Question

A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.

Answer 1

ANSWER

Given :-

A car covers a distance of 400 km at a certain speed.

If speed is 12 km/hr more than normal the time taken for the journey is 1hr 40 min less.

Required to find :-

Original speed ?

Formula used :-

Though this mathematics we need to use a bit Physics here.

Speed = Distance/Time

=> Time = Distance/Speed

Solution :-

Case - 1

Distance = 400 km

Time be 'T' hrs

Speed = x km/hr

Now,

Time = 400/x »»(1)

Case-2

Distance = 400 km

Time be 'T' - 1 hr 40 min

T - 100 min

= T - 5/3 hrs

Speed = x + 12 km/hr

Time = 400/x + 12

T - 5/3 = 400/x + 12

T = (400)/(x + 12) + 5/3 »»(2)

From 1 and 2 we have;

400/x = (400)/(x + 12) + 5/3

400/x - (400)/(x + 12) = 5/3

400[ 1/x - 1/(x + 12) ] = 5/3

400[ (x + 12 - x )/(x² + 12x) ] = 5/3

400[ (12)/(x² + 12x) ] = 5/3

[ 12/(x² + 12x) ] = 5/(3 x 400)

[12]/[x² + 12x] = 1/(3 x 80)

12/(x² + 12x) = 1/(240)

x² + 12x = 2880

x² + 12x - 2880 = 0

x² + 60x - 48x - 2880 = 0

x(x + 60) - 48(x + 60)

(x + 60) (x - 48)

x = -60 or 48

since, speed can't be negative .

→ x = 48 km/hr (✓)

Therefore,

• Speed of car = 48 km/hr

Answer 2

Given :-

A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less

To Find :-

Original speed

Solution :-

Let the original speed of car be s

In Case 1

We have

$\sf\begin{cases}\bf Distance = 400 \; km\\\bf Speed = s \; km/h\end{cases}$

As Time = Distance/Speed

Time = 400/s

In Case 2

We have

$\sf\begin{cases}\bf Distance = 400 \; km\\\bf Speed = s+12 \; km/h\end{cases}$

As Time = Distance/Speed

Time = 400/s + 12

Now,

As 1 hr = 60 min

1 hr 40 min = 60 + 40 = 100 min

ATQ

$\sf\dfrac{400}{s}-\dfrac{400}{s+12}=\dfrac{100}{60}$

$\sf \dfrac{400}{s}-\dfrac{400}{s+12} = \dfrac{5}{3}$

$\sf\dfrac{400s + 4800 - 400s}{s(s+12)} = \dfrac{5}{3}$

$\sf\dfrac{\cancel{400s} + 4800 - \cancel{400s}}{s(s+12)} = \dfrac{5}{3}$

$\sf \dfrac{4800}{s(s+12)}=\dfrac{5}{3}$

$\sf 3(4800) = 5[s(s+12)]$

$\sf 14400 = 5s^2 + 60s$

$\sf 5s^2+60s-14400=0$

Divide both sides by 5

$\sf\dfrac{5s^2+60s-14400}{5}=\dfrac{0}{5}$

$\sf s^2 + 12s - 2800 = 0$

Spilt the middle term

$\sf s^2 + (60s-48s) - 2880 = 0$

$\sf s^2 + 60s-48s-2880 = 0$

$\sf s(s + 60) - 48(s+60)=0$

$\sf (s+60)(s-48)=0$

So, Either

$\sf s+60=0$

$\sf s=0-60$

$\sf s=-60$

$\bf Or,$

$\sf s-48=0$

$\sf s=0+48$

$\sf s=48$

Since, we know that Speed can't be negative. Therefore, Original speed of car is 48 kmph