Question

A car left 3 minutes early than the scheduled time and in order to reach the destination 126 km away in time, it has to slow its speed by 6 km/h from the usual. what is the usual speed (in km/hr) of the car?

Answer 1

Hello Dear.

Here is the answer---

Let the usual speed of the Car be x km/hr.
Distance = 126 km.

If the Car will go with the Usual Speed, then the Time Covers will be given by---→
∵ Time = Distance/Speed
∴ Time(T₁) = 126/x hr.


Now, If Speed is increased by the 6 km/hr.
Then New Speed = (x + 6)km/hr.
Distance = 126 km.

∴ Time(T₂) = 126/(x + 6) hr.
      

According to the Question,

      T₁ - T₂ = 3 min.
⇒ 126/x - 126/(x + 6) = 3/60 hrs.
⇒ 126(x + 6 - x) × 60 = 3 × x(x + 6)
⇒ 126 × 6 × 60 = 3x² + 18x
⇒ 15120 = x² + 6x
⇒ x² + 6x - 15120 = 0

Splitting the Middle Term,
x² + 126x - 120x -  15120 = 0
⇒ x(x + 126) - 120(x + 126) = 0
⇒  (x + 126)(x - 120) = 0
⇒ x = -126  and x = 120.

∵, Speed cannot be negative, thus Rejecting x =  -126.
∴ Speed of the Car = 120 km/hr.

Thus, Usual Speed of the Car is 120 km/hr.


Hope it helps.

Answer 2

Answer:

for 3min of change speed has to be decreased by 6km/hr

then for 60min of change speed has to be decreased by... (60/3) *6=120kmph