Practice MCQ Questions and Answer on Speed Time and Distance

An aeroplane flies twice as fast as a train which covers 60 miles in 80 minutes. What distance will the aeroplane cover in 20 minutes ?

(A) 30 miles
(B) 35 miles
(C) 40 miles
(D) 50 miles

Solution: Time taken to cover 60 miles = 80 min = $\frac{4}{3}$ hrs ∴ Speed of the train = $(60 \times \frac{3}{4})$ mph = 45 mph Speed of the aeroplane = (2 × 45) mph = 90 mph Distance covered by the aeroplane in 60 min = 90 miles Distance covered by the aeroplane in 20 min : = $(\frac{90}{60} \times 20)$ miles = 30 miles

If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is :

(A) 50 km
(B) 56 km
(C) 70 km
(D) 80 km

Solution: Let the actual distance travelled be x km Then, $\begin{aligned} \Leftrightarrow \frac{x}{10} = \frac{x + 20}{14} \\ \Leftrightarrow 14 x = 10 x + 200 \\ \Leftrightarrow 4 x = 200 \\ \Leftrightarrow x = 50 km \end{aligned}$

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is :

(A) 100 kmph
(B) 110 kmph
(C) 120 kmph
(D) 130 kmph

Solution: Let speed of the car be x kmph Then, speed of the train : $= \frac{150}{100} x = (\frac{3}{2} x) kmph$ $\begin{aligned} ∴ \frac{75}{x} - \frac{75}{\frac{3}{2} x} = \frac{125}{10 \times 60} \\ \Leftrightarrow \frac{75}{x} - \frac{50}{x} = \frac{5}{24} \\ \Leftrightarrow x = (\frac{25 \times 24}{5}) \\ \Leftrightarrow x = 120 kmph \end{aligned}$

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

A man walking at the speed of 4 km/hr,cross a square field diagonally in 3 minutes.The area of the field is?

(A) 20000 sqm
(B) 25000 sqm
(C) 18000 sqm
(D) 19000 sqm

Solution: Speed of man, = 4 kmph = $\frac{4 \times 5}{18}$ m/sec In 3 min (180 sec) man will go = $\frac{20 \times 180}{8}$ = 200 m. That means the diagonal of the square field = 200 m. Diagonal of square, = Side of Square × $\sqrt{2}$ = 200 m. → Side of Square = $\frac{200}{\sqrt{2}}$ Area of Square = Side2 Area = $\frac{200 \times 200}{2}$ = 20000 sqm.

A bus covers three successive 3 km stretches at speed of 10 km/hr, 20 km/hr and 60 km/hr respectively. Its average speed over this distance is :

(A) 30 km/hr
(B) 25 km/hr
(C) 18 km/hr
(D) 10 km/hr

Solution: $\begin{aligned} Average speed = \frac{Total distance}{Total time} \\ Average speed = \frac{60 \times 3}{\frac{60}{10} + \frac{60}{20} + \frac{60}{60}} \\ Average speed = \frac{180 km}{(6 + 3 + 1) hr} \\ Average speed = \frac{180 km}{10 hr} \\ Average speed = 18 km/hr \end{aligned}$ Short trick formula : $(Average speed = \frac{3 x y z}{x y + y z + z x})$

Vinay and Versha run a race with their speed in the ratio of 5 : 3. They prefer to run on a circular track of circumference 1.5 km. What is the distance covered by Vinay when he passes Versha for the seventh time?

(A) 25.25 km
(B) 26.25 km
(C) 13.2 km
(D) 14.5 km

Solution: Since, the speeds of Vinay and Versha are in the ratio 5 : 3 i.e. when Vinay covers 5 rounds, then Versa covers 3 rounds, but first time Vinay and Versha meet when Vinay completes ${2 \frac{1}{2} = 2.5}$ round and Versha completes $\frac{1}{2}$ round. For Vinay to pass Versha seventh time, Vinay would have completed, = 7 × 2.5 rounds Since, each round is 1.5 km, the distance covered by Vinay is, = 7 × 2.5 × 1.5 = 26.25 km.

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.

A person walks a distance from point A to B at 15 km/h, and from point B to A at 30 km/h. If he takes 3 hours to complete the journey, then what is the distance from point A to B?

(A) 25 km
(B) 10 km
(C) 15 km
(D) 30 km

10.

The distance between two places A and B is 140 km. Two cars x and y start simultaneously from A and B respectively. If they move in the same direction, they meet after 7 hours. If they move towards each other, they meet after one hour. What is the speed (in km/h) or car y if its speed is more than that or car x?

(A) 100
(B) 80
(C) 90
(D) 60

Solution: Distance = 140 km y > x Same direction: y - x = $\frac{140}{7}$ = 20 . . . . . .(i) Opposite direction: y + x = $\frac{140}{1}$ = 140 . . . . . . (ii) Add equation (i) and (ii) 2y = 160 y = 80 km/hr

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Practice MCQ Questions and Answer on Speed Time and Distance

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