Practice MCQ Questions and Answer on Speed Time and Distance

431.

A thief steals a car at 1.30 pm and drive it off 40 km/hr. The theft is discovered at 2 pm and the owner sets off in another car at 50 km/hr he will catch the thief at :

(A) 5 pm
(B) 4 pm
(C) 4.30 pm
(D) 6 pm

Solution: Distance covered by thief in (2 pm - 1.30 pm) = $\frac{1}{2}$ hr at speed of 40 km/hr = 40 × $\frac{1}{2}$ = 20 kms Their relative speed in same direction : = (50 - 40) km/hr = 10 km/hr According to the question, 20 km, is the distance that has to be covered by owner to catch the thief. $\begin{aligned} Required time = \frac{20 km}{10 km/hr} \\ = 2 hours \end{aligned}$ Therefore, he will over take the thief at : = 2 pm + 2 hr = 4 pm

432.

The average speed of a bus is one-third of the speed of a train. The train covers 1125 km in 15 hours. How much distance will the bus cover in 36 minutes ?

(A) 12 km
(B) 18 km
(C) 21 km
(D) 75 km

Solution: Answer & Solution Answer: Option E Solution: Speed of the train : $\begin{aligned} = (\frac{1125}{15}) km/hr \\ = 75 km/hr \end{aligned}$ Speed of the bus : $\begin{aligned} = (\frac{1}{3} \times 75) km/hr \\ = 25 km/hr \end{aligned}$ Distance covered by the bus in 60 min = 25 km Distance covered by the bus in 36 min : $\begin{aligned} = (\frac{25}{60} \times 36) km \\ = 15 km \end{aligned}$

433.

A walk at a uniform rate of 4 km an hour; and 4 hr after his start, B bicycles after him at the uniform rate of 10 km an hour. How far from the starting point will B catch A ?

(A) 16.7 km
(B) 18.6 km
(C) 21.5 km
(D) 26.7 km

Solution: Distance covered by A in hrs = (4 × 4) km = 16 km Relative speed of B w.r.t. A = (10 - 4) km/hr = 6 km/hr Time taken to cover 16 km at relative speed : $\begin{aligned} = (\frac{16}{6}) hrs \\ = \frac{8}{3} hrs \end{aligned}$ Distance covered by in $\frac{8}{3}$ hrs $\begin{aligned} = (10 \times \frac{8}{3}) km \\ = (\frac{80}{3}) km \\ = 26 .7 km \end{aligned}$

434.

A man starts climbing a 11 m high wall at 5 pm. In each minute he climbs up 1 m but slips down 50 cm. At what time will he climb the wall?

(A) 5:30 pm
(B) 5:21 pm
(C) 5:25 pm
(D) 5:27 pm

Solution: 1st Method: Man climbs 1m and slips down 50 cm (0.5m) in one minute i.e. he climbs (1 - 0.5 = 0.5 m) in one minute. But in the last minute he will be climbing 1m as he gets on the top so no slip. Time taken to climb 11 meter = $\frac{10}{0.5} + 1$ = 21 minutes. He climbs the wall at 5:21 pm 2nd Method (short-cut): $= \frac{hight of pole - slipped distance}{climbed distance - slipped distance}$ $\begin{aligned} = \frac{x - z}{y - z} \times t \\ = \frac{11 - 0.5}{1 - 0.5} \times 1 \\ = 21 minutes \end{aligned}$

435.

A train 300 meres long is running at a speed of 25 metre per second. It will cross a bridge of 200 metres long in :

(A) 5 sec
(B) 10 sec
(C) 20 sec
(D) 25 sec

Solution: $\begin{aligned} Time = \frac{Distance}{Speed} \\ Time = \frac{300 + 200}{25} \\ Time = 20 \sec \end{aligned}$

436.

Two trains 105 meters and 90 meters long, run at the speeds of 45 kmph and 72 kmph respectively, in opposite directions on parallel tracks. The time which they take to cross each other, is:

(A) 8 seconds
(B) 6 seconds
(C) 7 seconds
(D) 5 seconds

Solution: Length of the 1st train = 105 m Length of the 2nd train = 90 m Relative speed of the trains,= 45 + 72 = 117 kmph= $\frac{117 \times 5}{18}$ = 32.5 m/sec Time taken to cross each other,= $\frac{Length of 1^{st} train + length of 2^{nd} train}{relative speed of the trains}$ ∴ Time taken = $\frac{195}{32.5}$ = 6 secs.

437.

Two cars start at the same time from in point and move alone two roads, at right angle to each other. The speeds are 36 km/hr and 48 km/hr respectively. After 15 sec distance between them will be :

(A) 400 m
(B) 150 m
(C) 300 m
(D) 250 m

Solution: Distance travelled by the 1st car in 15 seconds $\begin{aligned} OA = (36 \times \frac{5}{18}) \times 15 m \\ = 150 m \end{aligned}$ Distance travelled by the 2nd car in 15 seconds $\begin{aligned} OB = (48 \times \frac{5}{18}) \times 15 m \\ = 200 m \end{aligned}$ ∴ Distance between them after 15 seconds $\begin{aligned} AB = \sqrt{O A^{2} + O B^{2}} \\ = \sqrt{150^{2} + 200^{2}} \\ = \sqrt{22500 + 40000} \\ = \sqrt{62500} \\ = 250 m \end{aligned}$

438.

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 : $\begin{aligned} = (\frac{4}{7} \times 1540) miles \\ = 880 miles \end{aligned}$

439.

A man takes 100 minutes to cover a certain distance at a speed of 6 km/hr. If he walks with a speed of 10 km/hr, he covers the same distance in :

(A) 10 min
(B) 20 min
(C) 30 min
(D) 60 min

Solution: $\begin{aligned} Distance = Speed \times Time \\ Distance = (6 \times \frac{100}{60}) km \\ Distance = 10 km \end{aligned}$ $\begin{aligned} ∴ Required time = \frac{Distance}{Speed} \\ = \frac{10}{10} hrs \\ = 1 hr \\ = 60 min \end{aligned}$

440.

The average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and fro journey is 17 hours, covering a distance of 800 km. The speed of the train in the onward journey is :

(A) 45 km/hr
(B) 47.5 km/hr
(C) 52 km/hr
(D) 56.25 km/hr

Solution: Let the speed in return journey be x km/hr Then, speed in onward journey : $\begin{aligned} = (\frac{125}{100} x) km/hr \\ = (\frac{5}{4} x) km/hr \end{aligned}$ Average speed : $\begin{aligned} = (\frac{2 \times \frac{5}{4} x \times x}{\frac{5}{4} x + x}) km/hr \\ = (\frac{10 x}{9}) km/hr \\ ∴ (800 \times \frac{9}{10 x}) = 16 \\ \Leftrightarrow x = (\frac{800 \times 9}{16 \times 10}) \\ \Leftrightarrow x = 45 \end{aligned}$ So, speed in onward journey : $\begin{aligned} = (\frac{5}{4} \times 45) km/hr \\ = 56.25 km/hr \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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