Why are roads sloped inwards in hills? I recently noticed that the roads were sloped inwards at hills while taking a turn and a question jumped in my mind i understood that it is to provide centripetal force but if it isn't there why would the car skid off would it be something related to inertia of direction and it not being able to turn quickly enough
 A: The sloping of road on curves is known as 'Banking of road'.
Consider a level circular road(without banking) then if we find the minimum velocity to make a turn on that road is $\sqrt{\mu gr}$
So, to overcome this less speed problem roads are banked form curves.
When a road is curved one of the components of reaction force adds up with one component of friction. Thus, increasing the centripetal force which acts inwards the curve i.e. sum of frictional component and reaction component.
Thus we can see banking increases the velocity on a curved road.
V max on a curved banked road given in the diagram: 
Image source: Google
However some other reason can also be supported in this case.
A: 
Let this be our car.

When the car is moving on an inclined slope,
The net inward force= $ f + W cos \theta $

Net outward force=$ mv^2/r $
and $f$max= $\mu Wsin\theta$
thus max value of inward force, available to balance $mv^2/r$ is $\sqrt{1+\mu ^2}\space W$
But when road isn't slanted,

We can have,

$ mv^2/r$ is balanced by $ f $ only, whose max value is $\mu W $
Now you see the difference? 
The body can attain more velocity without skidding off the road when the road is constructed slanted inwards about the turns. (especially in the hilly regions)

Also, lets talk about some actual stuffs, where your car isn't just a point sized body. But one with its own dimensions. There, we can talk about rotation.

When the body is on inclined road,
for it to remain stable
Torque of $ mv^2/r -W sin \theta $ about the left wheel is to balance the torque of $ W cos \theta $. Say the distances of the centre of mass of car from sides be a and b (refer the image #1)
Torque equation:
$ (mv^2/r - W sin \theta)b= W a \space cos \theta$
$\Rightarrow (mv^2/r)*b = W (a\space cos \theta + b \space sin \theta)$ 
$\Rightarrow v^2$max$=\frac{Wr \sqrt{a^2+b^2}}{mb}$

Now lets check the straight road
Torque equation:
$ (mv^2/r)b= W a$
$\Rightarrow v^2$max$=\frac{Wra}{mb}$

Again here my friend, you can notice our not-so-beloved car can take more velocity on the slope without toppling away. 
Over-speeding in the turns of straight(not slanted) roads can make your car both "SLIP" and "TOPPLE". This is mostly what happens at the hilly road accidents.  So when you'll begin driving, better be slow than this... 
