Time and distance as a topic involves a variety of areas which include speed-time-distance concepts, relative speed, moving and stationary bodies, boats and streams, circular motion, and so on. **Time and Distance Practice Set 2** article is useful for candidates preparing for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.

**Answer**: Option C

**Explanation**: Raman’s speed = 20 km/hr = 20 × \(\frac{5}{18}\)= \(\frac{50}{9}\) m/sec

400 × \(\frac{9}{50}\) = 1\(\frac{1}{5}\)mins

**2. John travelled from his town to city. John went to city by bicycle at the speed of 25 km/h and came back at the speed of 4 km/h. If John took 5 hours and 48 min to complete his journey, what is the distance between town and city ? **

**Answer**: Option C

**Explanation**: Average speed of John = \(\frac{xy}{x+ y}\)= 2 × 25 × \(\frac{4}{25}\) + 4= \(\frac{200}{29}\) km/h

⇒ Distance traveled = Speed × Time = \(\frac{200}{29}\) × \(\frac{29}{5}\) = 40 Km

⇒ Distance between city and town = \(\frac{40}{2}\) = 20 km

**3. Speed of a train is 20 meters per second. It can cross a pole in 10 seconds. What is the length of train ?**

**Answer**: Option C

**Explanation**: Lenght of train = 20 × 10 = 200 meters

**4. Ram walks at a speed of 12 km/h. Today the day was very hot so walked at \(\frac{5}{6}\) of his average speed. He arrived his school 10 minutes late. Find the usual time he takes to cover the distance between his school and home?**

**C.** 54 meters

**D.** 60 meters

**Answer**: Option C

**Explanation**: If Ram is walking at \(\frac{5}{6}\) of his usual speed that means he is taking 6/5 of using time.

⇒ \(\frac{6}{5}\) of usual time – usual time = 10 mins

⇒ \(\frac{1}{5}\) of usual time = 10 mins

⇒ Usual time = 50 mins

**5. A car running at 65 km/h takes one hour to cover a distance. If the speed is reduced by 15 km/hour then in how much time it will cover the distance ?**

**Answer**: Option B

**Explanation**: ⇒Reduced speed = 65-15 = 50 km/h

⇒ Now car will take \(\frac{65}{50}\) × 60 mins = 78 mins

**D.** 1.5 m/s

**Answer**: Option B

**Explanation**:⇒Time taken by A to cover 100 meters = 60 seconds

⇒ Since A gives a start of 4 seconds then time takes by B = 72 seconds

⇒ B takes 72 seconds to cover 96 meters

⇒ Speed of B = \(\frac{96}{72}\) = 1.33 m/s

**2. In a kilometer race, A beats B by 100 meters. B beats C by 100 meters. By how much meters does A beat C in the same race ?
**

**Answer**: Option C

**Explanation**: ⇒ While A covers 1000 meters, B can cover 900 meters

⇒ While B covers 1000 meters, C can cover 900 meters

⇒ Lets assume that all three of them are running same race. So when B runs 900 meters, C can run 900 × \(\frac{9}{10}\) = 810

⇒ So A can beat C by 190 meters.

**3. Without any stoppage a person travels a certain distance at an average speed of 42 km/hr and with stoppages he covers the same distance at an average speed of 28 km/hr. How many minutes per hour does he stop?**

**B.** 30 minutes

**C.** 20 minutes

**D.** None of these

**Answer**: Option C

**Explanation**: Here, S1 = 42 ans S2 = 28

Stoppage Time/hr = \(\frac{S1 – S2}{S1}\) = \(\frac{41 – 28}{42}\)

= \(\frac{1}{3}\) hour = 30 Minutes

**4. A train passes through a telegraph post in 9 seconds moving with a speed of 54 km per hour. The length of the train is**

**Answer**: Option A

**Explanation**: Length of the train = Speed of the train x Time taken in crossing the post

54 x \(\frac{5}{18}\) x 9 = 135 m

**5. A train 50 m long passes a platform 100 m long in 10 seconds. The speed of the train in m/sec. is
**

**Answer**: Option B

**Explanation**:Speed of the train = \(\frac{Length of the train + Length of the platfrom}{Time taken in crossing the platform}\) = \(\frac{50 + 100}{10}\)= 15 sec

**B.** 50 sec

**C.** 60 sec

**D.** None of these

**Answer**: Option B

**Explanation**: Here L1 = 300m, L2= 200m

S1 = 90 km/hr and S2= 60 km/hr

S1 = S2 = 90 – 60 = 30 km/hr = 30 x\(\frac{5}{18}\) m/s

Time Taken = \(\frac{L1 + L2}{S1 – S2}\)

= \(\frac{300 + 200}{30 × 5/18}\)

= \(\frac{500 × 18 }{30 × 5}\) = 60 sec

**2. Two trains are running in opposite directions with the same speed. If the length of each train is 135 metres and they cross each other in 18 seconds, the speed of each train is**

**C.** 27 km/hr

**D.** None of these

**Answer**: Option C

**Explanation**: Let the speed of each tarin be x m/sec

We have L1 = L2 = 135 m

And S1 = S2 = xm/sec

Therefore time taken = \(\frac{L1+ L2}{S1+ S2}\)

18 = \(\frac{135 + 135}{x + x}\) = \(\frac{270}{2 × 18}\) x \(\frac{18}{5}\)km/hr

**3. A and B are two stations. A train goes from A to B A at 64 km/hr and returns to A at a slower speed. If its average speed for the whole journey is 56 km/hr, at what speed did it return? **

**Answer**: Option B

**Explanation**: Let the required speed by x km/hr

Then \(\frac{2 × 64 × x}{64 + X}\) = 60

128X = 64 X 56 + 56 X

X = \(\frac{64 × 56}{72}\) = 49.77km/hr

**4. Excluding stoppages, the speed of a bus is 54 km/hr and including stoppages, it is 45 km/hr. For how many minutes does the bus stop per hour?**

**Answer**: Option B

**Explanation**: Due to stoppages, it covers 9km less per hour.

Time is taken to cover 9 km = (\(\frac{9}{54}\)×60)min

= 10 min

So, the bus stops for 10 min. per hr.

**5. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?**

**Answer**: Option B

**Explanation**: Due to stoppages, it covers 9 km less.

Time taken to cover 9 km = [\(\frac{9}{54}\) x 60]min = 10 min.

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