ignoring overflow in two's complement addition of numbers with different signs[specific case]
Okay, let's take a look at your example:

 1001010
+0010101
--------
 1011111


Your numbers here only have 7 bits, so we have a smaller range than we typically deal with.  Nevertheless:

$1001010$ is 

$-64 + 8 + 2$, which is $-54$.

$0010101$ is 

$16 + 4 + 1$, which is $21$.

$21-54$ is $-33$.  Let's see whether our answer matches up:

$1011111$ is

$-64+16+8+4+2+1$, which is $-33$.

An intuitive way to think about this is that the number space is divided entirely in half (with $0$ acting as a positive number).  So, with the standard 8 bits, your numbers range from $-128$ to $127$.  There is no positive number from within that set that you could add to a negative number from that set and get a number either above $127$ or below $-128$.