A person walks from A to M at \(p\) km/h and from M to B at \(2q\) km/h. The time taken is the same as that in which, walking at a uniform speed of \(3r\) km/h, he could have walked from A to B and back again. If AM : MB = 1 : 2, then:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\frac{2}{r}=\frac{1}{p}+\frac{1}{q}\)
Time \(=\) distance \(\div\) speed. Write the two parts of the walk in terms of one length and set the two times equal.
Assuming AM \(=x\) km, MB \(=2x\) km, so AB \(=3x\) km:
A to M at \(p\) km/h: \(\displaystyle\frac{x}{p}\) h
M to B at \(2q\) km/h: \(\displaystyle\frac{2x}{2q}=\frac{x}{q}\) h
A to B and back (\(6x\) km) at \(3r\) km/h: \(\displaystyle\frac{6x}{3r}=\frac{2x}{r}\) h
\(\displaystyle\frac{x}{p}+\frac{x}{q}=\frac{2x}{r}\), and \(x\) cancels: \(\displaystyle\frac{2}{r}=\frac{1}{p}+\frac{1}{q}\)