Practice MCQ Questions and Answer on Speed Time and Distance

141.

A train covers a distance of 12 km in 12 minutes. If its speed is decreased by 5 km/h, then the time taken by it to cover the distance of 22 km will be:

(A) 24 minutes
(B) 20 minutes
(C) 22 minutes
(D) 18 minutes

Solution: Initial speed = $\frac{12 \times 60}{12}$ = 60 km/h New speed = 60 - 5 = 55 km/h Time taken to cover 22 km = $\frac{22 \times 60}{55}$ = 24 min

142.

A thief seeing a policeman at a distance of 150 metres starts running at 10 kmph and the policeman gives immediate chase at 12 kmph. When the thief is overtaken the thief has traveled a distance of:

(A) 750 m
(B) 900 m
(C) 800 m
(D) 1 km

Solution: P__150m___T______x m (Let)_______Q. Let Policeman caught thief at a distance (x + 150)m. And Thief has traveled x m. Speed of Policeman $\begin{aligned} = 12 kmph \\ = \frac{12 \times 5}{18} \\ = \frac{60}{18} m/sec \end{aligned}$ Speed of thief $\begin{aligned} = 10 km \\ = \frac{10 \times 5}{18} \\ = \frac{50}{18} m/sec \end{aligned}$ In this case time is constant means Policeman covered (x + 150)m in same time thief covered x m. Thus, $\begin{aligned} \frac{Speed of the thief}{Speed of Policeman} = \frac{x}{150 + x} \\ \Rightarrow \frac{50}{60} = \frac{x}{150 + x} \\ \Rightarrow 7500 + 50 x = 60 x \\ \Rightarrow 10 x = 7500 \\ \Rightarrow x = 750 m \end{aligned}$ So, Thief has traveled 750 m before the caught.

143.

A train, 110 m long is running at a speed of 60 km/hr. How many seconds does it to cross another train, 170 m long standing on parallel track ?

(A) 15.6 seconds
(B) 16.8 seconds
(C) 17.2 seconds
(D) 18 seconds

Solution: Given : Speed of running train = 60 km/hr Length of running train = 110 metres ⇒ Length of standing train = 170 metres ⇒ Speed of the standing train = 0 km/hr ⇒ Time taken by running train to cross the standing train : $\begin{aligned} \Rightarrow Time = \frac{(100 + 170) metres}{60 km/hr} \\ \Rightarrow Time = \frac{280 \times 18}{60 \times 5} \\ \Rightarrow Time = 16 .8 seconds \end{aligned}$

144.

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 $\begin{aligned} \Rightarrow \frac{x - 12}{x - 18} = \frac{x}{x - 8} \\ \Rightarrow (x - 12) (x - 8) = x (x - 18) \\ \Rightarrow x^{2} - 20 x + 96 = x^{2} - 18 x \\ \Rightarrow 96 = 2 x \\ \Rightarrow x = 48 metres \end{aligned}$

145.

A certain distance is covered by a cyclist at a certain speed. If a jogger covers half the distance in double the time, the ratio of the speed of the jogger to that of the cyclist is :

(A) 1 : 2
(B) 2 : 1
(C) 1 : 4
(D) 4 : 1

Solution: Let the distance covered by the cyclist be x and the time taken be y Then, required ratio : $\begin{aligned} = \frac{\frac{1}{2} x}{2 y} : \frac{x}{y} \\ = \frac{1}{4} : 1 \\ = 1 : 4 \end{aligned}$

146.

A car moves a distance of 600 km with uniform speed. The number of hours taken for the journey is $\frac{2}{3}$ of the number representing speed in km/h. The time taken to cover the distance is:

(A) 15 hours
(B) 18 hours
(C) 24 hours
(D) 20 hours

147.

A boy rides his bicycle 10 km at an average speed of 12 km/hr and travells 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) 11.0 km/hr
(D) 12.2 km/hr

Solution: $\begin{aligned} Average speed = \frac{Total distance}{Total time} \\ Average speed = \frac{10 + 12}{\frac{10}{12} + \frac{12}{10}} \\ Average speed = 10.8 km/hr \end{aligned}$

148.

A car traveled first 36 km at 6 km/h faster than the usual speed, but it returned the same distance at 6 km/h slower than usual speed. I the total time taken by car is 8 hours, for how many hours does it traveled at the faster speed ?

(A) 4 hours
(B) 3 hours
(C) 2 hours
(D) 1 hours

Solution: Let the original speed be S then,Total time taken, $\frac{36}{S - 6} + \frac{36}{S + 6}$ = 8 hours On solving the equation, we get S = 12, -3 Then possible value of S = 12 km/h. Thus, time taken by car at faster speed = $\frac{36}{12 + 6}$ = 2 hours

149.

In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is :

(A) 5 km/hr
(B) 6 km/hr
(C) 6.25 km/hr
(D) 7.5 km/hr

Solution: Let Abhay's speed be x km/hr Then, $\begin{aligned} \Rightarrow \frac{30}{x} - \frac{30}{2 x} = 3 \\ \Rightarrow 6 x = 30 \\ \Rightarrow x = 5 km/hr \end{aligned}$

150.

An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in $1 \frac{2}{3}$ hours, it must travel at a speed of:

(A) 300 kmph
(B) 360 kmph
(C) 600 kmph
(D) 720 kmph

Solution: $\begin{aligned} Distance = 240 \times 5 = 1200 km \\ Speed = \frac{Distance}{Time} \\ Speed = \frac{1200}{\frac{5}{3}} km/hr \\ [We can write 1 \frac{2}{3} hours as \frac{5}{3} hours] \\ ∴ Required speed \\ = 1200 \times \frac{3}{5} km/hr \\ = 720 km/hr \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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