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

1

A and B start moving from places X to Y and Y to X, respectively, at the same into on the same day. After crossing each other, A and B take $$5\frac{4}{9}$$ hours and 9 hours, respectively, to each their respective destinations. If the speed of A is 33 km/h, then the speed (in km/h) of B is:

A) 22 B) 2 C) $$25\frac{2}{3}$$ D) $$24\frac{1}{3}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & \frac{{{S_1}}}{{{S_2}}} = \sqrt {\frac{{{t_2}}}{{{t_1}}}} \cr & \frac{{33}}{{{S_2}}} = \sqrt {\frac{{9 \times 9}}{{49}}} \cr & \frac{{33}}{{{S_2}}} = \frac{9}{7} \cr & {S_2} = 25\frac{2}{3}{\text{ km/h}} \cr} $$

2

A train of length 287 m, running at 80 km/h, crosses another train moving in the opposite direction at 37 km/h in 18 second. What is the length of the other train?

A) 300 m B) 298 m C) 289 m D) 285 m

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & \frac{{\left( {287 + x} \right)}}{{\left( {80 + 37} \right) \times \frac{5}{{18}}}} = 18 \cr & \left( {287 + x} \right) = 18\left( {80 + 37} \right) \times \frac{5}{{18}} \cr & 287 + x = 117 \times 5 \cr & 287 + x = 585 \cr & x = 585 - 287 \cr & x = 298{\text{ m}} \cr} $$

3

A train x running at 74 km/h crosses another train y running at 52 km/h in the opposite direction in 12 seconds. If the length of y is two-thirds that of x, then what is the length of y (in m)?

A) 252 B) 200 C) 168 D) 210

Discuss

Answer & Solution

Answer: Option C

Solution:

According to question,
$$\eqalign{ & \frac{{x + y}}{{\left( {74 + 52} \right)\frac{{{\text{km}}}}{{{\text{hr}}}}}} = 12 \cr & x + y = 12 \times 126 \times \frac{5}{{18}} \cr & x + y = 420{\text{ m }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr} $$
y = 3a + 2a = 5a
x = 3a
So, from equation (i)
5a = 420
a = 84
Hence length of train y = 2 × 84 = 168

4

A takes 2 hours 30 minutes more than B to walk 40 km. If A doubles his speed, then he can make it in 1 hour less than B. What is the average time taken by A and B to walk a 40 km distance?

A) 5 hours 45 minutes B) 7 hours 15 minutes C) 5 hours 15 minutes D) 6 hours

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & \frac{{40}}{A} - \frac{{40}}{B} = 2\frac{1}{2} = \frac{5}{2}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \frac{{40}}{B} - \frac{{40}}{{2A}} = 1{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{Equation }}\left( {\text{i}} \right) + {\text{Equation }}\left( {{\text{ii}}} \right) \cr & \frac{{40}}{A} - \frac{{40}}{{2A}} = \frac{5}{2} + 1 \cr & \frac{{40}}{A} - \frac{{40}}{{2A}} = \frac{7}{2} \cr & \frac{{40}}{{2A}} = \frac{7}{2} \cr & A = \frac{{40}}{7} \cr & {\text{Put the value of A in equation }}\left( {\text{i}} \right) \cr & 7 - \frac{{40}}{B} = \frac{5}{2} \cr & 7 - \frac{5}{2} = \frac{{40}}{B} \cr & \frac{9}{2} = \frac{{40}}{B} \cr & B = \frac{{80}}{9} \cr & {\text{Now, required answer}} \cr & = \frac{{\frac{{40}}{{\frac{{40}}{7}}} + \frac{{40}}{{\frac{{80}}{9}}}}}{2} \cr & = \frac{{7 + \frac{9}{2}}}{2} \cr & = \frac{{23}}{4} \cr & = 5\frac{3}{4} \cr & = 5{\text{ hours }}45{\text{ minutes}} \cr} $$

5

Two cars start from the same place at the same time at right angles to each other. Their speeds are 54 km/hr and 72 km/hr, respectively. After 20 seconds the distance between them will be:

A) 480 m B) 540 m C) 720 m D) 500 m

Discuss

Answer & Solution

Answer: Option D

Solution:

54 km/h = 54 × $$\frac{5}{{18}}$$ = 15 m/s
72 km/h = 72 × $$\frac{5}{{18}}$$ = 20 m/s
Speed Time and Distance mcq question image

The distance of BC = 15 × 20 = 300 m
The distance of AB = 20 × 20 = 400
$${\text{AC}} = \sqrt {{{400}^2} + {{300}^2}} = 500$$

6

A person covers 40% of the distance from A to B at 8 km/h, 40% of the remaining distance at 9 km/h and the rest at 12 km/h. His average speed (in km/h) for the journey is:

A) $$9\frac{5}{8}$$ B) $$9\frac{1}{3}$$ C) $$9\frac{3}{8}$$ D) $$9\frac{2}{3}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Let total distance}} = 100 \cr & {\text{Average speed}} = \frac{{{\text{Total distance}}}}{{{\text{Total time}}}} \cr & = \frac{{100}}{{\frac{{40}}{8} + \frac{{24}}{9} + \frac{{36}}{{12}}}} \cr & = \frac{{100}}{{5 + \frac{8}{3} + 3}} \cr & = \frac{{100 \times 3}}{{32}} \cr & = \frac{{75}}{8} \cr & = 9\frac{3}{8} \cr} $$

7

Richa travels from A to B at the speed of 15 km/h, from B to C at 20 km/h, and from C to D at 30 km/h. If AB = BC = CD, then find the Richa's average speed.

A) 19 km/h B) 17 km/h C) 18 km/h D) 20 km/h

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Distance}} = {\text{ LCM of }}15,\,20,\,30 = 60{\text{ km}} \cr & {\text{Average speed}} = \frac{{{\text{Total distance}}}}{{{\text{Total time}}}} \cr & = \frac{{3 \times 60}}{{4 + 3 + 2}} \cr & = \frac{{180}}{9} \cr & = \boxed{20{\text{ km/h}}} \cr} $$

8

A man travelled a distance of 1200 km in 16 hours, He travelled partly by car at a speed of 40 km/h, and partly by train at a speed of 80 km/h. What is the distance traveled by car?

A) 100 km B) 120 km C) 96 km D) 80 km

Discuss

Answer & Solution

Answer: Option D

Solution:

8 unit ⟶ 16 hour
1 unit ⟶ 2 hour
Car = 2 hour = 2 × 40 = 80 km

9

A train travelling at the speed of x km/h crossed a 300 m long platform in 30 second, and overtook a man walking in the same direction at 6 km/h in 20 seconds. What is the value of x?

A) 60 B) 96 C) 48 D) 102

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{Train speed}} = x{\text{ km/h}} \cr & {\text{Length of train}} = L \cr & {\text{Length of platform}} = 300\,{\text{m}} \cr & {\text{Man's speed}} = 6{\text{ km/h}} \cr & \left( {x - 6} \right) \times \frac{5}{{18}} = \frac{L}{{20}} \cr & 100x - 600 = 18L{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & x \times \frac{5}{{18}} = \frac{{L + 300}}{{30}} \cr & 150x = 18L + 5400 \cr & 150x - 5400 = 18L{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{Equation }}\left( {\text{i}} \right)\,\& \,\left( {{\text{ii}}} \right) \cr & 100x - 600 = 150x - 5400 \cr & 50x = 4800 \cr & x = 96{\text{ km/ h}} \cr} $$

10

A man walks from point X to Y at a speed of 20 km/h, but comes back from point Y to X at a speed of 25 km/h. Find his average speed.

A) $$22\frac{2}{9}\,{\text{km/h}}$$ B) $$23\frac{2}{9}\,{\text{km/h}}$$ C) $$25\frac{2}{9}\,{\text{km/h}}$$ D) $$24\frac{2}{9}\,{\text{km/h}}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & {\text{Average speed}} = \frac{{2 \times 20 \times 25}}{{20 + 25}} \cr & = \frac{{2 \times 20 \times 25}}{{45}} \cr & = \frac{{200}}{9} \cr & = 22\frac{2}{9}\,{\text{km/h}} \cr} $$