If a number $N$ is added to both numerator and denominator of $\dfrac{1}{2}$, the fraction becomes $\dfrac{2}{3}$. Then $N$ equals—

Adding $N$ gives $\dfrac{1+N}{2+N} = \dfrac{2}{3}$, so $3(1+N) = 2(2+N) \Rightarrow 3 + 3N = 4 + 2N$, giving $N = 1$. (This is option a; the marked key c would require the result to be $\tfrac{4}{5}$.)