apart, they take :
$$\eqalign{ & {\text{ = }}\left( {\frac{1}{{38}} \times 47.5} \right){\text{ hrs}} \cr & {\text{ = 1}}\frac{1}{4}{\text{ hrs}} \cr} $$

Let the length AB = x

Then, if C is the position of the cat, we have AC = $$\frac{3}{8}$$x
When the cat runs towards the entrance, the train catches it at the entrance.
This means that when the train reaches the entrance, the cat has travelled a distance of $$\frac{3}{8}$$x.
Let us now consider the case when the cat runs towards the exit.
So, when the train reaches A, the cat reaches a point D such that CD = $$\frac{3}{8}$$x
Then,
$$\eqalign{ & \text{BD} = \left[ {x - \left( {\frac{3}{8}x + \frac{3}{8}x} \right)} \right] \cr & \text{BD} = \frac{x}{4} \cr} $$
Since the train catches the cat at the exit, so the train covers distance x (= AB) in the same time in which that cat covers distance $$\frac{x}{4}$$ (= BD)
∴ Required ratio :
$$\eqalign{ & = x:\frac{x}{4} \cr & = 4:1 \cr} $$

Let the speed of car be x km/hr
Distance = Speed × Time
Distance = 8x km
According to the question,
$$\eqalign{ & \Rightarrow \left( {x + 4} \right) \times 7.5 = 8x \cr & \Rightarrow 7.5x + 30 = 8x \cr & \Rightarrow 8x - 7.5x = 30 \cr & \Rightarrow 0.5x = 30 \cr & \Rightarrow x = \frac{{30}}{{0.5}} = 60{\text{ km/hr}} \cr} $$
Required distance :
= 8 × 60
= 480 km

Required time :
$$\eqalign{ & = \left( {\frac{{8.1 \times {{10}^{13}}}}{{3.0 \times {{10}^5}}}} \right){\text{seconds}} \cr & = 2.7 \times {10^8}{\text{ seconds}} \cr & = \left( {\frac{{2.7 \times {{10}^8}}}{{60 \times 60}}} \right){\text{hours}} \cr & = 7.5 \times {10^4}{\text{ hours}} \cr} $$

Ratio of speed = 3 : 4
Ratio of time taken = $$\frac{1}{3}$$ : $$\frac{1}{4}$$ = 4 : 3
Let A and B take 4x and 3x minutes respectively to reach a destination.
Then,
⇔ 4x - 3x = 20
⇔ x = 20
∴ Time taken by A
= 4x
= (4 × 20) min
= 80 min
= $$1\frac{1}{3}$$ hours

Let the speed in return journey be x km/hr
Then, speed in onward journey :
$$\eqalign{ & = \left( {\frac{{125}}{{100}}x} \right){\text{ km/hr}} \cr & = \left( {\frac{5}{4}x} \right){\text{ km/hr}} \cr} $$
Average speed :
$$\eqalign{ & {\text{ = }}\left( {\frac{{2 \times \frac{5}{4}x \times x}}{{\frac{5}{4}x + x}}} \right){\text{ km/hr}} \cr & {\text{ = }}\left( {\frac{{10x}}{9}} \right){\text{ km/hr}} \cr & \therefore \left( {800 \times \frac{9}{{10x}}} \right) = 16 \cr & \Leftrightarrow x = \left( {\frac{{800 \times 9}}{{16 \times 10}}} \right) \cr & \Leftrightarrow x = 45 \cr} $$
So, speed in onward journey :
$$\eqalign{ & = \left( {\frac{5}{4} \times 45} \right){\text{ km/hr}} \cr & = 56.25{\text{ km/hr}} \cr} $$

Let the to and fro distance to the mall be x km
Then,
$$\eqalign{ & \Rightarrow \frac{x}{{10}} - \frac{x}{{15}} = \frac{{30}}{{60}} \cr & \Rightarrow \frac{x}{{10}} - \frac{x}{{15}} = \frac{1}{2} \cr & \Rightarrow \frac{x}{{30}} = \frac{1}{2} \cr & \Rightarrow x = 15 \cr} $$
. Time taken to travel 15 km at 10 km/hr
$$\eqalign{ & = \left( {\frac{{15}}{{10}}} \right){\text{ hr}} \cr & = \frac{3}{2}{\text{ hr}} \cr & = 1\frac{1}{2}{\text{ hrs}} \cr} $$
Since 30 minutes were spent in shopping, so Ravi started for the mall 2 hours before 19.00 hrs i.e., at 17.00 hrs
Now, required time for to and fro journey :
$$\eqalign{ & {\text{ = }}\left( {18.15{\text{ hrs}} - 17.00{\text{ hrs}}} \right) - 30{\text{ min}} \cr & {\text{ = 45 min}} \cr & {\text{ = }}\frac{3}{4}{\text{ hrs}} \cr} $$
Hence, required speed :
$$\eqalign{ & {\text{ = }}\left( {15 \times \frac{4}{3}} \right){\text{ km/hr}} \cr & = 20{\text{ km/hr}} \cr} $$