Practice MCQ Questions and Answer on Speed Time and Distance

The ratio of speeds of A and B is 2 : 3 and therefore A takes 20 minutes more time than B. What is the ratio of time taken by A and B?

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

Solution: When distance is constant then Speed is inversely proportional to time. ST = D When distance constant, S ∝ $\frac{1}{T}$ So, ratio of time taken by A and B = 3 : 2

A car driver, driving in a fog, passes a pedestrian who was walking at the rate of 2 km/h in the same direction. The pedestrian could see the car for 6 minutes and it was visible to him up to distance of 0.6 km. what was speed of the car?

(A) 30 kmph
(B) 15 kmph
(C) 20 kmph
(D) 8 kmph

P and Q are 27 km away. Two trains with speed of 24 km/hr and 18 km/hr respectively start simultaneously from P and Q and travel in the same direction. They meet at a point R beyond Q. Distance QR is :

(A) 126 km
(B) 81 km
(C) 48 km
(D) 36 km

Solution: Relative speed = 24 - 18 = 6 km/hr Time required by faster train to overtake slower train : = $\frac{27}{6}$ = $4 \frac{1}{2}$ hr ∴ Distance between Q and R : = 18 × $4 \frac{1}{2}$ = 81 km

Shyam drives his car 30 km at a speed of 45 km/h and, for the next 1 h 20 m, he drives it at a speed of 51 km/h. Find his average speed (in km/h) for the entire journey.

(A) 49
(B) 48
(C) 48.5
(D) 47

Solution: $\begin{aligned} Time = 1 hr 20 m = \frac{4}{3} hr \\ Speed = 51 km/hr \\ ∴ Distance = \frac{4}{3} \times 51 = 68 \\ Average speed = \frac{Total Distance}{Total Time} \\ = \frac{30 + 68}{\frac{30}{45} + \frac{4}{3}} \\ = \frac{98}{\frac{2}{3} + \frac{4}{3}} \\ = 49 km/hr \end{aligned}$

The speed of a car increases by 2 kms after every one hours. If the distance travelled in the first one hour was 35 kms, what was the total distance travelled in 12 hours ?

(A) 456 kms
(B) 482 kms
(C) 552 kms
(D) 556 kms

Solution: Total distance travelled in 12 hours = (35 + 37 + 39 +...... upto 12 terms) This is an A.P. with first term, a = 35, Number of terms, n = 12, Common difference, d = 2 ∴ Required distance : $\begin{aligned} = \frac{12}{2} [2 \times 35 + (12 - 1) \times 2] \\ = 6 (70 + 22) \\ = 552 km \end{aligned}$

A train x running at 74 km/h crosses another train y running at 52 km/h in the opposite direction in 12 seconds. If the length of y is two-thirds that of x, then what is the length of y (in m)?

(A) 252
(B) 200
(C) 168
(D) 210

Solution: According to question, $\begin{aligned} \frac{x + y}{(74 + 52) \frac{km}{hr}} = 12 \\ x + y = 12 \times 126 \times \frac{5}{18} \\ x + y = 420 m . . . . . . (i) \end{aligned}$ y = 3a + 2a = 5a x = 3a So, from equation (i) 5a = 420 a = 84 Hence length of train y = 2 × 84 = 168

A bullock cart has to cover a distance of 120 km in 15 hours. If it covers half of the journey in $\frac{3}{5}$ th time, the second to cover the remaining distance in the time left has to be :

(A) 6.4 km/hr
(B) 6.67 km/hr
(C) 10 km/hr
(D) 15 km/hr

Solution: Total distance = 120 km Total time = 15 hours He covers half of the journey in $\frac{3}{5}$ th of the time = 15 × $\frac{3}{5}$ hours = 9 hours Now, remaining distance : = (120 - 60) km = 60 km And, remaining time : = (15 - 9) hours = 6 hours Average speed to cover a distance of 60 km will be : $\begin{aligned} = \frac{60 km}{6 hour} = 10 km/hr \\ {Speed = \frac{Distance}{Time}} \end{aligned}$

Sabarmati express takes 18 seconds to pass completely through a station 162 m long and 15 seconds through another station 120 m long. The length of the train is :

(A) 132 m
(B) 100 m
(C) 80 m
(D) 90 m

Solution: Let length of the train be x mSpeed of train, $\frac{x + 162}{18} = \frac{x + 120}{15}$ x = 90 m

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.

Two trains of equal length, running in opposite directions, pass a pole in 18 and 12 seconds. The trains will cross each other in:

(A) 14.4 seconds
(B) 15.5 seconds
(C) 18.8 seconds
(D) 20.2 seconds

Solution: Let length of each train be x meter. Then, speed of 1st train = $\frac{x}{18}$ m/sec Speed of 2nd train = $\frac{x}{12}$ m/sec Now, When both trains cross each other, time taken $\begin{aligned} = \frac{2 x}{\frac{x}{18} + \frac{x}{12}} \\ = \frac{2 x}{\frac{2 x + 3 x}{36}} \\ = \frac{2 x \times 36}{5 x} \\ = \frac{72}{5} \\ = 14.4 seconds \end{aligned}$

«
1
2
3
4
5
6
7
8
9
10
...
45
46
Next ›
Last »

Indian Polity Mcq IAS GK Question India GK World GK GK questions GK Quiz Biology GK Science GK Physics GK Chemistry GK General Awareness Computer GK Political GK Indian Political GK Economics GK History GK Sports GK Geography GK Teaching Aptitude GK CTET/TET GK B Ed Entrance GK Banking GK SSC GK UPSC GK Indian Army GK Hindi Grammar GK Hindi Grammar MCQ Reasoning questions Bihar GK Rajasthan GK Jharkhand GK MP GK UP GK Chhattisgarh GK Haryana GK HP GK Maharashtra GK Electronics GK Agriculture GK States/Capitals GK Books Name & Author Arthimetic Verbal Reasoning Non Verbal Reasoning English

Practice MCQ Questions and Answer on Speed Time and Distance

General Knowledge