Practice MCQ Questions and Answer on Speed Time and Distance

11.

Two trains starting at the same time from two-stations 300 km apart and going in opposite directions, cross each other at a distance of 160 km from one of them. The ratio of their speeds is :

(A) 7 : 9
(B) 16 : 20
(C) 8 : 7
(D) 8 : 12

Solution: A → ___160________Meeting____140_____← B Ratio of their distance covered = $\frac{160}{140}$ = 8 : 7 Then, Ratio of their speeds = 8 : 7 [When time is constant Speed is proportional to Distance covered]

12.

A man completed a certain journey by a car. If he covered 30% of the distance at the speed of 20 km/hr, 60% of the distance at 40 km/hr and the remaining distance at 10 km/hr, his average speed for the whole journey was :

(A) 25 km/hr
(B) 28 km/hr
(C) 30 km/hr
(D) 33 km/hr

Solution: Let 10% of journey's = 40 km Then, total journey = 400 kms And, $Average speed = \frac{Total distance}{Total time}$ $\begin{aligned} 30 % of journey = 400 \times \frac{30}{100} \\ = 120 km \\ 60 % of journey = 400 \times \frac{60}{100} \\ = 240 km \\ 10 % of journey = 400 \times \frac{10}{100} \\ = 40 km \\ Average speed = \frac{400}{\frac{120}{20} + \frac{240}{40} + \frac{40}{10}} \\ Average speed = \frac{400}{(6 + 6 + 4)} \\ Average speed = \frac{400}{16} \\ Average speed = 25 km/hr \end{aligned}$

13.

Two buses start from a bus terminal with a speed of 20 km/h at interval of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at interval of 8 minutes?

(A) 3 km/h
(B) 4 km/h
(C) 5 km/h
(D) 7 km/h

Solution: Let Speed of the man is x kmph. Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph. Or, $20 \times \frac{10}{60} = \frac{8}{60} \times (20 + x)$ Or, 200 = 160 + 8x Or, 8x = 40 Hence, x = 5kmph. Detailed Explanation: A _____________M_______________B A = Bus Terminal. B = Let meeting point of first bus and the man and this distance is covered by Bus in 10 minutes. I.e. Distance A to be is covered first bus in 10 min. As AB distance can be covered by second bus in 10 minutes as well. Distance Covered by Bus in 10 min = AB = $\frac{20}{60} \times 10$ = $\frac{10}{3}$ km. Now, M is the Meeting Point of Second Bus with Man. Man covered distance B to M in 8 minutes. Now, Relative distance of both Man and Bus will be same as both are traveling in opposite direction of each other. Let Speed of the man = x kmph. Relative speed = 20 + x To meet at Point M, bus and Man has covered the distance (AB) in 8 minutes with relative speed. And Same AB distance is covered by bus in 10 minutes. Thus, Distance covered in 8 minutes with relative speed (20 + x) kmph = distance covered by bus in 10 minuted with speed 20 kmph.

14.

Two boys A and B start at the same time to ride from Delhi to Meerut, 60 km away. A travels 4 km an hours slower than B, B reaches Meerut and at once turns back meeting A, 12 km from Meerut. A's rate was :

(A) 4 km/hr
(B) 8 km/hr
(C) 12 km/hr
(D) 16 km/hr

Solution: Let A's speed = x km/hr Then, B's speed = (x + 4) km/hr Clearly, time taken by B to cover (60 + 12) i.e., 72 km = Time taken by A to cover (60 - 12) i.e., 48 km $\begin{aligned} ∴ \frac{72}{x + 4} = \frac{48}{x} \\ \Rightarrow 72 x = 48 x + 192 \\ \Rightarrow 24 x = 192 \\ \Rightarrow x = 8 \end{aligned}$ Hence, A's speed = 8 km/hr

15.

A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

(A) 35 km
(B) $36 \frac{2}{3}$ km
(C) $37 \frac{1}{2}$ km
(D) 40 km

Solution: $\begin{aligned} Let distance = x k m and usual rate = y k m p h \\ Then, \frac{x}{y} - \frac{x}{y + 3} = \frac{40}{60} \\ \Rightarrow 2 y (y + 3) = 9 x . . . . . . . . . . (i) \\ and, \frac{x}{y - 2} - \frac{x}{y} = \frac{40}{60} \\ \Rightarrow y (y - 2) = 3 x . . . . . . . . . . . . . (i i) \\ On dividing (i) by (ii), \\ We get x = 40 km \end{aligned}$

16.

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

17.

I started on my bicycle at 7 am to reach a certain place. After going a certain distance, my bicycle went out of order. Consequently, I rested for 35 minutes and came back to my house walking all the way. I reached my house at 1 pm. If my cycling speed is 10 kmph and my walking speed is 1 kmph, then on my bicycle I covered a distance of :

(A) $4 \frac{61}{66}$ km
(B) $13 \frac{4}{9}$ km
(C) $14 \frac{3}{8}$ km
(D) $15 \frac{10}{21}$ km

Solution: Time taken : = 5 hrs 25 min = $\frac{65}{12}$ hrs Let the required distance be x km Then, $\begin{aligned} \Leftrightarrow \frac{x}{10} + \frac{x}{1} = \frac{65}{12} \\ \Leftrightarrow 11 x = \frac{650}{12} \\ \Leftrightarrow x = \frac{325}{66} \\ \Leftrightarrow x = 4 \frac{61}{66} km \end{aligned}$

18.

A is 10 miles west of B. C is 30 miles north of B. D is 20 miles east of C. What is the distance from A to D ?

(A) 10 miles
(B) 30 miles
(C) $10 \sqrt{10}$ miles
(D) $10 \sqrt{13}$ miles

Solution: Answer & Solution Answer: Option E Solution: $\begin{aligned} Required distance : \\ A D = \sqrt{{(A E)}^{2} + {(D E)}^{2}} \\ = \sqrt{{(30)}^{2} + {(30)}^{2}} \\ = \sqrt{900 + 900} \\ = \sqrt{1800} \\ = 30 \sqrt{2} miles \end{aligned}$

19.

Two cars start simultaneously from cities A and B, towards B and A respectively, on the same route. Once the two cars reach their destinations they turned around and move towards the other city without any loss of time. The two cars continue shuttling in this manner for exactly 20 hours. If the speed of the car starting from A is 60km/hr and the speed of the car starting from B is 40km/hr and the distance between the two cities is 120 km, find the number of times the two cars cross each other?

(A) 8
(B) 10
(C) 12
(D) 20

20.

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.

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Practice MCQ Questions and Answer on Speed Time and Distance

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