Practice MCQ Questions and Answer on Speed Time and Distance

71.

R jogs at twice the speed of walking and runs at twice the speed of jogging. From his home to office, he covers half of the distance by walking and the rest by jogging. From his office to home, he covers half the distance jogging and the rest by running. What is his average speed (in km/h) in a complete round from his home to office and back home if the distance between his office and home is 10 km and he walks at the speed of 5 km/h?

(A) $\frac{90}{8}$
(B) $\frac{60}{8}$
(C) $\frac{60}{9}$
(D) $\frac{80}{9}$

Solution: Walking speed = 5 kmph Jogging speed = 10 kmph Running speed = 20 kmph $\begin{aligned} Average speed = \frac{20}{\frac{5}{5} + \frac{5}{10} + \frac{5}{10} + \frac{5}{20}} \\ = \frac{20}{2 + \frac{5}{20}} \\ = \frac{20 \times 20}{45} \\ = \frac{80}{9} kmph \end{aligned}$

72.

In a flight of 6000 km, an aircraft was slowed down due to bad weather. The average speed for the trip was reduced by 400 kmph and the time of flight increased by 30 minutes. The original planned duration of the flight was :

(A) $2 \frac{1}{2}$ hours
(B) $3 \frac{1}{3}$ hours
(C) $4 \frac{1}{3}$ hours
(D) $5 \frac{1}{3}$ hours

Solution: Let the original planned duration of the flight be x hours Then, $\begin{aligned} \Leftrightarrow \frac{6000}{x} - \frac{600}{(x + \frac{1}{2})} = 400 \\ \Leftrightarrow \frac{6000}{x} - \frac{12000}{(2 x + 1)} = 400 \\ \Leftrightarrow \frac{15}{x} - \frac{30}{(2 x + 1)} = 1 \\ \Leftrightarrow 2 x^{2} + x - 15 = 0 \\ \Leftrightarrow (x + 3) (2 x - 5) = 0 \\ \Leftrightarrow x = \frac{5}{2} \\ \Leftrightarrow x = 2 \frac{1}{2} \end{aligned}$

73.

Each wheel of a car is making 5 revolutions per second. If the diameter of a wheel is 84 cm, then the speed of the car in cm/sec would be :

(A) 420 cm/sec
(B) 264 cm/sec
(C) 1000 cm/sec
(D) 1320 cm/sec

Solution: Circumference of the circle = 2πr = 2πr = 2 × $\frac{22}{7}$ × 42 = 264 cm Distance cover in 1 sec = 264 × 5 = 1320 cm/sec

74.

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

(A) 100 kmph
(B) 110 kmph
(C) 120 kmph
(D) 130 kmph

Solution: $\begin{aligned} Let speed of the car be x k m p h \\ Then, speed of the train \\ = \frac{150}{100} x = \frac{3}{2} x k m p h \\ ∴ \frac{75}{x} - \frac{75}{(3 / 2) x} = \frac{125}{10 \times 60} \\ \Rightarrow \frac{75}{x} - \frac{50}{x} = \frac{5}{24} \\ \Rightarrow x = \frac{25 \times 24}{5} \\ \Rightarrow x = 120 k m p h \end{aligned}$

75.

Two trains of equal length, running in opposite directions, pass a pole in 18 and 12 seconds. The trains will cross each other in:

(A) 14.4 seconds
(B) 15.5 seconds
(C) 18.8 seconds
(D) 20.2 seconds

Solution: Let length of each train be x meter. Then, speed of 1st train = $\frac{x}{18}$ m/sec Speed of 2nd train = $\frac{x}{12}$ m/sec Now, When both trains cross each other, time taken $\begin{aligned} = \frac{2 x}{\frac{x}{18} + \frac{x}{12}} \\ = \frac{2 x}{\frac{2 x + 3 x}{36}} \\ = \frac{2 x \times 36}{5 x} \\ = \frac{72}{5} \\ = 14.4 seconds \end{aligned}$

76.

A is twice fast as B and B is thrice fast as C is. The journey covered by C in $\frac{3}{2}$ hours will be covered by A is:

(A) 15 minutes
(B) 20 minutes
(C) 30 minutes
(D) 1 hour

Solution: Let speed of C = X Then Speed of B = 3X Then Speed of A = 6X Ratio of the speed of A and C = 1 : 6 So, Greater the speed less time taken in journey. C’s speed is 6 times less than A So A will take $\frac{1}{6}$ of the total time taken C to covered same distance. So, Time taken by A $\begin{aligned} = \frac{3}{2 \times 6} \\ = \frac{1}{4} hours \\ = 15 minutes \end{aligned}$

77.

Ashok left from place A for place B at 8 am and Rahul left place B for place A at 10.00 am the distance between place A and B is 637 km. If Ashok and Rahul are travelling at a uniform speed of 39 kmph and 47 kmph respectively, at what time will they meet ?

(A) 5.30 pm
(B) 4.30 pm
(C) 5 pm
(D) 4 pm

Solution: Speed of Ashok = 39 km/ph Speed of Rahul = 47 km/ph Distance between place A and place B = 637 km Distance covered by Ashok (from 8 am to 10 am) in 2 hours = 2 × 39 = 78 km ∴ Remaining distance = 637 - 78 = 559 Relative speed = 39 + 47 = 86 km/ph ∴ Time taken to travel 559 km = $\frac{559}{86}$ = 6.5 hours So, they meet at = (10 am + 6.5 hrs) = 4.30 pm

78.

A cyclist rides 24 km at 16 kmph and further 36 km at 15 kmph. Find his average speed for the journey ?

(A) 15.38 km/hr
(B) 15.5 km/hr
(C) 16 km/hr
(D) None of these

Solution: total distance travelled : = (24 + 36) km = 60 km Total time taken : $\begin{aligned} = (\frac{24}{16} + \frac{36}{15}) hrs \\ = (\frac{3}{2} + \frac{12}{5}) hrs \\ = \frac{39}{10} hrs \end{aligned}$ ∴ Average speed : $\begin{aligned} = (60 \times \frac{10}{39}) km/hr \\ = (\frac{200}{13}) km/hr \\ = 15.38 km/hr \end{aligned}$

79.

Shyam drives his car 30 km at a speed of 45 km/h and, for the next 1 h 20 m, he drives it at a speed of 51 km/h. Find his average speed (in km/h) for the entire journey.

(A) 49
(B) 48
(C) 48.5
(D) 47

Solution: $\begin{aligned} Time = 1 hr 20 m = \frac{4}{3} hr \\ Speed = 51 km/hr \\ ∴ Distance = \frac{4}{3} \times 51 = 68 \\ Average speed = \frac{Total Distance}{Total Time} \\ = \frac{30 + 68}{\frac{30}{45} + \frac{4}{3}} \\ = \frac{98}{\frac{2}{3} + \frac{4}{3}} \\ = 49 km/hr \end{aligned}$

80.

An individual is cycling at a speed of 25 km per hour. He catches his predecessor who had started earlier in two hours. What is the speed of his predecessor who had started 3 hours earlier ?

(A) 15 kmph
(B) 12 kmph
(C) 10 kmph
(D) 8 kmph

Solution: The distance covered in two hour, = 2 × 25 = 50 km Time taken by first individual = (3h + 2h) = 5h Then, the speed of predecessor, = $\frac{50}{5}$ = 10 kmph.

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Practice MCQ Questions and Answer on Speed Time and Distance

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