jmc

We can write the two digit number in the form of $10a+b$; reverse of $10a+b$ is $10b+a$. The sum of those numbers is: $$(10a+b)+(10b+a)=132$$$$11a+11b=132$$$$a+b=12$$ We can use brute force to find order pairs $(a,b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$. Thus, our ordered pairs are $(3,9);(4,8);(5,7);(6,6);(7,5);(8,4);(9,3)$; or $\boxed{7}$ ordered pairs.