Car P and Q started from point A and point B respectively running towards each other. Car P running at $4Z$ km/h and meets Q after 10 hours. After reaching point A, driver of car Q realized that car R (running towards point B) is $\left(\frac{3Y}{5} - 360\right)$ km ahead of car Q. So, car Q started chasing car R and overtakes it in 6 hours. Find speed of car P, if distance between point A and point B is $Y$ km and speed of car R is $2Z$ km/h.

### Concept & Relative Speed When two bodies move towards each other, relative speed is the sum of their speeds. When one chases another, it's the difference. $$ \text{Distance} = \text{Relative Speed} \times \text{Time} $$ ### Step-by-Step Solution * **Given:** P and Q meet in 10 hours. Speed of P = $4Z$. Let Q's speed be $V_q$. Total distance $AB = Y$. * $10 \times (4Z + V_q) = Y \implies 40Z + 10V_q = Y \implies 10V_q = Y - 40Z$. * Q reaches A, meaning Q travels distance $Y$ from B to A. * When Q is at A, Q starts chasing R. R is $\left(\frac{3Y}{5} - 360\right)$ km ahead. * Speed of R = $2Z$. Relative speed of Q chasing R = $V_q - 2Z$. * Q catches R in 6 hours: $6 \times (V_q - 2Z) = \frac{3Y}{5} - 360$. * Multiply by 5: $30(V_q - 2Z) = 3Y - 1800$. * Substitute $3Y = 3(40Z + 10V_q) = 120Z + 30V_q$. * $30V_q - 60Z = 120Z + 30V_q - 1800$. * The term $30V_q$ cancels out perfectly: $-60Z = 120Z - 1800 \implies 180Z = 1800 \implies Z = 10$. * Speed of car P = $4Z = 4 \times 10 = 40$ km/hr. ### Exam Strategy & Shortcut Recognize that the speed of Q ($V_q$) acts as an intermediate variable. By aligning the coefficients of $Y$ in both equations, $V_q$ completely cancels out, allowing for a direct solution for $Z$ without needing Q's actual speed. ### Common Pitfall Assuming Q meets P and *then immediately* chases R from the meeting point. The problem explicitly states Q chases R "After reaching point A", meaning Q completed the full distance $Y$. ### Final Answer Therefore, the correct answer is **40 km/hr**.