Practice MCQ Questions and Answer on Speed Time and Distance

81.

A man goes downstream with a boat to some destination and returns upstream to his original place in 5 hours. If the speed of the boat in still water and the stream are 10 kmph and 4 kmph respectively, the distance of the destination from the starting place is:

(A) 16 km
(B) 18 km
(C) 21 km
(D) 25 km

Solution: Let the distance of the destination from the starting point = x km. Speed downstream = (10 + 4) = 14 kmph Speed upstream = (10 - 4) = 6 kmph According to the question, Total time taken = 5 hours $\begin{aligned} \frac{x}{14} + \frac{x}{6} = 5 \\ \frac{3 x + 7 x}{42} = 5 \\ or, 10 x = 42 \times 5 \\ or, x = \frac{42 \times 5}{10} \\ So, x = 21 km \end{aligned}$

82.

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

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

Solution: Let the speed of Ravi = x km/hr Then Ajay's speed will be = (x + 4) km/hr Total distance, covered by Ajay : = (60 + 12) km = 72 km Total distance, covered by Ravi : = (60 - 12) km = 48 km According to the question, They run at same time $\begin{aligned} \Rightarrow \frac{72}{(x + 4)} = \frac{48}{x} \\ \Rightarrow 72 x = 48 x + 192 \\ \Rightarrow 24 x = 192 \\ \Rightarrow x = 8 km/hr \end{aligned}$ Therefore, Ravi's speed = 8 km/hr

83.

A cyclist moving on a circular track of radius 100 meters completes one revolution in 2 minutes. What is the average speed of cyclist (approx.)?

(A) 314 m/min
(B) 200 m/min
(C) 300 m/min
(D) 900 m/min

Solution: Distance covered in one complete revolution $\begin{aligned} = 2 π r \\ = \frac{2 \times 100 \times 22}{7} \\ \approx 628 m \\ Average speed \\ = \frac{628}{2} \\ = 314 m / min \end{aligned}$

84.

Length of a train is 330 metres and it is moving at the speed of 72 km/hr. In how much time will it takes cross a platform of length 710 metres?

(A) 72 seconds
(B) 56 seconds
(C) 64 seconds
(D) 52 seconds

Solution: $\begin{aligned} Time = \frac{330 + 710}{72 \times \frac{5}{18}} Sec . \\ = \frac{1040}{20} Sec . \\ = 52 Sec . \end{aligned}$

85.

A train without stopping travels 60 km/h and with stoppage 40 km/h. What is the time taken for stoppage on a route 300 km?

(A) 10 hours
(B) 20 hours
(C) 5 hours
(D) 2.5 hours

Solution: Since, the train travels at 60 km/h, it's speed per minute is 1 km per minute. Hence, if it's speed with stoppage is 40 km/h, it will travel 40 minutes per hour i.e. train stops 20 min per hour. Time taken to travel 300 km with stoppage, $= \frac{300}{40} = 7.5 hours$ Time taken for stoppage as it stops for 20 min per hour = 7 × 20 + 10 = 140 + 10 = 150 minutes. = $\frac{150}{60}$ hours = 2.5 hours.

86.

There are 8 equidistant points A, B, C, D, E, F, G and H in the clockwise direction on the periphery of a circle. In a time interval t, a person reaches from A to C with uniform motion while another person reaches the point E from the point B during the same time interval with uniform motion. Both the persons move in the same direction along the circumference of the circle and start at the same instant. How much time after the start, will the two persons meet each other ?

(A) 4t
(B) 7t
(C) 9t
(D) Never

Solution: Distance covered by first person in time t = $\frac{2}{8}$ round = $\frac{1}{4}$ round Distance covered by second person in time t = $\frac{3}{8}$ round Speed of the first person = $\frac{1}{4 t}$ Speed of the second person = $\frac{3}{8 t}$ Since the two persons start from A and B respectively, so they shall meet each other when there is a difference of $\frac{7}{8}$ round between the two. Relative speed of A and B : $\begin{aligned} = (\frac{3}{8 t} - \frac{1}{4 t}) \\ = \frac{1}{8 t} \end{aligned}$ Time taken to cover $\frac{7}{8}$ round at this speed : $\begin{aligned} = (\frac{7}{8} \times 8 t) \\ = 7 t \end{aligned}$

87.

A car goes 20 metres in a second. Find its speed in km/hr.

(A) 20 km/hr
(B) 18 km/hr
(C) 72 km/hr
(D) 36 km/hr

Solution: As we know : ⇒ 1 m/s = $\frac{18}{5}$ km/hr ⇒ 20 m/s = $\frac{18}{5}$ × 20 km/hr = 72 km/hr

88.

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

89.

If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:

(A) 50 km
(B) 56 km
(C) 70 km
(D) 80 km

Solution: $\begin{aligned} Let the actual distance travelled be x k m \\ Then, \frac{x}{10} = \frac{x + 20}{14} \\ \Rightarrow 14 x = 10 x + 200 \\ \Rightarrow 4 x = 200 \\ \Rightarrow x = 50 k m \end{aligned}$

90.

A person travels 5x distance at a speed of 5 km/h, x distance at a speed of 5 km/h, and 4x distance at a speed of 6 km/h, and takes a total of 112 minutes. What is the total distance (in km) travelled by the person?

(A) 8
(B) 10
(C) 9
(D) 12

Solution: $\begin{aligned} \frac{5 x}{5} + \frac{x}{5} + \frac{4 x}{6} = \frac{112}{60} \\ \Rightarrow \frac{30 x + 6 x + 20 x}{30} = \frac{112}{60} \\ \Rightarrow 56 x = 56 \\ \Rightarrow x = 1 \\ ∴ Total distance = 5 x + x + 4 x \\ = 10 x \\ = 10 km \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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