1. For a semiconductor, the drift current due to electrons and holes are:

$I_e = n e A v_e$ $I_h = p e A v_h$

where:

• $n$ = concentration of electrons
• $p$ = concentration of holes
• $e$ = electronic charge
• $A$ = cross-sectional area
• $v_e, v_h$ = drift velocities of electrons and holes

2. Given:

$\frac{n}{p} = \frac{7}{5}$

and the ratio of currents is

$\frac{I_e}{I_h} = \frac{7}{4}$

3. Using the current expressions,

$\frac{I_e}{I_h} = \frac{n e A v_e}{p e A v_h} = \frac{n}{p} \cdot \frac{v_e}{v_h}$

So,

$\frac{7}{4} = \frac{7}{5} \cdot \frac{v_e}{v_h}$

4. Solve for the drift velocity ratio:

$\frac{v_e}{v_h} = \frac{7}{4} \cdot \frac{5}{7} = \frac{5}{4}$

5. Therefore, the ratio of drift velocities is:

$\boxed{\frac{5}{4}}$

So, the correct option is C.