Question

Let initial speed of neutron is v₀ and kinetic energy is K.
1^st collision :

by momentum conservation
mv₀ = mv₁ + 2mv₂ $\Rightarrow$ v₁ + 2v₂ = v₀
by e = 1         v₂ - v₁ = v₀
$\Rightarrow$v₂  = $\frac{2v_0}{3} ; v_1=-\frac{v_0}{3}$
fractional loss = $\frac{{\displaystyle \frac{1}{2}m{v}_0^2-\frac{1}{2}m{\left(\frac{v_0}{3}\right)}^2}}{\frac{1}{2}m{v}_0^2}$
$\Rightarrow$P_d = $\frac{8}{9}\approx .89$
2^nd collision :

by momentum conservation
mv₀ = mv₁ +12mv₂
$\Rightarrow$ v₁ + 12v₂ = v₀
by e = 1       v₂ - v₁ = v_{0
$V_2=\frac{2v_0}{13} ; V_1=\frac{-11v_0}{13}$
Now fraction loss of energy
$\frac{P_c={\displaystyle \frac{1}{2}}m{v}_0^1-{\displaystyle \frac{1}{2}}m{\left({\displaystyle \frac{11v_0}{13}}\right)}^2}{{\displaystyle \frac{1}{2}m{v}_0^2}}=\frac{48}{169}\approx 0.28$}