What friction of a clockwise revolution does the hour hand of a clock turn through, when it goes from

The correct option is A $3$ to $9$

Let us consider that for one hour the fraction of revolution is $\frac{1}{12}$

then :

(i) $3$ to $9$ means $6$ hours

therefore; fraction of revolution $\frac{6}{12}=\frac{1}{2}$

(ii) $4$ to $7$ means $3$ hours

therefore; fraction of revolution $\frac{3}{12}=\frac{1}{4}$

(iii) $7$ to $10$ means $3$ hours

therefore; fraction of revolution $\frac{3}{12}=\frac{1}{4}$

(iv) $12$ to $9$ means $9$ hours

therefore; fraction of revolution $\frac{9}{12}=\frac{3}{4}$

(v) $1$ to $10$ means $9$ hours

therefore; fraction of revolution $\frac{9}{12}=\frac{3}{4}$

(vi) $6$ to $3$ means $9$ hours

therefore; fraction of revolution $\frac{9}{12}=\frac{3}{4}$