Q:

Having trouble with question involving percentiles and symmetry

A large university will begin a $13$-day period during which students may register
  for that semester’s courses. Of those $13$ days, the number of elapsed days
  before a randomly selected student registers has a continuous distribution
  with density function $f(t)$ that is symmetric about $t = 6.5$ and proportional
  to $1 \over t+1$ between days $0$ and $6.5.$ 
  A student registers at the $60$th percentile of this distribution. Calculate the
  number of elapsed days in the registration period for this student.

The constant of proportionality is $.5\over ln7.5$,
So I thought you should do $${\int^x_0}{.5\over ln7.5}{1\over t+1}dt=.6$$
$${.5\over ln7.5}ln(x+1)-{.5\over ln7.5}ln(0+1)=.6$$
Which comes out to $x+1=e^\left({.6{ln7.5\over.5}}\right)$
However the real solution says you should do
$${\int^{13-x}_0}{.5\over ln7.5}{1\over t+1}dt=.4$$
$${.5\over ln7.5}ln(13-x+1)-{.5\over ln7.5}ln(0+1)=.4$$
Which comes out to $14-x=e^\left({.4{ln7.5\over.5}}\right)$ (a different number than my solution.)
If I understand correctly, the rationale behind the correct solution is that since there is a symmetry, the first 4oth percentile is the same as the last, which is the 60th percentile.  I just don't understand why my way is not correct.

A:

The symmetry of the pdf means that after between 6.5 and 13 days the pdf is actually ..
$${.5\over ln7.5}{1\over (13-t)+1} $$
you could get the correct answer by noting that $x>6.5$
so 
$${\int^{6.5}_0}{.5\over ln7.5}{1\over t+1}dt + {\int^{x}_{6.5}}{.5\over ln7.5}{1\over 14-t}dt =.6 
\\ \implies  {\int^{x}_{6.5}}{.5\over ln7.5}{1\over 14-t}dt =.1 $$

