Quotient of free modules is torsion implies rank is the same?

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Let $N \subset M$ be free $A$-modules of finite rank. Suppose the quotient $M/N$ is torsion. Can I conclude the rank of $M$ and $N$ are the same?

For example, take $t k[t] \subset k[t]$.

Context: I arrived at this situation in the following way. Let $\mathcal E' \subset \mathcal E$ be locally free sheaves on a complex curve $X$. Suppose the quotient $\mathcal E / \mathcal E'$ is a torsion sheaf (in some stalk, there is torsion). Then I would like to show that $\mathcal E$ and $\mathcal E'$ have the same rank. Since $\mathcal E$ and $\mathcal E'$ are both locally free sheaves, I would imagine this comes down to a local computation. Either I should zoom into an affine subscheme on which both $\mathcal E$ and $\mathcal E'$ look like the sheafification of some module, or I should look at the stalks at some point. Either way, the above question seemed helpful. With the affirmative answer below by stewbasic, just zoom into stalks, and apply the answer to complete my problem.