Practice MCQ Questions and Answer on Speed Time and Distance

111.

The speeds of three cars are the ratio 2 : 3 : 4. The ratio of the taken by these cars to travel the same distance is :

(A) 2 : 3 : 4
(B) 4 : 3 : 2
(C) 4 : 3 : 6
(D) 6 : 4 : 3

Solution: Ratio of speeds = 2 : 3 : 4 ∴ Ratio of times taken = $\frac{1}{2}$ : $\frac{1}{3}$ : $\frac{1}{4}$ = 6 : 4 : 3

112.

If a train with a speed of 60 km/hr. crosses a pole in 30 seconds. The length of the train (in metres) is :

(A) 1000 metres
(B) 900 metres
(C) 750 metres
(D) 500 metres

Solution: The length of pole is considered as negligible i.e., = 0 i.e., when a train crosses the pole, it covers the distance equal to the length of train So, the time will be taken by the train = 30 sec And speed = 60 km/hr Then the length of the train : = 60 kmh × 30 sec = 60 × $\frac{5}{18}$ × 30 metres = 10 × $\frac{5}{3}$ × 30 = 500 metres

113.

A man travels the distance of his journey $\frac{3}{4}$ by bus, $\frac{1}{6}$ by rickshaw and remaining 2 km on foot. The total distance travelled by the man is :

(A) 12 km
(B) 18 km
(C) 20 km
(D) 24 km

Solution: Let the man travels 1 unit distance So, remaining distance $\begin{aligned} = 1 - (\frac{1}{6} + \frac{3}{4}) \\ = 1 - \frac{22}{24} \\ = \frac{1}{12} \\ ∵ \frac{1}{12} unit = 2 km \\ So, 1 unit = 24 km \end{aligned}$

114.

A man can row 15 km/h in still water and he finds that it takes him twice as much time to row up than as to row down the same distance in the river. The speed of current (in km/h) :

(A) 6 km/h
(B) 6.5 km/h
(C) 4.5 km/h
(D) 5 km/h

Solution: Let the speed of the current be x km/h. Now, according to the question, $\frac{15 + x}{15 - x} = \frac{2}{1}$ 15 + x = 30 - 2x 3x = 30 - 15 3x = 15 x = 5 km/h

115.

A train approaches a tunnel AB. Inside the tunnel a cat located at a point i.e. $\frac{5}{12}$ of the distance AB measured from the entrance A. When the train whistles the Cat runs. If the cat moves to the exit B, the train catches the cat exactly the exit. The speed of the train is greater than the speed of the cat by what order ?

(A) 1 : 6
(B) 3 : 5
(C) 6 : 1
(D) 5 : 4

Solution: Train(T)__________ A_____5k____CAT__________B T-------x-------------->----12k--------------------------> Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then $\begin{aligned} \Rightarrow \frac{v}{u} = \frac{x}{5 k} = \frac{x + 12 k}{7 k} \\ \Rightarrow 7 x = 5 (x + 12 k) \\ \Rightarrow \frac{x}{k} = \frac{30}{1} \\ Thus, \\ \Rightarrow \frac{u}{v} = \frac{30}{5} = \frac{6}{1} \end{aligned}$ $or, 6 : 1$

116.

A person has to cover a distance of 160 km in 15 hours. If he covers $\frac{4}{5}$ of the distance in $\frac{2}{3}$ of the time, then what should be his speed (in km/h) to cover the remaining distance in the remaining time?

(A) 6
(B) 8
(C) 6.4
(D) 6.5

Solution: ${\frac{4}{5}}^{th}$ part of distance $= 160 \times \frac{4}{5} = 128$ Remaining distance = 32 km ${\frac{2}{3}}^{th}$ of time $= 15 \times \frac{2}{3} = 10 hr$ Remaining time = 5 hr Speed $= \frac{32}{5} = 6.4 km/hr$

117.

A train without stoppage travels with an average speed of 80 km/h and with stoppage, it travels with an average speed of 64 km/h. For how many minutes does the train stop on an average per hour?

(A) 12
(B) 8
(C) 10
(D) 14

Solution: $\begin{aligned} (\frac{Faster speed - Slower speed}{Faster speed} \times 60) min/hr \\ = \frac{80 - 64}{80} \times 60 \\ = \frac{16}{80} \times 60 \\ = 12 min \end{aligned}$

118.

A train passes two bridges of length 800 m and 400 m in 100 seconds and 60 seconds respectively. The length of the train is:

(A) 80m
(B) 90m
(C) 200m
(D) 150m

Solution: 1st Method: Let length of the train be x m and speed of the train is s kmph. Speed, s = $\frac{x + 800}{100}$ . . . . . (i) Speed, s = $\frac{x + 400}{60}$ . . . . . (ii) Equating equation (i) and (ii), we get, Or, $\frac{x + 800}{100}$ = $\frac{x + 400}{60}$ Or, 5x + 200 = 3x + 2400 Or, 2x = 400 Or, x = 200m 2nd Method: As in both cases, the speed of the train is constant, and then we have; Time α distance $\begin{aligned} \frac{100}{60} = \frac{x + 800}{x + 400} \\ or, x = 200 m \end{aligned}$

119.

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.

120.

The speed of two trains are in the ratio 6 : 7. If the second train runs 364 km in 4 hours, then the speed of first train is :

(A) 60 km/hr
(B) 72 km/hr
(C) 78 km/hr
(D) 84 km/hr

Solution: Given, Ratio of speed of trains = 6 : 7 Second train covers 364 kms in 4 hours Then, its speed = $\frac{364}{4}$ = 91 km/hr In the question it is given that speed of the second train = 7 units But actual speed = 91 km/hr i.e., 7 units → 91 km 1 unit → 13 km Therefore, Speed of the first train is : = 6R = 6 × 13 = 78 km/hr

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Practice MCQ Questions and Answer on Speed Time and Distance

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