Practice MCQ Questions and Answer on Speed Time and Distance

281.

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.

282.

A and B travel the same distance at speed of 9 km/hr and 10 km/hr respectively. If A takes 36 min more than B, the distance travelled by each is :

(A) 48 km
(B) 54 km
(C) 60 km
(D) 66 km

Solution: Let tistance between A and B be $x$ km, then $\begin{aligned} \frac{x}{9} - \frac{x}{10} = \frac{36}{60} \\ \Rightarrow \frac{x}{90} = \frac{3}{5} \\ \Rightarrow x = \frac{3}{5} \times 90 \\ \Rightarrow x = 54 km \\ Hence option B is correct \end{aligned}$ $\begin{aligned} Alternate Solution: \\ Here, s_{1} = 9, t_{1} = x \\ s_{2} = 10, t_{2} = x - \frac{36}{60} \\ s_{1} t_{1} = s_{2} t_{2} \\ \Rightarrow 9 \times x = 10 \times (x - \frac{36}{60}) \\ \Rightarrow 99 x = 10 x - 6 \\ \Rightarrow 9 x = 6 \\ ∴ Distance travelled \\ = s_{1} \times t_{1} \\ = 9 \times 6 \\ = 54 km \end{aligned}$

283.

A person has to cover a distance of 160 km in 15 hours. If he covers $\frac{4}{5}$ of the distance in $\frac{2}{3}$ of the time, then what should be his speed (in km/h) to cover the remaining distance in the remaining time?

(A) 6
(B) 8
(C) 6.4
(D) 6.5

Solution: ${\frac{4}{5}}^{th}$ part of distance $= 160 \times \frac{4}{5} = 128$ Remaining distance = 32 km ${\frac{2}{3}}^{th}$ of time $= 15 \times \frac{2}{3} = 10 hr$ Remaining time = 5 hr Speed $= \frac{32}{5} = 6.4 km/hr$

284.

If A travels to his school from his house at the speed of 3 km/h, then he reaches the school 5 minutes late. If he travels at the speed of 4 km/h, he reaches the school 5 minutes earlier than school time. The distance of his school from his house is:

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

Solution: Let the distance between school and home be x km. The difference of time when A goes school to school with these two different speed is 10 min $\begin{aligned} = \frac{10}{60} hour \\ Now, \\ \frac{x}{3} - \frac{x}{4} = \frac{10}{60} \\ Or, \frac{x}{12} = \frac{1}{6} \\ Or, x = \frac{12}{6} \\ = 2 km \end{aligned}$

285.

Jane travelled $\frac{4}{7}$ as many miles on foot as by water and $\frac{2}{5}$ as many miles on horseback as by water. If she covered total of 3036 miles, how many miles did she travel on foot ?

(A) 1540 miles
(B) 880 miles
(C) 756 miles
(D) 616 miles

Solution: Suppose Jane travelled x miles by water, $\frac{4 x}{7}$ miles on foot and $\frac{2 x}{5}$ miles on horseback. Then, $\begin{aligned} \Leftrightarrow x + \frac{4 x}{7} + \frac{2 x}{5} = 3036 \\ \Leftrightarrow \frac{69 x}{35} = 3036 \\ \Leftrightarrow x = (\frac{3036 \times 35}{69}) \\ \Leftrightarrow x = 1540 \end{aligned}$ ∴ Distance travelled on foot : $\begin{aligned} = (\frac{4}{7} \times 1540) miles \\ = 880 miles \end{aligned}$

286.

Two persons A and B started from two different places towards each other. If the ratio of their speed be 3 : 5, then what is the ratio of distance covered by A and B respectively till the point of meeting?

(A) 1 : 2
(B) 3 : 4
(C) 3 : 5
(D) 5 : 3

287.

Anita travels from house at $3 \frac{1}{2}$ ÃÂÃÂ km/h and reaches her school 6 minutes late. The next day she travels at $4 \frac{1}{2}$ ÃÂÃÂ km/h and reaches her school 10 minutes early. What is the distance her house and the school?

(A) 5.6 km
(B) 4.8 km
(C) 5.4 km
(D) 4.2 km

Solution: Distance between Home & School $= \frac{\frac{7}{2} \times \frac{9}{2}}{\frac{9}{2} - \frac{7}{2}} \times \frac{16}{60} = 4.2 km$

288.

Two planes move along a circle of circumference 1.2 kms with constant speeds. When they move in different directions, they meet every 15 seconds and when they move in the same direction one plane overtakes the other every 60 seconds. The speed of the slower plane is :

(A) 0.02 km/sec
(B) 0.03 km/sec
(C) 0.04 km/sec
(D) 0.05 km/sec

Solution: Let their speeds be x m/sec and y m/sec respectively. Then, $\begin{aligned} \frac{1200}{x + y} = 15 \\ \Rightarrow x + y = 80. . . . . (i) \end{aligned}$ And, $\begin{aligned} \frac{1200}{x - y} = 60 \\ \Rightarrow x - y = 20. . . . . (i i) \end{aligned}$ Adding (i) and (ii), we get : 2x = 100 or x = 50 Putting x = 50 in (i), we get : y = 30 Hence, speed of slower plane : = 30 m/sec = 0.03 km/sec

289.

A person covers 40% of the distance from A to B at 8 km/h, 40% of the remaining distance at 9 km/h and the rest at 12 km/h. His average speed (in km/h) for the journey is:

(A) $9 \frac{5}{8}$
(B) $9 \frac{1}{3}$
(C) $9 \frac{3}{8}$
(D) $9 \frac{2}{3}$

Solution: $\begin{aligned} Let total distance = 100 \\ Average speed = \frac{Total Distance}{Total Time} \\ = \frac{100}{\frac{40}{8} + \frac{24}{9} + \frac{36}{12}} \\ = \frac{100}{5 + \frac{8}{3} + 3} \\ = \frac{100 \times 3}{32} \\ = \frac{75}{8} \\ = 9 \frac{3}{8} \end{aligned}$

290.

A bus covered a certain distance from village A to village B at the speed of 60 km/hr. However on its return journey it got stuck in traffic and covered the same distance at the speed of 40 km/hr and took 2 hours more to reach its destination. What is the distance covered between villages A and B ?

(A) 200 km
(B) 240 km
(C) 260 km
(D) Cannot be determined

Solution: Let the distance between village A and B be x km. Then, $\begin{aligned} \Rightarrow \frac{x}{40} - \frac{x}{60} = 2 \\ \Rightarrow x = 240 km \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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