Practice MCQ Questions and Answer on Speed Time and Distance

21.

Without stoppage, a train travels a certain distance with an average speed of 60 km/h, and with stoppage it covers the same distance with an average speed of 40 km/h. On an average, how many minutes per hour does train stop during the journey ?

(A) 20 min/hour
(B) 15 min/hour
(C) 10 min/hour
(D) 12 min/hour

22.

It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:

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

Solution: $\begin{aligned} Let the speed of the train be x km/hr \\ and that of the car be y km/hr \\ Then, \frac{120}{x} + \frac{480}{y} = 8 \\ \Rightarrow \frac{1}{x} + \frac{4}{y} = \frac{1}{15} . . . . . . . . . (i) \\ and, \frac{200}{x} + \frac{400}{y} = \frac{25}{3} \\ \Rightarrow \frac{1}{x} + \frac{2}{y} = \frac{1}{24} . . . . . . . . . (i i) \\ Solving (i) and (ii), \\ we get, x = 60 and y = 80 \\ ∴ Ratio of speeds \\ = 60 : 80 \\ = 3 : 4 \end{aligned}$

23.

A man travelled a certain distance train at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, the distance was :

(A) 25 km
(B) 30 km
(C) 20 km
(D) 15 km

Solution: $\begin{aligned} S_{average} = \frac{2 a b}{a + b} \\ = \frac{2 \times 25 \times 4}{25 + 4} \\ = \frac{200}{29} km/hr \\ Now, 2D = \frac{200}{29} \times (5 + \frac{4}{5}) \\ = \frac{200}{29} \times \frac{29}{5} \\ = 40 km \\ = D = 20 km \end{aligned}$

24.

In a race of 200 meters, B can gives a start of 10 meters to A, and C can gives a start of 20 meters to B. The starts that C can gives to A, in the same race is:

(A) 30 meters
(B) 25 meters
(C) 29 meters
(D) 27 meters

Solution: According to question, When B runs 200 m meters, A runs 190 meters; Hence, when B runs 180 meters, A runs = $\frac{190 \times 180}{200}$ = 171 meters; When C runs 200m, B runs 180 meters. Hence, C will give a start to A by = 200 - 171= 29 meters

25.

A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is:

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

Solution: $\begin{aligned} Let the total distance be 100 km . \\ Average speed \\ = \frac{total distance covered}{time taken} \\ = \frac{100}{\frac{30}{20} + \frac{60}{40} + \frac{10}{10}} \\ = \frac{100}{\frac{3}{2} + \frac{3}{2} + 1} \\ = \frac{100}{\frac{3 + 3 + 2}{2}} \\ = \frac{100 \times 2}{8} \\ = 25 kmph \end{aligned}$ Alternate 10% of journey's = 40 km Then, total journey = 400 kms $\begin{aligned} And, Average speed \\ = \frac{Total distance}{Total time} \\ 30 % of journey \\ = 400 \times \frac{30}{100} \\ = 120 km \\ 60 % of journey \\ = 400 \times \frac{60}{100} \\ = 240 km \\ 10 % of journey \\ = 400 \times \frac{10}{100} \\ = 40 km \\ Average speed \\ = \frac{400}{\frac{120}{20} + \frac{240}{40} + \frac{40}{10}} \\ = \frac{400}{6 + 6 + 4} \\ = \frac{400}{16} \\ ∴ Average speed = 25 km/hr \end{aligned}$

26.

An ant moved for several seconds and covered 3 mm in the first second and 4 mm more in each successive second than its predecessor. If the ant had covered 1 mm in the first second and 8 mm more in each successive second, then the difference between the path it would cover during the same time and actual path would be more than 6 mm but less than 30 mm. find the time for which the ant moved (in seconds).

(A) 5 s
(B) 4 s
(C) 6 s
(D) 2 s

27.

Points 'A' and 'B' are 70 km apart on a highway and two cars start at the same time. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other they meet in one hour. Find the speed of the two cars (in km/hr).

(A) 20, 30
(B) 40, 30
(C) 30, 50
(D) 20, 40

Solution: Let the speed of the cars be S1 and S2 And S1 - S2 = $\frac{70}{7}$ = 10 . . . . . (i) Also S1 + S2 = $\frac{70}{1}$ = 70 . . . . . (ii) From equation (i) and (ii) S1 = $\frac{10 + 70}{2}$ = 40 km /hr And S2 = $\frac{70 - 10}{2}$ = 30 km / hr ∴ Required speeds are 40 km/hr and 30 km/hr

28.

A ship leaves on a long voyage. When it is 18 miles from shore, a sea plane, whose speed is ten times that of the ship, is sent to deliver mail. How far from the shore does the sea plane catch up with the ship?

(A) 24 miles
(B) 25 miles
(C) 22 miles
(D) 20 miles

Solution: A _______18miles_____B ___x(let)___C Let the speed of the ship be 1 mile/h and then speed of the plane becomes 10 miles/h. Let time t is taken to cover the distance x from point B to point C by ship. Hence, in time t plane will cover (18 + x) miles. Now, $\frac{x}{1} = \frac{18 + x}{10}$ Or, 10x = 18 + x Or, x = 2 miles. Hence, ship will cover 2 miles in 2 hours and plane will cover 20 miles in same time. Short method: Relative speed = (10 - 1) = 9 miles/h. hence, time taken to catch up the ship is 2 hours so distance covered by the ship = 20 miles.

29.

How long must a driver take to drive the final 70 miles of a trip if he wants to average 50 miles an hour for the entire trip and during the first part of the trip he drove 50 miles in $1 \frac{1}{2}$ hours ?

(A) 54 min
(B) 1 hr
(C) 66 min
(D) 70 min

Solution: Total distance = (57 + 50) miles = 120 miles Average speed = 50 miles/hour Required time of journey : $\begin{aligned} = (\frac{120}{50}) hr \\ = \frac{12}{5} hr \\ = 2 \frac{2}{5} hr \\ = 2 hr 24 min \end{aligned}$ Time taken to cover 50 miles : $\begin{aligned} = 1 \frac{1}{2} hr \\ = 1 hr 30 min \end{aligned}$ ∴ Remaining time : = (2 hr 24 min - 1 hr 30 min) = 54 min

30.

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

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Practice MCQ Questions and Answer on Speed Time and Distance

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