At 5.45 p.m., Kathy and Peter left Hotel Y for Hotel Z at average speeds of 96 km/h and 68 km/h respectively. Upon reaching Hotel Z, Kathy rested for 10 minutes. She then headed back for Hotel Y at an average speed of 96 km/h along the same route. Peter and Kathy met each other on their way at 9 p.m.

1. How much more distance had Kathy covered than Peter when they met on their way?
2. How far apart were Hotel Y and Hotel Z? Express your answer in mixed number.

(a)
Duration from 5.45 p.m. to 9 p.m. = 3 h 15 min = 3¹₄ h
Duration from 5.45 p.m. to 9 p.m. without 10 min rest = 3 h 5 min = 3¹₁₂ h

60 min = 1 h
15 min = ¹⁵₆₀ = ¹₄ h
5 min = ⁵₆₀ = ¹₁₂ h


Distance covered by Kathy
= 96 x 3¹₁₂
= 96 x ³⁷₁₂ 
= 296 km

Distance covered by Peter
= 68 x 3¹₁₂
= 68 x ³⁷₁₂
= 221 km

Distance that Kathy covered more than Peter
= 296 - 221
= 75 km

(b)
Distance between Hotel Y and Hotel Z
= (296 + 221) ÷ 2
= 517 ÷ 2
= 258¹₂ km

Answer(s): (a) 75 km; (b) 258¹₂ km