Question

2. A woman travelled from Lagos to Ondo a distance of 350km. Her average speed on the outward journey was ykm. On the return journey, her average speed was 10km/hr faster and therefore she completed the return journey 37.5minutes earlier. Find A. The value of y B. Get average speed on the return journey in km/hr C. The total time taken for the whole journey in hrs. Good luck

Answer 1

70 km/h

9 hr 22.5 min

Explanation:

Given

• Distance = 350 km
• Speed = y
• Return speed = y + 10
• Time difference on return = 37.5 min = 37.5/60 hr = 5/8 hr

Difference in time as equation:

• 350/y - 5/8 = 350/(y+10)
• (8*350 -5y)/8y = 350/(y+10)
• (y+10)(560 - y)= 8y*70
• 560y - y² +5600 - 10y = 560y
• y² + 10y - 5600=0
• y = (-10 ± √(100 +4*5600))/2
• y = (-10 ±150)/2
• y= 70
• y = -80 not considered as negative value

Average speed was 70 km/h

Total time = 350/70 + 350/(70+10) = 5 + 4.375 = 9.375 hr = 9 hr 22.5 min