Practice MCQ Questions and Answer on Speed Time and Distance

121.

A man takes 6 hours 15 minutes in walking a distance and riding back to starting place. He could walk both ways in 7 hours 45 minutes. The time taken by him to ride back both ways is:

(A) 4 hours
(B) 4 hours 30 min.
(C) 4 hours 45 min.
(D) 5 hours

122.

A car runs first 275 km at an average speed of 50 km/h and the next 315 km at an average speed of 70 km/h. What is the average speed (in km/h) for the entire journey?

(A) 58.5
(B) 60
(C) 59
(D) 62

Solution: Time taken to cover 275 km = $\frac{275}{50}$ h = 5.5 h Time taken to cover 315 km = $\frac{315}{70}$ h = 4.5 h Average speed of entire journey $\begin{aligned} = \frac{Total distance}{Total time taken} \\ = \frac{590}{10} \\ = 59 Answer \end{aligned}$

123.

The distance between the places H and O is D units. The average speed that gets a person from H to O in a stipulated time is S units. He takes 20 minutes more time than usual if he travels at 60 km/h, and reaches 44 minutes early if he travels at 75 km/h. The sum of the numerical values of D and S is:

(A) 358
(B) 384
(C) 376
(D) 344

Solution: Let right time for the train to cover its journey (in minutes) be $t$ hours, then As we know, Distance = Speed × Time According to the question, $\begin{aligned} 60 \times (t + \frac{20}{60}) = 75 \times (t - \frac{44}{60}) \\ \Rightarrow 4 (t + \frac{1}{3}) = 5 (t - \frac{11}{15}) \\ \Rightarrow 4 t + \frac{4}{3} = 5 t - \frac{11}{3} \\ \Rightarrow 5 t - 4 t = \frac{11}{3} + \frac{4}{3} \\ \Rightarrow t = \frac{11 + 4}{3} \\ \Rightarrow t = 5 hours \\ Distance (D) = 60 \times (5 + \frac{20}{60}) \\ \Rightarrow D = 60 \times (5 + \frac{1}{3}) \\ \Rightarrow D = 60 \times \frac{16}{3} \\ \Rightarrow D = 320 km \end{aligned}$ Distance between H and O is 320 km Average speed (S) = $\frac{320}{5}$ = 64 km/hr ∴ Sum of numerical value of D and S = 320 + 64 = 384

124.

Two trains are running with speed 30 km/hr and 58 km/hr in the same direction, a man in the slower train passes the faster train in 18 sec. The length (in metres) of the faster train is :

(A) 70 metres
(B) 100 metres
(C) 128 metres
(D) 140 metres

Solution: Int e question it is given that, a man who sit in the slower train cross the faster train it means faster train cross the man 18 sec. ⇒ Relative speed of faster train and man in the same direction = (58 - 30) = 28 kmph So, the distance covered by faster train in 18 sec : = 28 kmph × 18 sec = 28 × $\frac{5}{18}$ × 18 = 140 metres

125.

The average speed of a train is 20% less on the return journey than on onward journey. The train halts for half an hour at the destination station before starting on the return journey. If the total time taken for the to and fro journey is 23 hours, covering a distance of 1000 km, the speed of the train on the return journey is:

(A) 60 km/hr
(B) 50 km/hr
(C) 40 km/hr
(D) 55 km/hr

Solution: Train was halted for half an hour So, total time taken in Journey = 23 - $\frac{1}{2}$ = 22.5 hours Average speed in Whole Journey = $\frac{1000}{22.5}$ = 44.5 km/hr The average speed on return journey is 20% less than onward journey. Therefore, ratio of average speed of onward and return journey, $\begin{aligned} = \frac{100}{80} \\ = \frac{5}{4} \end{aligned}$ Let average speed of onward journey = 5x Average speed on return journey = 4x Average speed on whole journey = $\frac{5 x + 4 x}{2}$ 44.5 = $\frac{5 x + 4 x}{2}$ 89 = 9x x = 9.88 Average speed on return = 9.88 × 4 = 39.52 = 40 km/hr (Approx.)

126.

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, $\begin{aligned} \Leftrightarrow \frac{3 x}{5} + \frac{7 x}{20} + 6.5 = x \\ \Leftrightarrow \frac{12 x + 7 x + 20 \times 6.5}{20} = x \\ \Leftrightarrow 12 x + 7 x + 20 \times 6.5 = 20 x \\ \Leftrightarrow x = 130 km \end{aligned}$

127.

Three runners A, B and C run a race, with runner A finishing 12 metres ahead of runner B and 18 metres ahead of runner C, in another race of same type runner B finished 8 metres ahead of runner C. Each runner travels the entire distance at a constant speed. The length of the race is :

(A) 36 metres
(B) 48 metres
(C) 60 metres
(D) 72 metres

Solution: Let A finish the race of $x$ metres B finish the race of = $x$ - 12 C finish the race of = $x$ - 18 ..... (i) In another race of B & C B finish race of = $x$ C finish race of = $x$ - 8 ..... (ii) Ratio of speeds of B & C $\begin{aligned} \Rightarrow \frac{x - 12}{x - 18} = \frac{x}{x - 8} \\ \Rightarrow (x - 12) (x - 8) = x (x - 18) \\ \Rightarrow x^{2} - 20 x + 96 = x^{2} - 18 x \\ \Rightarrow 96 = 2 x \\ \Rightarrow x = 48 metres \end{aligned}$

128.

Anna left for city A from city B at 5.20 am. She travelled at the speed of 80 km/hr for 2 hours 15 minutes. After that the speed was reduced to 60 km/hr. If the distance between two cities is 350 kms, at what time did Anna reach reach city A ?

(A) 9.20 am
(B) 9.25 am
(C) 9.35 am
(D) 10.05 am

Solution: Answer & Solution Answer: Option E Solution: Distance covered in 2 hrs 15 min, i.e., = $2 \frac{1}{4}$ hrs = $(80 \times \frac{9}{4})$ hrs = 180 hrs Time taken to cover remaining distance : $\begin{aligned} = (\frac{350 - 180}{60}) hrs \\ = \frac{17}{6} hrs \\ = 2 \frac{5}{6} hrs \\ = 2 hrs 50 min \end{aligned}$ Total time taken : (2 hrs 15 min + 2 hrs 50 min) = 5 hrs 5 min So, Anna reached city A at 10.25 am

129.

The ratio of the distance between two place A and B to the distance between places B and C is 3 : 5. A man travels from A to B at a speed of x km/h and from B to C at a speed of 50 km/h. If his average speed for the entire journey is 40 km/h, then what is the value of (x - 10) : (x + 1)?

(A) 31 : 20
(B) 20 : 31
(C) 11 : 10
(D) 10 : 11

Solution: Let the distance between A and B to the distance between B and C be 3a and 5a respectively According to the question Time taken by man to travel from A and B at a speed of x km/hr = $\frac{3 a}{x}$ Time taken by man to travel from B and C at a speed of 50 km/hr = $\frac{5 a}{50} \Rightarrow \frac{a}{10}$ Average speed of the entire journey = $\frac{Total distance}{Total time taken}$ $\begin{aligned} \Rightarrow \frac{3 a + 5 a}{\frac{3 a}{x} + \frac{a}{10}} = 40 \\ \Rightarrow \frac{8 a}{\frac{3 a}{x} + \frac{a}{10}} = 40 \\ \Rightarrow 8 a = 40 a (\frac{3}{x} + \frac{1}{10}) \\ \Rightarrow \frac{1}{5} = \frac{30 + x}{10 x} \\ \Rightarrow 2 x = 30 + x \\ \Rightarrow 2 x - x = 30 \\ \Rightarrow x = 30 km/hr \end{aligned}$ Now, The value of (x - 10) : (x + 1) = (30 - 10) : (30 + 1) ⇒ 20 : 31 ∴ The required value is 20 : 31

130.

A moving train crosses a man standing on a platform and the platform 300 metres long in 10 seconds and 25 seconds respectively. What will be the time taken by the train to cross a platform 200 metre long?

(A) $16 \frac{2}{3} sec .$
(B) 18 sec.
(C) 20 sec.
(D) 22 sec.

Solution: If train crosses the platform i.e. it covers the distance equal to the length of train and platform. In the question train crosses the man who stands on the platform in 10 seconds and crosses the man + platform in 25 seconds i.e. train crosses the platform whose length is 300 metres in 25 - 10 = 15 seconds, here train's length is not added. So speed of the train $= \frac{300}{5} \Rightarrow 20 m/sec$ Length of the train = 10 × 20 = 200 metres (If train crosses the only man in 10 seconds) Time taken by the train to cross a platform 200 m long $\begin{aligned} = \frac{Length of train + Platform}{Speed} \\ = \frac{200 + 200}{20} \\ = \frac{400}{20} \\ = 20 \end{aligned}$ Time taken by train = 20 seconds

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Practice MCQ Questions and Answer on Speed Time and Distance

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