Practice MCQ Questions and Answer on Speed Time and Distance

51.

I started on my bicycle at 7 am to reach a certain place. After going a certain distance, my bicycle went out of order. Consequently, I rested for 35 minutes and came back to my house walking all the way. I reached my house at 1 pm. If my cycling speed is 10 kmph and my walking speed is 1 kmph, then on my bicycle I covered a distance of :

(A) $4 \frac{61}{66}$ km
(B) $13 \frac{4}{9}$ km
(C) $14 \frac{3}{8}$ km
(D) $15 \frac{10}{21}$ km

Solution: Time taken : = 5 hrs 25 min = $\frac{65}{12}$ hrs Let the required distance be x km Then, $\begin{aligned} \Leftrightarrow \frac{x}{10} + \frac{x}{1} = \frac{65}{12} \\ \Leftrightarrow 11 x = \frac{650}{12} \\ \Leftrightarrow x = \frac{325}{66} \\ \Leftrightarrow x = 4 \frac{61}{66} km \end{aligned}$

52.

An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in $1 \frac{2}{3}$ hours, it must travel at a speed of :

(A) 300 kmph
(B) 360 kmph
(C) 600 kmph
(D) 720 kmph

Solution: $\begin{aligned} Distance = (240 \times 5) km \\ = 1200 km \\ ∴ Required Speed: \\ = \frac{Distance}{Time} \\ = \frac{1200}{\frac{5}{3}} \\ = (1200 \times \frac{3}{5}) km/hr \\ = 720 kmph \end{aligned}$

53.

A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is :

(A) $35$ km
(B) $36 \frac{2}{3}$ km
(C) $37 \frac{1}{2}$ km
(D) $40$ km

Solution: Let distance = x km and usual rate = y kmph $\begin{aligned} \Rightarrow \frac{x}{y} - \frac{x}{y + 3} = \frac{40}{60} \\ \Rightarrow 2 y (y + 3) = 9 x . . . . . (i) \end{aligned}$ And, $\begin{aligned} \Rightarrow \frac{x}{y - 2} - \frac{x}{y} = \frac{40}{60} \\ \Rightarrow y (y - 2) = 3 x . . . . . (i i) \end{aligned}$ On dividing (i) by (ii), we get : $\begin{aligned} \Rightarrow 2 (y + 3) = 3 (y - 2) \\ \Rightarrow y = 12 \end{aligned}$ ∴ Distance : $\begin{aligned} = x km \\ = (\frac{2 y (y + 3)}{9}) km \\ = (\frac{2 \times 12 \times 15}{9}) km \\ = 40 km \end{aligned}$

54.

A takes 2 hours more than B to walk d km, but if A doubles his speed, then he can make it in 1 hour less than B. How much times does B required for walking d km ?

(A) $\frac{d}{2}$ hours
(B) 3 hours
(C) 4 hours
(D) $\frac{2 d}{3}$ hours

Solution: Suppose B takes x hours to walk d km Then, A takes (x + 2) hours to walk d km A's speed = $(\frac{d}{x + 2})$ km/hr and B's speed = $(\frac{d}{x})$ km/hr A's new speed = $(\frac{2 d}{x + 2})$ km/hr $\begin{aligned} ∴ \frac{d}{(\frac{d}{x})} - \frac{d}{(\frac{2 d}{x + 2})} = 1 \\ \Leftrightarrow x - (\frac{x + 2}{2}) = 1 \\ \Leftrightarrow x - 2 = 2 \\ \Leftrightarrow x = 4 hours \end{aligned}$

55.

A boy rides his bicycle 10 km at an average speed of 12 km/hr and travells 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately :

(A) 10.4 km/hr
(B) 10.8 km/hr
(C) 11.0 km/hr
(D) 12.2 km/hr

Solution: $\begin{aligned} Average speed = \frac{Total distance}{Total time} \\ Average speed = \frac{10 + 12}{\frac{10}{12} + \frac{12}{10}} \\ Average speed = 10.8 km/hr \end{aligned}$

56.

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) $9 \frac{2}{3}$

Solution: $\begin{aligned} Let total distance = 100 \\ Average speed = \frac{Total Distance}{Total Time} \\ = \frac{100}{\frac{40}{8} + \frac{24}{9} + \frac{36}{12}} \\ = \frac{100}{5 + \frac{8}{3} + 3} \\ = \frac{100 \times 3}{32} \\ = \frac{75}{8} \\ = 9 \frac{3}{8} \end{aligned}$

57.

A dog start chasing to a cat 2 hour later. It takes 2 hours to dog to catch the cat. If the speed of the dog is 30 km/h. what is the speed of cat?

(A) 10 km/h
(B) 12 km/h
(C) 15 km/h
(D) 20 km/h

Solution: Let speed of the cat be S km/h. Time taken = $\frac{Distanced advanced}{Relative speed}$ 2 = $\frac{2 \times S}{30 - S}$ S = 15 km/h

58.

2 km 5 metres is equal to ?

(A) 2.5 km
(B) 2.005 km
(C) 2.0005 km
(D) 2.05 km

Solution: We know, 1 km = 1000 metres ⇒ 2 km 5 metres = 2 km + $\frac{5}{1000}$ km = 2 km + 0.005 km = 2.005 km

59.

Without stoppage, a train travels a certain distance with an average speed of 60 km/h, and with stoppage it covers the same distance with an average speed of 40 km/h. On an average, how many minutes per hour does train stop during the journey ?

(A) 20 min/hour
(B) 15 min/hour
(C) 10 min/hour
(D) 12 min/hour

60.

A is twice as fast as B and B is thrice as fast as C. The journey covered by C in $1 \frac{1}{2}$ hours will be covered by A in :

(A) 15 min
(B) 2 min
(C) 30 min
(D) 1 hour

Solution: Distance cover by C in 90 mintue will cover by B $\frac{90}{3}$ = 30 minute Distance cover by B in 30 mintue will cover by A $\frac{30}{2}$ = 15 minute

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Practice MCQ Questions and Answer on Speed Time and Distance

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