# Time and Distance Questions Answers

1) How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/

- 1) \begin{aligned} 1\frac{1}{5} min\end{aligned}
- 2) \begin{aligned} 2\frac{1}{5} min\end{aligned}
- 3) \begin{aligned} 3\frac{1}{5} min\end{aligned}
- 4) \begin{aligned} 4\frac{1}{5} min\end{aligned}

**Ans.**A

**Explanation :**

We know that, \begin{aligned} Time = \frac{Distance}{Speed} \ Speed = 20\text{ km/hr} = 20*\frac{5}{18}{ m/sec} \= \frac{50}{9}{ m/sec} \ \text{ Time =} \left(400*\frac{9}{50}\right) \= 72 {sec} = 1\frac{1}{5}{ min} \end{aligned}

2) A cyclist covers a distance of 750 meter in 2 minutes 30 seconds. What is the speed in km/hr of cycli

- 1) 16 km/hr
- 2) 17 km/hr
- 3) 18 km/hr
- 4) 19 km/hr

**Ans.**C

**Explanation :**

\begin{aligned} Speed = \frac{Distance}{Time} \Distance = 750 meter \Time = \text{2 min 30 sec} = 150 sec \Speed = \frac{750}{150} = 5 m/sec \ => 5*\frac{18}{5} km/hr = 18 km/hr \end{aligned}

3) An athlete runs 200 meters in 24 seconds. His speed is

- 1) 10 km/hr
- 2) 17 k/hr
- 3) 27 km/hr
- 4) 30 km/hr

**Ans.**D

**Explanation :**

\begin{aligned} Speed = \frac{Distance}{Time} \ = \frac{200}{24}m/sec = \frac{25}{3} m/sec \\frac{25}{3}*\frac{18}{5} km/hr = 30 km/hr \end{aligned}

4) A person crosses a 600 meter long street in 5 minutes. What is the speed in Km/

- 1) 6.2 km/hr
- 2) 7.2 km/hr
- 3) 8.2 km/hr
- 4) 9.2 km/hr

**Ans.**B

**Explanation :**

Two things to give attention on this question. First time is in minutes, we need to change it to seconds to get speed in m/sec, then we need to get the final answer in km/hr. So lets solve this. \begin{aligned} Speed = \frac{Distance}{Time} \\ Distance = 650 meter \\ Time = \text{5 minutes} = 300 sec \\ Speed = \frac{600}{300} = 2 m/sec \\ => 2*\frac{18}{5} km/hr = 7.2 km/hr \end{aligned}

5) A man is walking at the rate of 5 km/hr crosses a bridge in 15 minutes. The length of the bridge

- 1) 1000 meters
- 2) 1050 meters
- 3) 1200 meters
- 4) 1250 meters

**Ans.**D

**Explanation :**

We need to get the answer in meters. So we will first of change distance from km/hour to meter/sec by multiplying it with 5/18 and also change 15 minutes to seconds by multiplying it with 60. \begin{aligned} Speed = 5* \frac{5}{18} = \frac{25}{18}m/sec \ Time = 15*60 seconds = 900 seconds \ Distance = Time * Speed \ Distance = \frac{25}{18}*900 = 1250 meter \end{aligned}

6) A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. Find the speed at which the train must run to reduce the time of journey to 40 minute

- 1) 50 km/hr
- 2) 60 km/hr
- 3) 65 km/hr
- 4) 70 km/hr

**Ans.**B

**Explanation :**

We are having time and speed given, so first we will calculate the distance. Then we can get new speed for given time and distance. Lets solve it. Time = 50/60 hr = 5/6 hr Speed = 48 mph Distance = S*T = 48 * 5/6 = 40 km New time will be 40 minutes so, Time = 40/60 hr = 2/3 hr Now we know, Speed = Distance/Time New speed = 40*3/2 kmph = 60kmph

7) A car moves at 80 km/hr. What is the speed of the car in meters per second

- 1) \begin{aligned} 20\frac{2}{9} m\sec \end{aligned}
- 2) \begin{aligned} 22\frac{2}{9} m\sec \end{aligned}
- 3) \begin{aligned} 24\frac{2}{9} m\sec \end{aligned}
- 4) \begin{aligned} 26\frac{2}{9} m\sec \end{aligned}

**Ans.**B

**Explanation :**

\begin{aligned} Speed = \left(80*\frac{5}{18}\right) m/sec \= \frac{200}{9} m/sec \= 22\frac{2}{9} m\sec \end{aligned}

8) Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.

- 1) 9 km/hour
- 2) 10 km/hour
- 3) 11 km/hour
- 4) 12 km/hour

**Ans.**D

**Explanation :**

We need to calculate the distance, then we can calculate the time and finally our answer. Lets solve this, Let the distance travelled by x km Time = Distance/Speed \begin{aligned} \frac{x}{10} - \frac{x}{15} = 2 \\text{[because, 2 pm - 12 noon = 2 hours]} \ 3x - 2x = 60 \ x = 60. Time = \frac{Distance}{Speed} \ Time @ 10km/hr = \frac{60}{10} = 6 hours \end{aligned} So 2 P.M. - 6 = 8 A.M Robert starts at 8 A.M. He have to reach at 1 P.M. i.e, in 5 hours So, Speed = 60/5 = 12 km/hr

9) A person travels from P to Q at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trip

- 1) 44 km/hour
- 2) 46 km/hour
- 3) 48 km/hour
- 4) 50 km/hour

**Ans.**C

**Explanation :**

Speed while going = 40 km/hr Speed while returning = 150% of 40 = 60 km/hr Average speed = \begin{aligned} \frac{2xy}{x+y} \= \frac{2*40*60}{40+60} = \frac{4800}{100} \= 48 Km/hr \end{aligned}

10) The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m and travels towards A at 75 Km/hr. At what time do they me

- 1) 10 am
- 2) 11 am
- 3) 12 pm
- 4) 1 pm

**Ans.**B

**Explanation :**

Suppose they meet x hrs after 8 a.m. then (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330 So, 60x+75(x-1) = 330 x=3. So,they meet at (8+3). i.e 11a.m.