When two cells of different emf’s are connected in series to an external resistance the current is 5A. When the poles of one cell are interchanged, the current is 3A. The ratio of emf’s of two cell is

Let the emfs be $E_1 and E_2$ and internal resistances be $r_1 and r_2$

External resistance be R.

Case 1 :

Current in circuit is 

$i = \frac{E_1+E_2}{r_1+r_2+R}=5A$

Case 2 :

Current in circuit is 

$i = \frac{E_1-E_2}{r_1+r_2+R}=3A$

Dividing above equations gives

$\frac{E_1+E_2}{E_1-E_2}=\frac{5}{3}$

$\Rightarrow \frac{E_1}{E_2}=\frac{4}{1}$