Practice MCQ Questions and Answer on Speed Time and Distance

91.

Rani goes to school from her house in 30 minutes. Raja takes 45 minutes in covering the same distance. Find the ratio between time taken by Rani and Raja ?

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

92.

In a race of one kilometre. A gives B a start of 100 metres and still wins by 20 seconds. But if A gives B a start of 25 seconds, B wins by 50 metres. The time taken by A to run kilometre is :

(A) 17 sec
(B) $\frac{500}{29} sec$
(C) $\frac{1200}{29} sec$
(D) $\frac{700}{29} sec$

Solution: Let the time taken by A to run 1000 metres = x seconds Time taken by B to sun 900 metres = x + 20 seconds Speed of A and B : $= \frac{1000}{x}, \frac{900}{x + 20}$ respectively According to the question, $\frac{950}{\frac{1000}{x}} - \frac{1000}{\frac{900}{x + 20}} = 25$ (As B is the winner now) $\begin{aligned} \Rightarrow \frac{950 x}{1000} - \frac{1000 (x + 20)}{900} = 25 \\ \Rightarrow \frac{19 x}{20} - \frac{10 x + 200}{9} = 25 \\ \Rightarrow 171 x - 200 x - 4000 = 25 \times 20 \times 9 \\ \Rightarrow 29 x = 500 \\ \Rightarrow x = \frac{500}{29} sec \end{aligned}$

93.

A plane left half an hour later than the scheduled time and in order to reach its destination 1500 kilometers away in time, it had to increase its speed by 33.33 percent over its usual speed. Find its increased speed.

(A) 250 kmph
(B) 500 kmph
(C) 750 kmph
(D) 1000 kmph

94.

Rohan, Shikha's boyfriend, had to pick her from her home for a live concert on her 23^rd birthday. The venue of the concert and Shikha's home were in opposite directions from Rohan's office. He got late because of some work at office and realised that if he goes to pick Shikha from her home, which was a 48-minute drive from his office, they would be late for the show by 16 minutes. He asked her to start from her home towards his office in an auto-rickshaw and himself started driving towards her home. Both of them started simultaneously, he picked her as soon as they met and they managed to reach the venue just in time for the concert. If Rohan drives at an average speed of 60 km/hr, find the speed (in km/hr) of the auto-rickshaw.

(A) 16 km
(B) 20 km
(C) 12 km
(D) 15 km

Solution: Let the speed of the rickshaw = S km/h. Let after time T they meet. So, 60t + ST = $\frac{4 \times 60}{5}$ $[Since, 48 min = \frac{4}{5} hour]$ Rohan saves 16 min. by making Shikha move as well. So, she saves 16 min by moving for distance X (Let). 2X = $\frac{60 \times 16}{5}$ X = 8 km Speed of Auto rickshaw = 12 km. $(ST = 8 km, T = \frac{2}{3})$

95.

The area of square park is 25 sq. Km. Time taken to complete a round of the field once, at a speed of 3 kmph is:

(A) 4 hours 60 minutes
(B) 4 hours 50 minutes
(C) 6 hours 40 minutes
(D) 5 hours 40 minutes

Solution: Area of square = side × side= 25 sq. km. Side = 5 km. perimeter = 4 × 5 = 20 km. Time taken to complete one round with speed 3 kmph, = $\frac{20}{3}$ = 6.66 hours = 6 hour 40 minutes

96.

Two boats A and B start towards each other from two places, 108 km apart. Speed of the boats A and B in still water are 12 km/h and 15 km/h respectively. If A proceeds down and B up the stream, they will meet after:

(A) 4.5 hours
(B) 4 hours
(C) 5.4 hours
(D) 6 hours

Solution: Let the speed of the stream be x kmph and both the boats meet after t hour. According to the question, (12 + x) × t + (15 - x) × t = 108 Or, 12t + 15t = 108 Or, 27t = 108 ∴ t = $\frac{108}{27}$ = 4 hours

97.

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 sec, B in 308 sec and C in 198 sec, all starting from the same point. After what time will they again meet at the stating point again ?

(A) 46 min 12 sec
(B) 45 min
(C) 42 min 36 sec
(D) 26 min 18 sec

Solution: Time taken by A = 252 sec = 22 ×32 ×7 Time taken by B = 308 sec = 22 × 7 × 11 Time taken by C = 198 sec = 2 × 9 × 11 Together will meet at starting point : = L.C.M. (252, 308, 198) = 22 × 32 × 7 × 11 sec So, required time (minutes) $\begin{aligned} = \frac{4 \times 9 \times 7 \times 11}{60} \\ = 46 min 12 \sec \end{aligned}$

98.

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.

99.

Walking at 60% of his usual speed, a man reaches his destination 1 hour 40 minutes late. His usual time (in hours) to reach the destination is:

(A) $3 \frac{1}{4}$
(B) $2 \frac{1}{2}$
(C) $3 \frac{1}{8}$
(D) $2 \frac{1}{4}$

Solution: $60 % = \frac{3}{5}$ Let the speed of the man be 5x 60% of the speed = 5x × $\frac{3}{5}$ = 3x Ratio of speed of man before and after = 5x : 3x As we know, speed is inversely proportional to time. Time ratio of man before and after = 3x : 5x According to the question 5x - 3x = 1 hr 40 min 5x - 3x = $(1 + \frac{40}{60}) hr$ 2x = $\frac{5}{3}$ x = $\frac{5}{3 \times 2}$ x = $\frac{5}{6}$ hr Required time = 3x = 3 × $\frac{5}{6}$ = $2 \frac{1}{2}$ hr

100.

A man walk a certain distance and rides back in 4 hours 30 minutes. He could ride both ways in 3 hours. The time required by the man to walk both ways is :

(A) 4 hours 30 minutes
(B) 4 hours 45 minutes
(C) 5 hours
(D) 6 hours

Solution: Time taken to ride one way : = $\frac{3}{2}$ = 1.5 hrs Time taken to walk one way : = 4.5 - 1.5 = 3 hrs Time taken to walk both way : = 3 × 2 = 6 hours

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Practice MCQ Questions and Answer on Speed Time and Distance

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