Numerical answer
P and Q start at the same time from towns X and Y respectively, walking towards each other at constant speeds. After they cross paths, P takes 16 hours to reach Y, and Q takes 4 hours to reach X. If Q's speed is 45 km/h, what is the distance XY, in km?
Show answer and solution
Answer: 540
Let them cross after t hours, with speeds vP and vQ. The remaining stretch for P (to Y) is what Q covered in t hours, and vice versa: vQ·t = vP·16 and vP·t = vQ·4.
Dividing these: t/4 = 16/t (from rearranging) → t² = 16 × 4 = 64 → t = 8.
From vP·t = vQ·4: vP = vQ × 4/t = 45 × 4/8 = 22.5 km/h.
Distance XY = (vP + vQ) × t = (22.5 + 45) × 8 = 540 km.
Check: P covers 22.5 × 8 = 180 km by the crossing point, leaving 360 km at 22.5 km/h = 16 h ✓; Q covers 45 × 8 = 360 km, leaving 180 km at 45 km/h = 4 h ✓.
Tip: Meeting time = √(product of the two post-crossing times); the speed ratio is the inverse square root of that same ratio.
Dividing these: t/4 = 16/t (from rearranging) → t² = 16 × 4 = 64 → t = 8.
From vP·t = vQ·4: vP = vQ × 4/t = 45 × 4/8 = 22.5 km/h.
Distance XY = (vP + vQ) × t = (22.5 + 45) × 8 = 540 km.
Check: P covers 22.5 × 8 = 180 km by the crossing point, leaving 360 km at 22.5 km/h = 16 h ✓; Q covers 45 × 8 = 360 km, leaving 180 km at 45 km/h = 4 h ✓.
Tip: Meeting time = √(product of the two post-crossing times); the speed ratio is the inverse square root of that same ratio.
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