Two masses of 3.4 kg and 2.5 kg are accelerated from an initial...

The distance traversed in the $n^{\text{th}}$ second is given by $S_n = u + \dfrac{a}{2}(2n - 1)$.

For the first mass ($m_1 = 3.4$ kg, $u_1 = 5$ m/s):
$104 = 5 + \dfrac{a_1}{2}(2 \times 5 - 1)$
$99 = \dfrac{9a_1}{2} \Rightarrow a_1 = 22$ m/s$^2$

For the second mass ($m_2 = 2.5$ kg, $u_2 = 12$ m/s):
$129 = 12 + \dfrac{a_2}{2}(2 \times 5 - 1)$
$117 = \dfrac{9a_2}{2} \Rightarrow a_2 = 26$ m/s$^2$

Velocities of the masses after $10$ s:
$v_1 = u_1 + a_1 t = 5 + 22 \times 10 = 225$ m/s
$v_2 = u_2 + a_2 t = 12 + 26 \times 10 = 272$ m/s

Momenta of the masses after $10$ s:
$p_1 = m_1 v_1 = 3.4 \times 225 = 765$ kg m/s
$p_2 = m_2 v_2 = 2.5 \times 272 = 680$ kg m/s

Ratio of their momenta:
$\dfrac{p_1}{p_2} = \dfrac{765}{680} = \dfrac{9}{8}$

Comparing with $\dfrac{x}{8}$, we get $x = 9$.

Answer: $9$