# Why is it against common sense?

The question is this. A man can walk at speeds of 6kmph uphill, 7.5kmph along level surface and 10kmph downhill. He travels from A to B in 3 hours and from B to A in 1 hour. What is distance AB? I assumed uphill distance as $x$, level distance $y$ and downhill distance $z$ from A to B so that $AB=x+y+z$. Given $\frac{x}{6}+\frac{y}{7.5}+\frac{z}{10}=3,\quad \frac{x}{10}+\frac{y}{7.5}+\frac{z}{6}=1$ which gives $x-z=30 \iff x=z+30$. When substituted in the equation for A to B and simplified, I obtained $2z+y=-15\iff z+z+y=x-30+z+y=-15\iff AB=x+y+z=30-15=15$. This result appears senseless because 1)in 3 hours, even if it is completely uphill, he would travel 18km, 2)he cannot travel 15km in 1 hour even if it is completely downhill during his return journey. I cannot detect the error.

• Obviously if his max speed is only 1.7 times his min speed, he cannot have his journey time vary by a factor 3. Once you allow him to take breaks the problem becomes indeterminate. – almagest Jun 16 '16 at 14:25
• It appears that there is something wrong with the question itself, not so much with your work. You correctly row reduced, and wound up at the equation $2z+y=-15$. This directly implies that at least one of $z$ or $y$ is negative, which doesn't make sense. Who travels a negative distance uphill, thereby shaving time off of their total trip length. It would appear as written with the speeds and durations given it is an impossible scenario. – JMoravitz Jun 16 '16 at 14:28
• Perhaps the problem was intended to have been from $B$ to $A$ in two hours. That would get you to $x-z=15$ and $y+2z=3.75$, implying $x+y+z=18.75$ – JMoravitz Jun 16 '16 at 14:33

You obtained $2z+y=-15$, which appears correct. But this implies that at least one of $z$, $y$ is negative. The problem as stated is physically impossible.