1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – almagest Jun 16 '16 at 14:25
  • $\begingroup$ 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. $\endgroup$ – JMoravitz Jun 16 '16 at 14:28
  • 1
    $\begingroup$ 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 $x+y+z=18.75$ $\endgroup$ – JMoravitz Jun 16 '16 at 14:33
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.