Practice MCQ Questions and Answer on Speed Time and Distance

351.

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.

352.

A thief running at 8 km/hr is chased by a policeman whose speed is 10 km/hr. If the thief is 100 metres ahead of the policeman, then the time required for the policeman to catch the thief will be :

(A) 2 minutes
(B) 3 minutes
(C) 6 minutes
(D) 10 minutes

Solution: Relative speed = (10 - 8) km/hr = 2 km/hr Required time = Time taken to cover 100 m at relative speed : $\begin{aligned} = (\frac{100}{2000}) hr \\ = (\frac{1}{20}) hr \\ = (\frac{1}{20} \times 60) min \\ = 3 minutes \end{aligned}$

353.

A car goes 10 meters in a second. Find its speed in km/hour.

(A) 40 km/hr
(B) 32 km/hr
(C) 48 km/hr
(D) 36 km/hr

Solution: Speed = 10/sec Speed = 10 × $\frac{18}{5}$ km/hr Speed = 36 km/hr

354.

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

355.

A postman goes with a speed of 36 km/h what is the speed of postman in m/s?

(A) 4.5 m/s
(B) 6 m/s
(C) 5 m/s
(D) 10 m/s

Solution: Speed = 36 km/h = $\frac{36 \times 5}{18}$ = 10 m/s

356.

Two buses start from a bus terminal with a speed of 20 km/h at interval 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 interval of 8 minutes?

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

Solution: Let Speed of the man is x kmph. Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph. Or, $20 \times \frac{10}{60} = \frac{8}{60} \times (20 + x)$ Or, 200 = 160 + 8x Or, 8x = 40 Hence, x = 5kmph. Detailed Explanation: A _____________M_______________B A = Bus Terminal. B = Let meeting point of first bus and the man and this distance is covered by Bus in 10 minutes. I.e. Distance A to be is covered first bus in 10 min. As AB distance can be covered by second bus in 10 minutes as well. Distance Covered by Bus in 10 min = AB = $\frac{20}{60} \times 10$ = $\frac{10}{3}$ km. Now, M is the Meeting Point of Second Bus with Man. Man covered distance B to M in 8 minutes. Now, Relative distance of both Man and Bus will be same as both are traveling in opposite direction of each other. Let Speed of the man = x kmph. Relative speed = 20 + x To meet at Point M, bus and Man has covered the distance (AB) in 8 minutes with relative speed. And Same AB distance is covered by bus in 10 minutes. Thus, Distance covered in 8 minutes with relative speed (20 + x) kmph = distance covered by bus in 10 minuted with speed 20 kmph.

357.

A train passes two persons walking in the same direction at a speed of 3 km/hr and 5 km/hr respectively in 10 seconds and 11 seconds respectively. The speed of the train is :

(A) 28 km/hr
(B) 27 km/hr
(C) 25 km/hr
(D) 24 km/hr

Solution: Let the train's speed = x km/hr $∵$ When train will cross a man then it covers only its length. $\begin{aligned} ∴ (x - 3) \times \frac{10}{60} = (x - 5) \times \frac{11}{60} \\ x = 25 km/hr \end{aligned}$ Note : Products of time × speed is always subtracted if both the men are running in the same direction and the products of time × speed is added only if the men are running opposite direction. ⇒ Here train's direction is not considered. But attention please ⇒ Always divided by the difference of time.

358.

A constable is 114 metre behind a thief. The constable runs 21 metres per minute and the thief runs 15 metres in a minute. In what time will the constable catch the thief ?

(A) 19 min
(B) 18 min
(C) 17 min
(D) 16 min

Solution: $\begin{aligned} V_{rel .} = (21 - 15) m/min \\ = 6 m/min \end{aligned}$ Time taken to catch the thief : = $\frac{114}{6}$ min = 19 min

359.

P and Q are 27 km away. Two trains with speed of 24 km/hr and 18 km/hr respectively start simultaneously from P and Q and travel in the same direction. They meet at a point R beyond Q. Distance QR is :

(A) 126 km
(B) 81 km
(C) 48 km
(D) 36 km

Solution: Relative speed = 24 - 18 = 6 km/hr Time required by faster train to overtake slower train : = $\frac{27}{6}$ = $4 \frac{1}{2}$ hr ∴ Distance between Q and R : = 18 × $4 \frac{1}{2}$ = 81 km

360.

A man performs $\frac{3}{5}$ of the total journey by rail, $\frac{7}{20}$ by bus and the remaining 6.5 km on foot. His total journey is :

(A) 65 km
(B) 100 km
(C) 120 km
(D) 130 km

Solution: Let the total journey be x km Then, $\begin{aligned} \Leftrightarrow \frac{3 x}{5} + \frac{7 x}{20} + 6.5 = x \\ \Leftrightarrow \frac{12 x + 7 x + 20 \times 6.5}{20} = x \\ \Leftrightarrow 12 x + 7 x + 20 \times 6.5 = 20 x \\ \Leftrightarrow x = 130 km \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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