## Quant Question Of The Day: 274

**Time Speed and Distance**

**Two cars P and Q start from two points A and B towards each other simultaneously. They meet for the first time 40 km from B. After meeting they exchange their speeds as well as directions and proceed to their respective starting points. On reaching their starting points, they turn back with the same speeds and meet at a point 20 km from A. Find the distancebetween A and B.A. 110 kmB. 120 kmC. 130km D. 100 km. **

**See our previous ‘Questions of the Day’:**

Quant Question Of The Day: 273

Quant Question Of The Day: 272

In such questions if first meeting happens at A km from one end and second meeting at B km from another end then distance between two ends is equal to 3A-B.

In this question 3×40-20=100

Provided that the speed ratio (faster is to slower) should be less than or equal to 2.

Absolutely Correct Ashish! 🙂

But, how to derived that it is always 3A – B ?

Take distance = d

Now time would be equal for covering d-40 and 40 with speeds Va and Vb so (d-40)/Va=40/Vb

Va/Vb=d-40/40

In the same wat for second meeting Va/Vb=(d+d-20)/d+20

Equate the above 2 equations solve the quadratic.

In the same way if we take 40=a and 20=b … you will get d=3a-b

But the interesting thing is the ratio of sppeds of faster is to slower should be less than or equal to 2 otherwise second meeting may take place as overtaking not by way of face to face … coming from oppostite directions.

answer is d(100km)