Two persons P and Q started travelling towards each other at two places A and B respectively at 7 AM. The speeds of P and Q are in the ratio 4:5. Somewhere between A and B they met at C and spent some time together. From C at 8.07 AM P and Q started travelling towards B and A respectively. If P reaches B by 9.12 AM, the time they spent at C together is

Given:

P and Q started from A and B towards each other at 7:00 AM

Speeds of P : Q = 4 : 5

They meet at C and from C, they start again at 8:07 AM

P reaches B at 9:12 AM

Formula used:

Time = Distance / Speed

Speed Ratio = Inverse of Time Ratio

Calculation:

Let distance from A to B = D

Let speed of P = 4x, speed of Q = 5x

When they move towards each other, their relative speed = 4x + 5x = 9x

Let they meet at time t (in hours) after 7 AM

Then Distance = 9x × t ⇒ D = 9x × t

From C to B, P takes time = 9:12 AM − 8:07 AM = 1 hour 5 minutes = 65 minutes = 65/60 hrs

Distance from C to B = 5x × t (as Q would’ve covered 5x × t from B to C)

P covers this distance from C to B in 65/60 hours at 4x speed:

\(\frac{5xt}{4x} = \frac{65}{60}\)

⇒ \(\frac{5t}{4} = \frac{65}{60}\)

⇒ \(5t = \frac{65}{60} × 4 = \frac{260}{60}\)

⇒ \(t = \frac{260}{300} = \frac{13}{15}\) hours

⇒ \(\frac{13}{15} × 60 = 52\) minutes

So, they met at 7:00 AM + 52 minutes = 7:52 AM

They resumed at 8:07 AM

∴ Time spent at C = 8:07 − 7:52 = 15 minutes