24. In the sum $K A N$ each letter stands for a different digit.
$$
\frac{+G A}{R O O}
$$

What is the answer to the subtraction $R N$ ?
$$
\underline{-K G}
$$

A 10
B 11
C 12
D 21
E 22
24. B Since $N \neq G$, to obtain the same value of $O$ for both the units and tens digits of the addition implies that there has been a 'carry' of 1 from the addition $N+A$ and therefore $N=G+1$. Similarly, since $K \neq R$, there has been a 'carry' of 1 from the addition $A+G$ and hence $R=K+1$. Therefore
$$
10 R+N-(10 K+G)=10(K+1)+G+1-(10 K+G)=10+1=11 .
$$