Practice MCQ Questions and Answer on Speed Time and Distance

341.

Running at $\frac{5}{4}$ of his usual speed, an athlete improves his timing by 5 minutes. The time he usually takes to run the same distance is:

(A) 30 min.
(B) 28 min.
(C) 25 min.
(D) 23 min.

Solution: When the athlete walks at $\frac{5}{4}$ of his usual speed then he takes $\frac{4}{5}$ of his usual time $(As Speed \propto \frac{1}{Time})$ and he saves 5 minutes. → Usual time - $\frac{4}{5}$ × usual time = 5 minutes → $\frac{1}{5}$ × Usual time = 5 minutes Usual time = 25 minutes.

342.

A boy rides his bicycle 10 km at an average speed of 12 km/hr and again travels 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 km/hr
(D) 12.2 km/hr

Solution: Total distance travelled : = (10 + 12) km = 22 km Total time taken : $\begin{aligned} = (\frac{10}{12} + \frac{12}{10}) hrs \\ = \frac{61}{30} hrs \end{aligned}$ ∴ Average speed : $\begin{aligned} = (22 \times \frac{30}{61}) km/hr \\ = 10.8 km/hr \end{aligned}$

343.

A train 270 metres long is running at a speed of 36 km per hour then it will cross a bridge of length 180 metres in :

(A) 40 seconds
(B) 45 seconds
(C) 50 seconds
(D) 35 seconds

Solution: $\begin{aligned} Crossing time = \frac{l_{1} + l_{2}}{Speed} \\ = \frac{270 + 180}{36 \times \frac{5}{18}} \\ = \frac{450}{10} \\ = 45 seconds \end{aligned}$

344.

Travelling at 60 km/h, a person reaches his destination in a certain time. He covers 60% of his journey in ${\frac{2}{5}}^{th}$ of the time. At what speed (in km/h) should he travel to cover the remaining journey so that he reaches the destination right on time?

(A) 48
(B) 40
(C) 42
(D) 36

Solution: Let the total time = 5 hours The distance covered in 5 hours = 5 × 60 = 300 km Speed = $\frac{120}{3}$ = 40 km/hr

345.

Two runner start running together for a certain distance, one at 5 km/h and another at 8 km/h. The former arrives one and half an hour before the latter. The distance in Km is :

(A) 12 km
(B) 20 km
(C) 25 km
(D) 36 km

Solution: $\begin{aligned} Let x be the distance, then \\ \frac{x}{5} - \frac{x}{8} = \frac{3}{2} \\ \Rightarrow \frac{8 x - 5 x}{40} = \frac{3}{2} \\ \Rightarrow \frac{3 x}{40} = \frac{3}{2} \\ ∴ x = 20 km \end{aligned}$

346.

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

347.

In a race of one kilometre. A gives B a start of 100 metres and still wins by 20 seconds. But if A gives B a start of 25 seconds, B wins by 50 metres. The time taken by A to run kilometre is :

(A) 17 sec
(B) $\frac{500}{29} sec$
(C) $\frac{1200}{29} sec$
(D) $\frac{700}{29} sec$

Solution: Let the time taken by A to run 1000 metres = x seconds Time taken by B to sun 900 metres = x + 20 seconds Speed of A and B : $= \frac{1000}{x}, \frac{900}{x + 20}$ respectively According to the question, $\frac{950}{\frac{1000}{x}} - \frac{1000}{\frac{900}{x + 20}} = 25$ (As B is the winner now) $\begin{aligned} \Rightarrow \frac{950 x}{1000} - \frac{1000 (x + 20)}{900} = 25 \\ \Rightarrow \frac{19 x}{20} - \frac{10 x + 200}{9} = 25 \\ \Rightarrow 171 x - 200 x - 4000 = 25 \times 20 \times 9 \\ \Rightarrow 29 x = 500 \\ \Rightarrow x = \frac{500}{29} sec \end{aligned}$

348.

Due to inclement weather, an Aeroplane reduce its speed by 300 km/hr and reached the destination of 1200 km late by 2 hours. Then the schedule duration of the flight was :

(A) 1 hour
(B) 1.5 hours
(C) 2 hours
(D) 2.5 hours

Solution: Let the original speed of Aeroplane be x km/hr $\begin{aligned} \frac{1200}{(x - 300)} - \frac{1200}{x} = 2 \\ \frac{x - x + 300}{(x - 300) x} = \frac{2}{1200} \end{aligned}$ Original time : $\begin{aligned} t = \frac{1200}{600} \\ t = 2 hours \end{aligned}$

349.

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, $\begin{aligned} = \frac{2 \times 36 \times 45}{36 + 45} \\ = \frac{36 \times 90}{81} \\ = 40 kmph \end{aligned}$

350.

A man covers half of his journey at 6 km/h and the remaining half at 3 km/h. His average speed is

(A) 9 km/h
(B) 4.5 km/h
(C) 4 km/h
(D) 3 km/h

Solution: $\begin{aligned} Average speed \\ = \frac{2 x y}{x + y} \\ = \frac{2 \times 6 \times 3}{6 + 3} \\ = \frac{36}{9} \\ = 4 kmph \end{aligned}$

«
1
2
...
32
33
34
35
36
37
38
...
45
46
Next ›
Last »

Indian Polity Mcq IAS GK Question India GK World GK GK questions GK Quiz Biology GK Science GK Physics GK Chemistry GK General Awareness Computer GK Political GK Indian Political GK Economics GK History GK Sports GK Geography GK Teaching Aptitude GK CTET/TET GK B Ed Entrance GK Banking GK SSC GK UPSC GK Indian Army GK Hindi Grammar GK Hindi Grammar MCQ Reasoning questions Bihar GK Rajasthan GK Jharkhand GK MP GK UP GK Chhattisgarh GK Haryana GK HP GK Maharashtra GK Electronics GK Agriculture GK States/Capitals GK Books Name & Author Arthimetic Verbal Reasoning Non Verbal Reasoning English

Practice MCQ Questions and Answer on Speed Time and Distance

General Knowledge