Nuclear Chemistry

An (n, p) reaction means the target absorbs a neutron $^{1}_{0}n$ and emits a proton $^{1}_{1}p$.
Target $^{A}_{Z}X + ^{1}_{0}n \rightarrow ^{60}_{27}Co + ^{1}_{1}p$
Balancing mass numbers: $A + 1 = 60 + 1 \Rightarrow A = 60$
Balancing atomic numbers: $Z + 0 = 27 + 1 \Rightarrow Z = 28$
The element with Z = 28 is nickel, so the target is $^{60}_{28}Ni$.
$^{60}_{28}Ni + ^{1}_{0}n \rightarrow ^{60}_{27}Co + ^{1}_{1}p$

Fission reaction: ${}^{235}_{92}U + {}^{1}_{0}n \rightarrow {}^{139}_{54}Xe + {}^{94}_{38}Sr + x$
Atomic numbers: left side $92 + 0 = 92$; right side $54 + 38 = 92$. So x carries no charge (Z = 0).
Mass numbers: left side $235 + 1 = 236$; right side $139 + 94 = 233$. So x has a total mass number of $236 - 233 = 3$.
Particles with Z = 0 and A = 1 are neutrons, so x = 3 neutrons: ${}^{235}_{92}U + {}^{1}_{0}n \rightarrow {}^{139}_{54}Xe + {}^{94}_{38}Sr + 3\,{}^{1}_{0}n$
So the product x is 3 neutrons.
Note: the answer key of the source gives 2 neutrons; it is changed here because the mass numbers balance only with 3 neutrons.

Here b is the mass number and a is the atomic number of X; c and d are the mass number and atomic number of Y.
Effect of each emission: positron ($\beta^+$): A unchanged, Z decreases by 1. $\alpha$: A decreases by 4, Z decreases by 2. $\beta^-$: A unchanged, Z increases by 1.
Change in mass number: three $\alpha$ particles in total, so $c = b - 3 \times 4 = b - 12$.
Change in atomic number: $-1$ (positron) $- 2 \times 2$ (two $\alpha$) $+ 2 \times 1$ (two $\beta$) $- 2$ (last $\alpha$) $= -5$, so $d = a - 5$.
So the correct relation is c = b – 12 and d = a – 5.