A man left town am at 10:00 AM and traveled by car to town N at an average speed of 72 km/h. He spent 2hours for a meeting and returned through town M by bu...
Question 1 Report
A man left town am at 10:00 AM and traveled by car to town N at an average speed of 72 km/h.
He spent 2hours for a meeting and returned through town M by bus at an average speed of 40KM/H.
If the distance covered by the bus was 2km longer than that of the car and he arrived at town M at 1 :55PM.
calculate distance from M to N.
Answer Details
Setting up the time equation
Total time from 10:00 a.m. to 1:55 p.m. \(=3\text{ h }55\text{ min}=3\tfrac{55}{60}\text{ h}=3\tfrac{11}{12}\text{ h}\).
Removing the 2-hour meeting, the driving+bus time is
\[3\tfrac{11}{12}-2=1\tfrac{11}{12}\text{ h}=\tfrac{23}{12}\text{ h}.\]
Let the car distance \(M\!\to\!N\) be \(d\) km (at 72 km/h). The bus distance is \((d+2)\) km (at 40 km/h). Then
\[\frac{d}{72}+\frac{d+2}{40}=\frac{23}{12}.\]
Multiply through by 360:
\[5d+9(d+2)=690\Rightarrow 14d+18=690\Rightarrow 14d=672\Rightarrow d=48.\]
Check: car \(=48/72=40\) min, bus \(=50/40=1\text{ h }15\text{ min}\); driving \(=1\text{ h }55\text{ min}\); with the 2 h meeting the total is 3 h 55 min, arriving 1:55 p.m. \(\checkmark\)
Distance from \(M\) to \(N\) \(= 48\) km.
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