Practice MCQ Questions and Answer on Speed Time and Distance

371.

A man covered a distance of 180 km in 4 hours on a bike. How much distance will be cover on a bicycle in 8 hours if he rides the bicycle at one-sixth the speed of the bike ?

(A) 54 km
(B) 60 km
(C) 72 km
(D) 84 km

Solution: Speed of the bike : $\begin{aligned} = (\frac{180}{4}) km/hr \\ = 45 km/hr \end{aligned}$ Speed of the bicycle : $\begin{aligned} = (\frac{1}{6} \times 45) km/hr \\ = 7. 5 km/hr \end{aligned}$ ∴ Required distance : = (7.5 × 8) km = 60 km

372.

A man takes 50 minutes to cover a certain distance at a speed of 6 km/hr. If he walks with a speed of 10 km/hr, he covers the same distance in :

(A) 10 min
(B) 20 min
(C) 30 min
(D) 40 min

Solution: $\begin{aligned} Distance = Speed \times Time \\ Distance = (6 \times \frac{50}{60}) km \\ Distance = 5 km \\ ∴ Required time \\ = \frac{Distance}{Speed} \\ = \frac{5}{10} hrs \\ = \frac{1}{2} hrs \\ = 30 min \end{aligned}$

373.

Walking $\frac{3}{4}$ of his normal speed, Rabi is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and office:

(A) 48 min.
(B) 60 min.
(C) 42 min.
(D) 62 min.

Solution: 1st method: $\frac{4}{3}$ of usual time = Usual time + 16 minutes; Hence, $\frac{1}{3} rd$ of usual time = 16 minutes; Thus, Usual time = 16 × 3 = 48 minutes. 2nd method: When speed goes down to $\frac{3}{4} th$ (i.e. 75%) time will go up to $\frac{4}{3} rd$ (or 133.33%) of the original time. Since, the extra time required is 16 minutes; it should be equated to $\frac{1}{3} rd$ of the normal time. Hence, the usual time required will be 48 minutes.

374.

If a train runs at 40 kmph, it reaches its destination late by 11 minutes but if it runs at 50 kmph, it is late by 5 min only. The correct time for the train to complete its journey is :

(A) 13 min
(B) 15 min
(C) 19 min
(D) 21 min

Solution: Let the correct time to complete the journey be x min Distance covered in (x + 11) min at 40 kmph = Distance covered in (x + 5) min at 50 kmph $\begin{aligned} ∴ \frac{(x + 11)}{60} \times 40 = \frac{(x + 5)}{60} \times 50 \\ \Leftrightarrow x = 19 min \end{aligned}$

375.

Ramesh travels 760 km to his home, partly by train and partly by car. He takes 8 hours, if he travels 160 km by train and the rest by car. He takes 12 minutes more, if he travels 240 km by train and the rest by car. What are the spends of the car and the train respectively ?

(A) 90 km/hr, 60 km/hr
(B) 100 km/hr, 80 km/hr
(C) 80 km/hr, 70 km/hr
(D) 100 km/hr, 90 km/hr

Solution: Let the speeds of the train and the car be x km/hr and y km/hr respectively. Then, $\begin{aligned} \Rightarrow \frac{160}{x} + \frac{600}{y} = 8 \\ \Rightarrow \frac{20}{x} + \frac{75}{y} = 1. . . . . (i) \end{aligned}$ And, $\begin{aligned} \Rightarrow \frac{240}{x} + \frac{520}{y} = 8 \frac{1}{5} \\ \Rightarrow \frac{240}{x} + \frac{520}{y} = \frac{41}{5} . . . . . (i i) \end{aligned}$ Multiplying (i) by 12 and subtracting (ii) from it, we get : $\begin{aligned} \Rightarrow \frac{380}{y} = 12 - \frac{41}{5} \\ \Rightarrow \frac{380}{y} = \frac{19}{5} \\ \Rightarrow y = (380 \times \frac{5}{19}) \\ \Rightarrow y = 100 \end{aligned}$ Putting y = 100 in equation (i), we get : $\begin{aligned} \Rightarrow \frac{20}{x} + \frac{3}{4} = 1 \\ \Rightarrow \frac{20}{x} = \frac{1}{4} \\ \Rightarrow x = 80 \end{aligned}$ ∴ 100 km/hr, 80 km/hr

376.

The average speed of a bus is one-third of the speed of a train. The train covers 1125 km in 15 hours. How much distance will the bus cover in 36 minutes ?

(A) 12 km
(B) 18 km
(C) 21 km
(D) 75 km

Solution: Answer & Solution Answer: Option E Solution: Speed of the train : $\begin{aligned} = (\frac{1125}{15}) km/hr \\ = 75 km/hr \end{aligned}$ Speed of the bus : $\begin{aligned} = (\frac{1}{3} \times 75) km/hr \\ = 25 km/hr \end{aligned}$ Distance covered by the bus in 60 min = 25 km Distance covered by the bus in 36 min : $\begin{aligned} = (\frac{25}{60} \times 36) km \\ = 15 km \end{aligned}$

377.

A train covers a distance in 30 minutes. If it runs at a speed of 54 km/h on an average. The speed at which the train must run to reduce the time of the journey to 20 minutes is:

(A) None of these
(B) 81 km/h
(C) 60 km/h
(D) 75 km/h

Solution: $\begin{aligned} Distance = 5 a \times \frac{1}{2} = 27 km \\ Time = 20 min = \frac{1}{3} hr \\ Speed = \frac{27}{\frac{1}{3}} = 81 km/hr \end{aligned}$

378.

The distance between place A and B is 999 km. An express train leaves place A at 6 am and runs at a speed of 55.5 km/hr. The train stops on the way for 1 hour 20 minutes. It reaches B at :

(A) 1.20 am
(B) 12 pm
(C) 6 am
(D) 11 pm

Solution: Time will be taken by train if it does not stop : $\begin{aligned} = \frac{Distance}{Speed} \\ = \frac{999 kms}{55.5 km/hr} \end{aligned}$ Without stop = 18 hr But if stops on the way for 1 hour 20 minutes before reaching B. Total time : = 18 hr + 1 hr 20 min = 19 hour 20 minutes Reaching time at B : = 6 am + 19 hr 20 min = 1.20 am

379.

In a race of 200 metres, B can give a start of 10 metres to A and C can give a start of 20 metres to B. The start that C can give to A, in the same race, is :

(A) 30 metres
(B) 25 metres
(C) 29 metres
(D) 27 metres

Solution: $\begin{aligned} A B C \\ 190 200 \\ 180 200 \\ A B C \\ 171 180 200 \end{aligned}$ ⇒ C can give a start of 200 - 171 = 29 metres to A

380.

A boy goes three equal distance, each of length x km, with speed of y km/hr, $\frac{3 y}{5}$ km/hr and $\frac{2 y}{5}$ respectively. If the total time taken is 1 hour, then x : y is equal to :

(A) 6 : 13
(B) 6 : 23
(C) 6 : 31
(D) 6 : 37

Solution: Total time taken : $\begin{aligned} = [\frac{x}{y} + \frac{x}{(\frac{3 y}{5})} + \frac{x}{(\frac{2 y}{5})}] hours \\ = (\frac{x}{y} + \frac{5 x}{3 y} + \frac{5 x}{2 y}) hours \\ = (\frac{6 x + 10 x + 15 x}{6 y}) hours \\ = (\frac{31 x}{6 y}) hours \\ ∴ \frac{31 x}{6 y} = 1 \\ \Leftrightarrow \frac{x}{y} = \frac{6}{31} \\ \Leftrightarrow x : y = 6 : 31 \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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