Rafiq and Abir start simultaneously from a place A towards B 50 km apart. Rafiq's speed is 3km/h less than that of Abir. Abir, after reaching B, turns back and meets Rafiq at a places 10 km away from B. Rafiq's speed is-

Question: Rafiq and Abir start simultaneously from a place A towards B 50 km apart. Rafiq's speed is 3km/h less than that of Abir. Abir, after reaching B, turns back and meets Rafiq at a places 10 km away from B. Rafiq's speed is-

Question:
Let, the speed of Rafiq be = a kmph
Now,
Abir's speed = (a + 3) kmph

∴ Distance covered by Abir = (50 + 10) = 60 km
∴ Distance covered by Rafiq = (50 - 10) = 40 km.

ATQ,
60/(a + 3) = 40/a
⇒ 60a = 40a + 120
⇒ 60a - 40a = 120
⇒ 20a = 120
⇒ a = 120/20
∴ a = 6 kmph