Question

An automobile covered the distance of 240 km between A and B with a certain speed. On its way back, the automobile covered half of the distance at the same speed and for the rest of the trip he increased his speed by 10 km/hour. As a result, the drive back took him 2 5 of an hour less than the drive from A to B. What was the automobile’s speed when it was driving from A to B?

Answer 1

Answer:

Therefore the speed of automobile when it was driving from A to B was 50 km/ h.

Explanation:

Given that,

An automobile covered the distance of 240 km between A and B with a certain speed.

Let the speed of the automobile be x km/hour.

We know that,

$Time =\frac{distance}{speed}$

Time taken to travel from A to B point is $=\frac{240}{x}$ hours.

On its back, the automobile covered half of distance= 120 km at the same speed and for the trip he increased his speed 10 km/hour.

Time taken to complete the trip is

$=(\frac{120}{x}+\frac{120}{x+10})$ hours

The driver took $\frac25$ of an hour less when he come back.

According to problem,

$\frac{240}{x}-(\frac{120}{x}+\frac{120}{x+10})=\frac25$

$\Rightarrow\frac{240}{x}- \frac{120}{x}-\frac{120}{x+10}=\frac25$

$\Rightarrow\frac{240-120}{x}-\frac{120}{x+10}=\frac25$

$\Rightarrow\frac{120}{x}-\frac{120}{x+10}=\frac25$

$\Rightarrow\frac{120(x+10)-120x}{x(x+10)}=\frac25$

$\Rightarrow\frac{120x+1200-120x}{x^2+10x}=\frac25$

$\Rightarrow\frac{1200}{x^2+10x}=\frac25$

$\Rightarrow 2(x^2+10x)=1200\times 5$

$\Rightarrow (x^2+10x)=\frac{1200\times 5}{2}$

$\Rightarrow (x^2+10x)=3000$

$\Rightarrow x^2+10x-3000=0$

$\Rightarrow x^2+60x-50x-3000=0$

⇒x(x+60)-50(x+60)=0

⇒(x+60)(x-50)=0

⇒x= -60, 50

The speed could not negative.

So, x=50

Therefore the speed of automobile when it was driving from A to B was 50 km/ h.