JEE Advanced 2018 — Paper 1 — Question 24

$\mathrm{x}_{\mathrm{A}}=\frac{1}{2}, \mathrm{x}_{\mathrm{B}}=\frac{1}{2}$

$\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{A}}^{0} \frac{1}{2}+\mathrm{P}_{\mathrm{B}}^{0} \times \frac{1}{2}$

(Given $\mathrm{P}_{\mathrm{A}}^{0}=20$ )

$90=45 \times 2=P_{A}^{0}+P_{B}^{0} …(1)$

$\mathrm{P}_{\mathrm{B}}^{0}=90-20=70$

$22.5=\mathrm{P}_{\mathrm{A}}^{0} \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0}\left(1-\mathrm{x}_{\mathrm{A}}\right)$

$=20 \mathrm{x}_{\mathrm{A}}+70\left(1-\mathrm{x}_{\mathrm{A}}\right)$

$22.5=20 \mathrm{x}_{\mathrm{A}}+70-70 \mathrm{x}_{\mathrm{A}}=70-50 \mathrm{x}_{\mathrm{A}}$

$\mathrm{x}_{\mathrm{A}}=\frac{47.5}{50}=\frac{19}{20}$ $\mathrm{X}_{\mathrm{B}}=\frac{1}{20}$

$\frac{\mathrm{x}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{B}}}=\frac{\frac{19}{20}}{\frac{1}{20}}=19$