Practice MCQ Questions and Answer on Speed Time and Distance

101.

A and B start at the same time with speed of 40 km/hr. and 50 km/hr respectively. If in covering the journey A takes 15 min longer than B, the total distance of the journey is :

(A) 40 km
(B) 48 km
(C) 50 km
(D) 52 km

Solution: $\begin{aligned} A : B \\ Ratio of speed = 4 : 5 \\ Ratio of time = 5 : 4 \\ (5 - 4) unit = 15 unit \\ 1 unit = 15 min \\ So, time taken by B : \\ = 4 \times 15 \\ = 60 min \\ = 1 hr \\ Distance = S \times T \\ = 50 \times 1 \\ = 50 km \end{aligned}$

102.

The speeds of three cars are the ratio 2 : 3 : 4. The ratio of the taken by these cars to travel the same distance is :

(A) 2 : 3 : 4
(B) 4 : 3 : 2
(C) 4 : 3 : 6
(D) 6 : 4 : 3

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

103.

If a distance of 50 m is covered in 1 minute, 90 m in 2 minutes and 130 m 3 minutes. Find the distance covered in 15^th minute.

(A) 610 minutes
(B) 750 minutes
(C) 1000 minutes
(D) 650 minutes

104.

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

Solution: Let AB = BC = CD = 60 km (LCM of 15, 20 and 30) Total distance = 3 × 60 = 180 km Time taken by Richa to travels from A to B = $\frac{60}{15}$ = 4 hrs. Time taken by Richa to travels from B to C = $\frac{60}{20}$ = 3 hrs. Time taken by Richa to travels from C to D = $\frac{60}{30}$ = 2 hrs. Total time taken by Richa to travels from A to D = 4 + 3 + 2 = 9 hrs. ∴ Average speed = $\frac{180}{9}$ = 20 km/hr.

105.

In a 100 metres race, Kamal defeats Bimal by 5 seconds. If the speed of Kamal is 18 kmph, then the speed of Bimal is :

(A) 15.4 kmph
(B) 14.5 kmph
(C) 14.4 kmph
(D) 14 kmph

Solution: Time taken by Kamal : $= \frac{100}{18 \times \frac{5}{18}} = 20 \sec$ Time taken by Bimal : $\begin{aligned} = 20 + 5 \\ = 25 \sec \end{aligned}$ Speed of Bimal : $\begin{aligned} = \frac{100}{25} \times \frac{18}{5} \\ = 14.4 kmph \end{aligned}$

106.

If a train with a speed of 60 km/hr. crosses a pole in 30 seconds. The length of the train (in metres) is :

(A) 1000 metres
(B) 900 metres
(C) 750 metres
(D) 500 metres

Solution: The length of pole is considered as negligible i.e., = 0 i.e., when a train crosses the pole, it covers the distance equal to the length of train So, the time will be taken by the train = 30 sec And speed = 60 km/hr Then the length of the train : = 60 kmh × 30 sec = 60 × $\frac{5}{18}$ × 30 metres = 10 × $\frac{5}{3}$ × 30 = 500 metres

107.

The distance between two stations A and B is 200 km. A train runs from A to B at a speed of 75 km/h, while another train runs from B to A at a speed of 85 km/h. What will be the distance between the two trains (in km) 3 minutes before they meet?

(A) 5
(B) 8
(C) 10
(D) 6

Solution: $\begin{aligned} Required distance = (75 + 85) \times 3 min \\ = 160 \times \frac{3}{60} \\ = 8 km \end{aligned}$

108.

Ravi and Ajay start simultaneously from a place A towards B 60 km apart. Ravi's speed is 4km/h less than that of Ajay. Ajay, after reaching B, turns back and meets Ravi at a places 12 km away from B. Ravi's speed is:

(A) 12 km/h
(B) 10 km/h
(C) 8 km/h
(D) 6 km/h

Solution: Ajay → (x + 4) kmph. A ________ 60 km _________ B Ravi → x kmph. Let the speed of Ravi be x kmph; Hence, Ajay's speed = (x + 4) kmph; Distance covered by Ajay = 60 + 12 = 72 km; Distance covered by Ravi = 60 - 12 = 48 km. According to question, $\begin{aligned} \frac{72}{x + 4} = \frac{48}{x} \\ or, \frac{3}{x + 4} = \frac{2}{x} \\ or, 3 x = 2 x + 8 \\ or, x = 8 kmph \end{aligned}$

109.

A bus covered a certain distance from village A to village B at the speed of 60 km/hr. However on its return journey it got stuck in traffic and covered the same distance at the speed of 40 km/hr and took 2 hours more to reach its destination. What is the distance covered between villages A and B ?

(A) 200 km
(B) 240 km
(C) 260 km
(D) Cannot be determined

Solution: Let the distance between village A and B be x km. Then, $\begin{aligned} \Rightarrow \frac{x}{40} - \frac{x}{60} = 2 \\ \Rightarrow x = 240 km \end{aligned}$

110.

A tiger is 50 of its own leaps behinds a deer. The tiger takes 5 leaps and per minutes to the deer's 4. If the tiger and the deer cover 8 m and 5 m per leap respectively, what distance will the tiger have to run before it caches the deer?

(A) 600 m
(B) 700 m
(C) 800 m
(D) 1000 m

Solution: Speed of tiger = 40 m/min Speed of deer = 20 m/min. Relative speed = 40 - 20 = 20 m/min. Initial difference in distance = 50 × 8 = 400 m Time take to catch = $\frac{400}{20}$ = 20 min. Distance traveled in 20 min,= 20 × 40 = 800 m

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Practice MCQ Questions and Answer on Speed Time and Distance

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