Consider the combination of two capacitors C _ 1 and C _ 2 , with C...

When connected in parallel

${C}_{eq} = {C}_{1} + {C}_{2}$

When in series

${C}_{eq}^{'} = \dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}$

${C}_{1} + {C}_{2} = 10\left(\dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}\right)$

${\left({C}_{1} + {C}_{2}\right)}^{2} = 10{C}_{1}{C}_{2}$

${C}_{1}^{2} + {C}_{2}^{2} - 8{C}_{1}{C}_{2} = 0$

dividing by ${C}_{1}^{2}$

${\left(\dfrac{{C}_{2}}{{C}_{1}}\right)}^{2} - \dfrac{8{C}_{2}}{{C}_{1}} + 1 = 0$

Let $\dfrac{{C}_{2}}{{C}_{1}} = x$

${x}^{2} - 8x + 1 = 0$

$x = \dfrac{8 \pm \sqrt{64 - 4}}{2} = 4 + \sqrt{15}$

This question was Bonus in JEE Main 2021 exam because of incorrect data in original question.