My wife is very kind, she always picks me up at work by car and drives me home. Today, I finished at work 30 minutes earlier! So I decided to walk home... on the way I met my wife. She was on her way to pick me up, so I sat in the car and she drove me home. Today I was home 10 minutes earlier than usual.
How long did I walk?
Does anyone know how to solve it using maths and not just guessing? :)
Say the drive to work reliably takes $d$ minutes in each direction, and let's imagine that your wife, punctual soul, always turns up exactly as you finish work.
So normally you get home $d$ minutes after finishing time $f$, that is $f+d$, with your wife having left to pick you at $f-d$, driving for $2d$ minutes.
But this time, you started walking at $f-30$, and after leaving at $f-d$, your wife got home at $f+d-10$ and thus she only drove for $2d-10$ minutes - which must be $(d-5)$ minutes out and $(d-5)$ minutes back.
And that means - as others have already mentioned - that she picked you up $5$ minutes earlier than usual, that is at $(f-d)+(d-5) = f-5$.
So you were walking from $f-30$ to $f-5$; that is, $25$ minutes.