Question

A man driving his moped at 24 kmph reaches his destination 5 minutes later to an appointment. If he had driven at 30 kmph he would have reached his destination 4 minutes before time. How far is his destination?

Answer 1

Hint: This is a simple problem of time and distance. First of all we have to know basic concepts of time, distance and speed .Distance means a gap between two fixed destinations or points which can be measured in any unit of length like cm,m km,etc. Then , time is total duration which can be measured in seconds, minutes, hours,etc.Speed is the ratio of total distance travelled to the total time taken to reach from an initial point to final point or destination.
So, this is related to solving systems of equations and trying to get a common solution to the equation. In certain problems where we deal with unknown/s for finding their solution which comes under the Linear equation.

Complete step-by-step answer:
GIVEN:
Case 1- Has he travelled with the speed at 24 kmph, reaches his destination 5 minutes later . It means if the time taken is ‘t’ hours in nominal speed, it would take 5 minutes more if it goes with a speed of 24 kmph.
Case 2- Has he travelled with the speed at 30 kmph, reaching his destination 4 minutes earlier . It means if the time taken is ‘t’ hours in nominal speed, it would take 4 minutes less if it goes with a speed of 30 kmph.
Let the distance between two fixed points be ‘d’ km .
So, d= 24×(t+\[\dfrac{5}{{60}}\] ) ………..…..5 min= \[\dfrac{5}{{60}}\]hour,
And,d=30×(t-$\dfrac{4}{{60}}$ ) ………..…..4 min= $\dfrac{4}{{60}}$ hour.
Equating them we got,
24×(t+\[\dfrac{5}{{60}}\] ) = 30×(t-$\dfrac{4}{{60}}$)
24t +2 = 30t-2
30t-24t = 4
So,t =$\dfrac{4}{6}$ hour
Hence, d= = 24×(t+\[\dfrac{5}{{60}}\] ) =24 ×($\dfrac{4}{6}$+\[\dfrac{5}{{60}}\]) = 24×$\dfrac{{45}}{{60}}$ =18 km.
Answer-Required distance = 18 kms

Note: Proper conversion should be made i.e. from minutes to hours or vice-versa while solving problems. And if speed is in kmph then time should be in hours and distance should be in km.