If the system of equations 2x + 3y = 5, 4x + ky = 10

The given system of equations

$2 x+3 y=5$

$4 x+k y=10$

$\frac{a_{1}}{a_{2}}=\frac{2}{4}, \frac{b_{1}}{b_{2}}=\frac{3}{k}, \frac{c_{1}}{c_{2}}=\frac{5}{10}$

For the equations to have infinite number of solutions

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\frac{2}{4}=\frac{3}{k}=\frac{5}{10}$

If we take

$\frac{2}{4}=\frac{3}{4}$

$2 k=12$

$k=\frac{12}{2}$

$k=6$

And

$\frac{3}{k}=\frac{5}{10}$

$30=5 k$

$\frac{30}{5}=k$

$6=k$

Therefore, the value of k is 6.

Hence, the correct choice is $d$