Practice MCQ Questions and Answer on Speed Time and Distance

331.

A and B start at the same time with speed of 40 km/hr. and 50 km/hr respectively. If in covering the journey A takes 15 min longer than B, the total distance of the journey is :

(A) 40 km
(B) 48 km
(C) 50 km
(D) 52 km

Solution: $\begin{aligned} A : B \\ Ratio of speed = 4 : 5 \\ Ratio of time = 5 : 4 \\ (5 - 4) unit = 15 unit \\ 1 unit = 15 min \\ So, time taken by B : \\ = 4 \times 15 \\ = 60 min \\ = 1 hr \\ Distance = S \times T \\ = 50 \times 1 \\ = 50 km \end{aligned}$

332.

With an uniform speed, a car covers a distance in 8 hours. had the speed been increased by 4 km/hr, the same distance could have been covered in 7 hr and 30 min. What is the distance covered ?

(A) 420 km
(B) 480 km
(C) 520 km
(D) 640 km

Solution: Let the speed of car be x km/hr Distance = Speed × Time Distance = 8x km According to the question, $\begin{aligned} \Rightarrow (x + 4) \times 7.5 = 8 x \\ \Rightarrow 7.5 x + 30 = 8 x \\ \Rightarrow 8 x - 7.5 x = 30 \\ \Rightarrow 0.5 x = 30 \\ \Rightarrow x = \frac{30}{0.5} = 60 km/hr \end{aligned}$ Required distance : = 8 × 60 = 480 km

333.

A tiger is 50 of its own leaps behinds a deer. The tiger takes 5 leaps and per minutes to the deer's 4. If the tiger and the deer cover 8 m and 5 m per leap respectively, what distance will the tiger have to run before it caches the deer?

(A) 600 m
(B) 700 m
(C) 800 m
(D) 1000 m

Solution: Speed of tiger = 40 m/min Speed of deer = 20 m/min. Relative speed = 40 - 20 = 20 m/min. Initial difference in distance = 50 × 8 = 400 m Time take to catch = $\frac{400}{20}$ = 20 min. Distance traveled in 20 min,= 20 × 40 = 800 m

334.

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.)

335.

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

336.

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

337.

A student walks from his house at a speed of $2 \frac{1}{2}$ km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hours and reaches 6 minutes before school time. How far is the school from his house ?

(A) $\frac{5}{4}$ km
(B) $\frac{7}{4}$ km
(C) $\frac{9}{4}$ km
(D) $\frac{11}{4}$ km

Solution: Difference between his reaching time : = (14 - 10) hrs = 4 hrs = 4 hrs → 6m + 6m (late + before) = 4 hrs → 12 minutes = 1 unit = $\frac{12}{4 \times 60}$ km ( $∵$ 1 m = 60 seconds) 1 unit = $\frac{1}{20}$ km Then, 35 units : = 35 × $\frac{1}{20}$ km = $\frac{7}{4}$ km Then the distance between his house and school is = $\frac{7}{4}$ km

338.

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 $\begin{aligned} \Rightarrow \frac{v}{u} = \frac{x}{5 k} = \frac{x + 12 k}{7 k} \\ \Rightarrow 7 x = 5 (x + 12 k) \\ \Rightarrow \frac{x}{k} = \frac{30}{1} \\ Thus, \\ \Rightarrow \frac{u}{v} = \frac{30}{5} = \frac{6}{1} \end{aligned}$ $or, 6 : 1$

339.

Two trains start at same time from stations A and B, 1800 km apart, and proceed towards each other at an average speed of 44 km/h and 46 km/h, respectively. Where will the trains meet?

(A) 920 km from station A
(B) 900 km from station B
(C) 880 km from station A
(D) 880 km from station B

Solution: Time = $\frac{1800}{44 + 46}$ Time = 20 hours 20 × 44 = 880 km from station A

340.

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

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Practice MCQ Questions and Answer on Speed Time and Distance

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