Practice MCQ Questions and Answer on Speed Time and Distance

401.

Between 5 am and 5 pm of a particular day for how many times are the minute and the hour hands together?

(A) 11
(B) 22
(C) 33
(D) 44

402.

A plane left half an hour later than the scheduled time and in order to reach its destination 1500 kilometers away in time, it had to increase its speed by 33.33 percent over its usual speed. Find its increased speed.

(A) 250 kmph
(B) 500 kmph
(C) 750 kmph
(D) 1000 kmph

403.

Mary jogs 9 km at a speed of 6 km per hour. At what speed would she need to jog during the next 1.5 hours to have an average of 9 km per hour for the entire jogging session ?

(A) 9 kmph
(B) 10 kmph
(C) 12 kmph
(D) 14 kmph

Solution: Let speed of jogging be x km/hr Total time taken : $\begin{aligned} = (\frac{9}{6} hrs + 1.5 hrs) \\ = 3 hrs \end{aligned}$ Total distance covered = $(9 + 1.5 x) km$ $\begin{aligned} ∴ \frac{9 + 1.5 x}{3} = 9 \\ \Leftrightarrow 9 + 1.5 x = 27 \\ \Leftrightarrow \frac{3}{2} x = 18 \\ \Leftrightarrow x = (18 \times \frac{2}{3}) \\ \Leftrightarrow x = 12 kmph \end{aligned}$

404.

Kim and Om are travelling from points A to B, which are 400 km apart. Travelling at a certain speed Kim takes one hour more than Om to reach point B. If Kim doubles her speed she will take 1 hour 30 mins less than Om to reach point B. At what speed was Kim driving from point A to B ? (In kmph)

(A) 90 kmph
(B) 70 kmph
(C) 160 kmph
(D) 80 kmph

Solution: Let the speed of Kim be a and that of Om be b Distance between point A and B = 400 km Then, $\frac{400}{a} - \frac{400}{b} = 1$ Let, $\begin{aligned} \frac{1}{a} = x and \frac{1}{b} = y \\ 400 x - 400 y = 1. . . . . (i) \end{aligned}$ Speed of km doubles and she will take lets time i.e., 1 hr 30 min than 0 metre Again, $\begin{aligned} \frac{400}{b} - \frac{400}{2 b} = \frac{3}{2} \\ ∴ 400 y - 200 x = \frac{3}{2} \\ 800 y - 400 x = 3 \end{aligned}$ Solving (i) and (ii), we get $\begin{aligned} 400 x - 400 y = 1 \\ \frac{- 400 x + 800 y = 3}{400 y = 4} \\ ∴ y = \frac{4}{100} = \frac{1}{100} km \\ ∴ b = 100 km \\ Now, \\ \frac{400}{a} - \frac{400}{100} = 4 \\ or, \frac{400}{a} \\ ∴ a = 80 kmph \end{aligned}$

405.

A coolie standing on a railway platform observe that a train going in one direction takes 4 second to pass him. Another train of same length going in opposite direction takes 5 seconds to pass him. The time taken (in seconds) by two train to cross each other will be:

(A) 35 seconds
(B) 36.5 seconds
(C) $\frac{40}{9}$ seconds
(D) 36 seconds

Solution: Let length of each train be X m, then $\begin{aligned} S_{1} = \frac{X}{4} and \\ S_{2} = \frac{X}{5} \\ Required time to cross each other, \\ = \frac{2 X}{\frac{X}{4} + \frac{X}{5}} \\ = \frac{40}{9} seconds \end{aligned}$

406.

Two boats A and B start towards each other from two places, 108 km apart. Speed of the boats A and B in still water are 12 km/h and 15 km/h respectively. If A proceeds down and B up the stream, they will meet after:

(A) 4.5 hours
(B) 4 hours
(C) 5.4 hours
(D) 6 hours

Solution: Let the speed of the stream be x kmph and both the boats meet after t hour. According to the question, (12 + x) × t + (15 - x) × t = 108 Or, 12t + 15t = 108 Or, 27t = 108 ∴ t = $\frac{108}{27}$ = 4 hours

407.

Two horses started simultaneously towards each other and meet each other 3 hr 20 mins later. How much time will it take the slower horse to cover the whole distance if the first arrived at the place of departure of the second 5 hours later than the second arrived at the point of departure of the first?

(A) 10 hours
(B) 5 hours
(C) 15 hours
(D) 6 hours

408.

The diameter of each wheel of car is 70 cm. If each wheel rotates 400 times per minutes, then the speed of the car (in km/hr) if : $(Take π = \frac{22}{7})$

(A) 5.28 km/hr
(B) 528 km/hr
(C) 52.8 km/hr
(D) 0.528 km/hr

Solution: Circumference of wheel = 2πr ⇒ 2πr ⇒ 2 × $\frac{22}{7}$ × $\frac{70}{2}$ ⇒ 220 cm Speed per hours : = $\frac{220 \times 400 \times 60}{1000 \times 100}$ = 52.8 km/hr

409.

A car driver, driving in a fog, passes a pedestrian who was walking at the rate of 2 km/h in the same direction. The pedestrian could see the car for 6 minutes and it was visible to him up to distance of 0.6 km. what was speed of the car?

(A) 30 kmph
(B) 15 kmph
(C) 20 kmph
(D) 8 kmph

410.

A thief sees a jeep at a distance of 250 m, coming towards him at 36 km/h. Thief takes 5 seconds to realize that there is nothing but the police is approaching him by the jeep and start running away from police at 54 km/h. But police realise after 10 second, when the thief starts running away, that he is actually a thief and gives chase at 72 km/h. How long after thief saw police did catchup with him and what is the distance police had travel to do so?

(A) 50 s, 1000 m
(B) 65 s, 1150 m
(C) 65 s, 1300 m
(D) 45 s, 1050 m

Solution: Initial speed of the police = 10 m/s Increased speed of the Police = 20 m/s Speed of the thief = 15 m/s Initial difference between speed of thief and police = 250 m After 5 seconds difference between thief and police = 200 + (5 × 10) = 250 m Now, the time required by police to catch the thief = $\frac{250}{5}$ = 50 s Total time = 50 + 15 = 65 s Total distance = 1000 + (15 × 10) = 1150 m.

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Practice MCQ Questions and Answer on Speed Time and Distance

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