Aptitude - Time and Distance - Discussion

Thanks @Kinnari.

@Ilesh @Jayesh.

You guys made it simpler to understand. Thank you very much.

Thank you for explaining @Kinnari.

Let distance travelled by person at 14km/h be d1 and distance travelled at 10km/h be d2.

d1-d2=20km (given in Q that he walks 20km more)
D=speed * time.

{Time is constant and reason is;

It is because supposed for example if a person walks 10km/h for 7 hrs means he would have covered 70km. If speed changes to 14km/h and if we assume time is not constant then to cover 70km distance at 14km/h 5hrs would be needed, then in such case there would be no case of extra distance covered like in present Q which is 20km.

If time is constant like 7hrs so at distance at 10km/h =70km & 14km/h=98km and only then we can say it travelled 28km extra)}

Now coming to the present Q,
d1-d2=20km,
14t-10t=20.
t(14-10)=20.
4t=20.
T=5hr.

Q says if a person walks 14km/h that means 14km/h is not the original speed...
D=10*5=50km.

D= (S1*T ) - (S2*T)
20= (14*T) - (10*T)
20= 14T - 10T
T = 20/4
T= 5 hrs.

Therefore Actual distance where actual speed was 10km/ hr.

D= S*T
= 10km/hr * 5hr.
= 50km.

I hope this was helpful.

Explain it clearly.

Thanks @Kinnari.

@All.

According to me;

Let the actual distance = y.

At speed of 10KMPH, the distance is y.

So, 10kmph = y.
This implies, 1kmph=y/10 ------> (1).

When he travels at a speed, of 14kmph, the distance is y+20.
So, 14kmph = y+20.
This implies, 1kmph=(y+20)/14 -------> (2)
EQUATE (1)AND (2).

y/10 = (y+20)/14
14y = 10y+200
4y = 200
y = 50km.

Distance = a/x-y *y.
20/14 - 10 * 10,
20/4 * 10 = 5 * 10 = 50km.

According to me, by using the simple formula we can iterate lucidly like this way,

Let the time be x hours

According to the question;

14x - 10x = 20
4x = 20
x = 5 hours.

His actual speed is 10 km/h.

So, the actual distance is 10 * 5 = 50 km.