Practice MCQ Questions and Answer on Speed Time and Distance

311.

A soldier fired two bullets at an interval of 335 seconds moving at a uniform speed V₁. A terrorist who was running ahead of the soldier in same direction, hears the two shots at an interval of 330 seconds. If the speed of sound is 1188 km/h, then who is the faster and how much?

(A) Soldier, 22 km/h
(B) Terrorist, 25 km/h
(C) Soldier, 18 km/h
(D) Terrorist, 20 km/h

Solution: $\frac{Speed of sound}{Relative speed of soldier and terrorist} =$ $\frac{Time taken}{Difference in time}$ $\frac{1188}{S} = \frac{330}{5}$ Soldier = 18 km/h

312.

A, B and C start together from the same place to walk round a circular path of length 12km. A walks at the rate of 4 km/h, B 3 km/h and C $\frac{3}{2}$ km/h. They will meet together at the starting place at the end of:

(A) 10 hours
(B) 12 hours
(C) 15 hours
(D) 24 hours

Solution: Time taken to complete the revolution: A → $\frac{12}{4}$ = 3 hours B → $\frac{12}{3}$ = 4 hours C → 12 × $\frac{2}{3}$ = 8 hours Required time,= LCM of 3, 4, 8. = 24 hours.

313.

A train passes two persons walking in the same direction at a speed of 3 kmph and 5 kmph respectively in 10 seconds and 11 seconds respectively. The speed of the train is

(A) 28 kmph
(B) 27 kmph
(C) 25 kmph
(D) 24 kmph

Solution: 1st method: Let the speed of the train be S. And length of the train be x. When a train crosses a man, its travels its own distance. $\begin{aligned} According to question; \\ \frac{x}{(s - 3) \times \frac{5}{18}} = 10 \\ or, 18 x = 50 \times s - 150. . . . . (i) \\ and \\ \frac{x}{(x - 5) \times \frac{5}{18}} = 11 \\ 18 x = 55 \times s - 275. . . . . . (ii) \\ Equating equation (i) and (ii) \\ 50 \times s - 150 = 55 \times s - 275 \\ or, 5 \times s = 125 \\ or, s = 25 kmph \end{aligned}$

314.

Aryan runs at a speed of 40 meters/minute. Rahul follows him after an interval of 5 minutes and runs at a speed of 50 metres/minute. Rahul's dog runs at a speed of 60 metres/minute and starts along with Rahul. The dog reaches Aryan and then comesback to Rahul, and continues to do so till Rahul reaches Aryan. What is the total distance covered by the dog ?

(A) 600 m
(B) 750 m
(C) 980 m
(D) 1200 m

Solution: Distance covered by Aryan in 5 min : = (40 × 5) m = 200 m Relative speed of Rahul w.r.t. Aryan : = (50 - 40) m/min = 10 m/min Time taken to cover 200 m at relative speed : $\begin{aligned} = (\frac{200}{10}) min \\ = 20 min \end{aligned}$ Distance covered by the dog in 20 min : = (60 × 20) m = 1200 m

315.

Each wheel of a car is making 5 revolutions per second. If the diameter of a wheel is 84 cm, then the speed of the car in cm/sec would be :

(A) 420 cm/sec
(B) 264 cm/sec
(C) 1000 cm/sec
(D) 1320 cm/sec

Solution: Circumference of the circle = 2πr = 2πr = 2 × $\frac{22}{7}$ × 42 = 264 cm Distance cover in 1 sec = 264 × 5 = 1320 cm/sec

316.

A train running at a speed of 54 km/hr crosses a platform in 30 seconds. The platform is renovated and its length is doubled. Now, the same train running at same speed crosses the platform in 46 seconds. Find the length of the train.

(A) 180 metres
(B) 200 metres
(C) 210 metres
(D) 240 metres

Solution: Let length of the Platform is X m and Train is Y m. Speed of the train = 54 kmph = $\frac{54 \times 5}{18}$ = 15 m/sec. To cross the platform, train needs to travel (X + Y) m in 30 sec. $\begin{aligned} Speed = \frac{Distance}{Time} \\ 15 = \frac{X + Y}{30} \\ X + Y = 450 . . . . . . . . . . . (1) \end{aligned}$ Now Platform is renovated and its length is doubled. So, Now, train need to travel (2X + Y) m to cross the platform. Thus, $\begin{aligned} Speed = \frac{Distance}{Time} \\ 15 = \frac{2X + Y}{46} \\ 2X + Y = 690 . . . . . . . . . . . (2) \end{aligned}$ Multiplying equation (1) by (2) 2X + 2Y = 900 ------------------ (3) Now, equation (2) - (3) 2X + Y - 2X - 2Y = 690 - 900 - Y = - 210 Y = 210 Length of the train = 210 metres

317.

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 $= \frac{\frac{7}{2} \times \frac{9}{2}}{\frac{9}{2} - \frac{7}{2}} \times \frac{16}{60} = 4.2 km$

318.

A train travels at a speed of 30 km/hr for 12 minutes and at a speed of 45 km/hr for the next 8 minutes. The average speed of the train for this journey is :

(A) 30 km/hr
(B) 36 km/hr
(C) 37.5 km/hr
(D) 48 km/hr

Solution: Total distance travelled : $\begin{aligned} = (30 \times \frac{12}{60} + 45 \times \frac{8}{60}) km \\ = 12 km \end{aligned}$ Total time taken : = (12 + 8) min = 20 min = $\frac{1}{3}$ hr ∴ Average speed : = (12 × 3) km/hr = 36 km/hr

319.

Rani goes to school from her house in 30 minutes. Raja takes 45 minutes in covering the same distance. Find the ratio between time taken by Rani and Raja ?

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

320.

A man travels for 5 hours 15 minutes. If he covers the first half of the journey at 60 km/h and rest at 45 km/h. Find the total distance travelled by him.

(A) 189 km
(B) 378 km
(C) 270 km
(D) $1028 \frac{6}{7}$ km

Solution: Average speed : $\begin{aligned} = \frac{2 \times 60 \times 45}{60 + 45} \\ = \frac{1080}{21} km/hr \end{aligned}$ Time : $\begin{aligned} = 5 + \frac{1}{4} hr \\ = \frac{21}{4} hr \end{aligned}$ Distance travelled : $\begin{aligned} = \frac{1080}{21} \times \frac{21}{4} \\ = 270 km \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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