# Maths Questions Related to PSC Time and Distance

Here we are listed some of the questions from Time and Distance section for Kerala PSC/ UPSC/ IBPS exams. You can solve those problems and which will help you secure one mark for sure.

Important Terms

Speed = distance/time |

Distance = (Speed * Time) |

Time = distance/ Speed |

Formulas

Convert from kph (km/h) to mps(m/sec) | x km/=(hrx∗5/18) m/sec |

Convert from mps(m/sec) to kph(km/h) | x m/= X *(18/5) seckm/h |

#### Problem:

**1)** Raghav had to walk a certain distance. He walked half the distance at a speed of 4 km/hour and the remaining half at a speed of 5 km/hour. If the total time taken was 9 hour, the distance was

Solution:

*Let half of the distance traveled be d.*

*Time taken at the speed of 4 km/hr = (d/4)*

*Time taken at the speed of 5 km/hr = (d/5)*

*Therefore, total time = (d/4) + (d/5) = 9 hrs.*

*⇒ (5d + 4d)/20 = 9*

*⇒ 9d = 180*

*⇒ d = 20.*

*Therefore total distance = *d x* 2 = 40 *kms

**2**. Kartik rides a bicycle 8 km/hr and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/hr and reaches the school 5 minutes early. How far is the school from his house?

Solution:

*Let the total distance be *d km*. We have speed = 8km/hr and 10km/hr.*

*We know, time (t) = distance (d)/speed (s).*

*⇒ t = d/8 (for the first day)*

*⇒ t = d/10 (for the second day)*

*Difference between the time taken on first day and second day is 2.5 + 5 = 7.5 mins.*

*Therefore, (d/8) – (d/10) = 7.5/60. (7.5 mins converted to hours)*

*⇒ (10d – 8d)/80 = 75/600*

*⇒ d = (75 x 80)/1200.*

*⇒ d = 5 kms.*

*Therefore, the distance between his school and house 5 km.*

**3**. Virat travelled 75 kms in 7 hours. He went some distance at the rate of 12 km/hr and the rest at 10 km/hr. How far did he travel at the rate of 12 km/hr?

Solution:

*Let the distance travelled at the rate of 12 kmph be x km*

*Time taken to travel this distance = x/12 …….(1)*

*Then the distance travelled at the rate of 10 km/hr will be 75 – x*

*Time taken to travel this distance = (75 – x)/10 …….(2)*

*Total time taken to cover the total distance = 7 hrs*

*⇒ (x/12) + (75 – x)/10 = 7 [sum of (1) and (2)]*

*⇒ 10x + 900 – 12x = 840*

*⇒ –2x = –60*

*⇒ x = 30.*

*Therefore, the distance travelled at the rate of 12 km/hr = 30km.*

**4**. Sanu travelled a distance of 20 kms at the speed of 5 km/hr, he reached 40 minutes late. If he had walked at the speed of 8 km/hr, how early from fixed time will he reach?

Solution:

*Time *taken* to travel 20 *kms* at the speed of 5 *kmph*.*

*⇒ t = 20/5 = 4 hrs.*

*Actual time of reaching = 4 hrs – 40 mins = 3 hrs 20 mins ……(1)*

*Time taken to travel 20 kms at the speed of 8 kmph.*

*⇒ t = 20/8 = 2.5 hrs. ……(2)*

*Difference between (1) and (2) = 3 hrs 20 mins – 2 hrs 30 mins*

*200 mins – 150 mins = 50 mins.*

*Sanu would have reached 50 mins earlier*.

**5**. A boy rides his bicycle 10 kms at an average speed of 12 kmph and again travels 12 kms at an average speed of 10 kmph. His average speed for the entire trip is approximately:

Solution:

*Average speed is represented by total distance divided by the time taken.*

*Total distance = 10 kms + 12 kms = 22 kms*

*Total Time = (10/12) + (12/10)*

*⇒ (10/12) + (12/10) = (100 + 144)/120 = 244/120 hrs*

*Average speed = 22/(244/120) = 10.8 kmph.*

*Average speed for the entire trip is 10.8 *kmph*.*

**6**. Two cars A and B start simultaneously from a certain place at the speed of 30 km/hr and 45 km/hr respectively. The car B reaches the destination 2 hours earlier than A. What is the distance between the starting point and destination?

Solution:

*Let *as assume* the distance to be common multiple of 30 and 45 for e.g. 90 km*

*At 30 *kmph* he takes 3 hours and at 45 *kmph* he takes 2 hours*

*The difference between the two times is 1 hour*

*Since as per the question the difference in time is 2 hours the distance is double of that assumed i.e. 180 km*

**7**. Raghubir after travelling 84 km, found that if he travelled at 5 km an hour more, he would have taken 5 hrs less, he actually travelled at a rate of :

Solution:

*Let his original speed be S *kmph

*Time taken = 84/S*

*If his speed is 5 *kmph* more, time is 5 hrs less*

*∴ 84/S – 5 = 84/(S + 5)*

*(S + 5)(84 – 5S) = 84S*

*5S ^{2} + 25S – 420 = 0*

*S = 7*

**Simpler method**

*84 has the following factors – 2, 3, 4, 6, 7, 12, 14, 21, 28, 42*

*Of the above *factors* 7 and 12 are at a difference of 5 (note the use of 5 *kmph* in speed in the question)*

*Through simple trial and error we can find that to travel 84 km he takes 12 hours at 7 *kmph

*and at 12 *kmph* he takes 7 hours both of which are at a difference of 5.*

We hope this maths questions, answers related to Time and Distance is helping you for your PSC preparations. If you have any doubts please feel free to contact us under the comment section.