Question

Two cars P and Q start from 2 points A and B towards each other simultaneously. They meet for the first time 40km from B. After meeting they exchange their speeds and directions and proceed to their corresponding starting points. On reaching their starting points they turn back with the same speeds and meet at a point 20km from A. Find distance between A and B.
Options: 130, 100, 120, 110​

Answer 1

The distance between A and B is 100 km.

Step-by-step explanation:

Let the distance between points A and B be $x$ km.

Case 1: When the cars P and Q meet at a point 40 km from B.

Let the speed of car P be $p$ km/h and speed of car Q be $q$ km/h.

Now, time taken by both cars to reach the meeting point will be same.

Time taken by car P is equal to, $t_p=\frac{x-40}{p}$

where, $x-40$ is the distance traveled by car P. As the meeting point is 40 km from point B and total distance is $x$. So, distance covered by car P is $x-40$ km.

Time taken by car Q is equal to, $t_q=\frac{40}{q}$

Now, equating the time of both cars, we get

$\frac{x-40}{p}=\frac{40}{q}\\\frac{p}{q}=\frac{x-40}{40}------------- 1$

Case 2: When the cars P and Q meet at a point 20 km from A.

Given:

Cars exchange their speeds. So, speed of car P is $q$ and car Q is $p$.

Now, total distance covered by car P is equal to sum of the return distance and an additional distance of 20 km to reach the meeting point. Return distance is $x-40$ km. So, total distance covered by car P is given as:

$d_p=x-40+20=x-20$

Similarly, total distance covered by car Q is given as:

$d_q=40+(x-20)=40+x-20=x+20$

Again equating time for both the cars, we get

$\frac{x-20}{q}=\frac{x+20}{p}\\\frac{p}{q}=\frac{x+20}{x-20}------------ 2$

Now, equating equations 1 and 2, we get

$\frac{x-40}{40}=\frac{x+20}{x-20}\\(x-40)(x-20)=40(x+20)\\x^2-20x-40x+800=40x+800\\x^2-60x=40x\\x^2=60x+40x\\x^2=100x\\x=100\ km$

Therefore, the distance between points A and B is 100 km.