Practice MCQ Questions and Answer on Speed Time and Distance

61.

Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?

(A) 8 kmph
(B) 11 kmph
(C) 12 kmph
(D) 14 kmph

Solution: $\begin{aligned} Let the distance travelled by x km \\ Then, \frac{x}{10} - \frac{x}{15} = 2 \\ \Rightarrow 3 x - 2 x = 60 \\ \Rightarrow x = 60 k m \\ Time taken to travel 60 k m at 10 km/hr \\ = \frac{60}{10} h r s = 6 h r s \\ So, Robert started 6 hours before 2 P . M . i . e ., a t A . M . \\ ∴ Required speed \\ = \frac{60}{5} k m p h = 12 k m p h \end{aligned}$

62.

In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less then Sameer. Abhay's speed (in km/hr) is :

(A) 5
(B) 6
(C) 6.25
(D) 7.5

Solution: In these type of questions go through options to save your valuable time : Option (A) Abhay's speed = 5 km/hr Abhay's time = $\frac{30}{5}$ = 6 hr Sameer's time = 6 - 2 = 4 hr Abhay's new time : = $\frac{30}{5 \times 2}$ = 3 hr Hence option (A) is correct as it satisfies all the conditions.

63.

A train travelling at 48 kmph crosses another train, having half of its length and travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. The length of railway platform is:

(A) 200m
(B) 300m
(C) 350m
(D) 400m

Solution: Let the length of the train traveling at 48 kmph be 2x meters. And length of the platform is y meters. $\begin{aligned} Relative speed of train \\ = (48 + 42) kmph \\ = \frac{90 \times 5}{18} \\ = 25 m / \sec \\ And 48 kmph \\ = \frac{48 \times 5}{18} \\ = \frac{40}{3} m / \sec \\ According to the question, \\ \frac{2 x + x}{25} = 12; \\ O r, 3 x = 12 \times 25 = 300 \\ O r, x = \frac{300}{3} = 100 m \\ Then, length of the train \\ = 2 x \\ = 100 \times 2 = 200 m \\ \frac{200 + y}{\frac{40}{3}} = 45 \\ 600 + 3 y = 40 \times 45 \\ Or, 3 y = 1800 - 600 \\ = 1200 \\ Or, y = \frac{1200}{3} = 400 m \\ Length of the platform \\ = 400 m \end{aligned}$

64.

Raj and Prem walk in opposite direction at the rate 3 km/hr and 2 km per hour respectively. How far will they be from each other after 2 hrs ?

(A) 8 km
(B) 10 km
(C) 2 km
(D) 61 km

65.

Two men start together to walk a certain distance, one at 4 km/hr and another at 3 km/hr. The former arrives half an hour before the latter. Find the distance :

(A) 8 km
(B) 7 km
(C) 6 km
(D) 9 km

Solution: If the required distance be x km, then $\begin{aligned} \frac{x}{3} - \frac{x}{4} = \frac{1}{2} \\ \Rightarrow \frac{4 x - 3 x}{12} = \frac{1}{2} \\ \Rightarrow \frac{x}{12} = \frac{1}{2} \\ \Rightarrow x = 6 km \end{aligned}$

66.

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: $\begin{aligned} Let distance = x k m and usual rate = y k m p h \\ Then, \frac{x}{y} - \frac{x}{y + 3} = \frac{40}{60} \\ \Rightarrow 2 y (y + 3) = 9 x . . . . . . . . . . (i) \\ and, \frac{x}{y - 2} - \frac{x}{y} = \frac{40}{60} \\ \Rightarrow y (y - 2) = 3 x . . . . . . . . . . . . . (i i) \\ On dividing (i) by (ii), \\ We get x = 40 km \end{aligned}$

67.

A man walking at the rate of 5 km/hr crosses a bridge in 15 min. The length of the bridge (in metres) is :

(A) 600 metres
(B) 750 metres
(C) 1000 metres
(D) 1250 metres

Solution: $\begin{aligned} 15 min = \frac{1}{4} hrs \\ 1 hr \to 5 kms \\ \frac{1}{4} hr \to \frac{5}{4} kms \\ So, length of the bridge \\ = \frac{5}{4} km \\ = 1250 metres \end{aligned}$

68.

Buses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes ?

(A) 3 km/hr
(B) 4 km/hr
(C) 5 km/hr
(D) 7 km/hr

Solution: Let the speed of the man be x km/hr Then, Distance covered by the bus in 2 minutes = Distance covered by the man in 8 min $\begin{aligned} \Rightarrow 20 \times \frac{2}{60} = x \times \frac{8}{60} \\ \Rightarrow x = (\frac{2}{3} \times \frac{60}{8}) \\ \Rightarrow x = 5 km/hr \end{aligned}$

69.

Vinay and Mahesh are 250 metres apart from each other. They are moving towards each other with the speed of 36 km/hr and 54 km/hr respectively. In how much time will they meet each other?

(A) 15 seconds
(B) 10 seconds
(C) 20 seconds
(D) 12 seconds

Solution: $\begin{aligned} Time = \frac{250}{(36 + 54) \times \frac{5}{18}} \\ = \frac{250 \times 18}{90 \times 5} \\ = 10 second \end{aligned}$

70.

A man covers the journey from station A to station B at a uniform speed of 36 kmph and returns to A with a uniform speed of 45 kmph, his average speed for the whole journey is:

(A) 40 kmph
(B) 40.5 kmph
(C) 41 kmph
(D) 42 kmph

Solution: Average speed for this case is given by, $\frac{2 x y}{x + y},$ where X and Y are two different speeds. Average speed, $\begin{aligned} = \frac{2 \times 36 \times 45}{36 + 45} \\ = \frac{36 \times 90}{81} \\ = 40 kmph \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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