The sales for a company are normally distributed with mean $\mu$ and variance $\sigma^2$. The difference between the $90$th and $40$th percentile is $500$. The $70$th percentile is $1700$. What is the $60$th percentile?
I'm not quite sure how to proceed. [Subsequent to comment, see OP answer.]
On
First,
$Pr( Z\leq\frac{\pi_{.9}-\mu}{\sigma})=.9\rightarrow \frac{\pi_{.9}-\mu}{\sigma} = .1285$
$Pr( Z\leq\frac{\pi_{.4}-\mu}{\sigma})=.4 \rightarrow \frac{\pi_{.4}-\mu}{\sigma} = -.25$
$\Large \frac{\pi_{.9}-\mu}{\sigma} - \frac{\pi_{.4}-\mu}{\sigma} = \frac{\pi_{.9}-\pi_{.4} }{\sigma}$ = $.3785 \rightarrow .3785\sigma = 500 \rightarrow \sigma = 1321.004$
Then,
$\large \frac{1700 - \mu}{1321.004} =$ $.5244 \rightarrow \mu = 1007.266$
Finally,
$\large \frac{\pi_{.6}-1007.266}{1321.004}$ $= .255 \rightarrow \pi_{.6} = 1344.122$
You need to formulate several equations in several unknowns, and then solve. For now, I'll try to get you started without just giving the solution. Use the fact that $Z = (x - \mu)/\sigma$ has a standard normal distribution. I think this is fundamentally a question on using tables of the standard normal distribution.
From "The 70th percentile is 1700," one can see that $$P\{X \le 1700\} = P\{Z \le (1700 - \mu)/\sigma\}= 0.7$$ But also $P\{Z \le 0.5244\} = 0.7$ from tables, so $(1700 - \mu)/\sigma = 0.5244$ is one of your equations.
With that clue in mind, please try so show some progress toward another equation, and maybe a solution. Most of us on this site resist posting answers to homework problems without some participation from the student.