NIMCET 2018 Question Paper with solutions

${\left(235\right)}_{{\mathit{R}}_1}$ = ${\left(565\right)}_{10}$ = ${\left(1065\right)}_{{\mathit{R}}_2}$
⇒ ${\left(235\right)}_{{\mathit{R}}_1}$ = ${\left(565\right)}_{10}$
⇒ $2{\mathit{R}}_1^2+3{\mathit{R}}_1$ + 5 = 565
⇒ $2{\mathit{R}}_1^2+3{\mathit{R}}_1$ - 560 = 0
If we put ${\mathit{R}}_1$ = 16 , then
$2{\left(16\right)}^2$ + 3 (16) - 560 = 0
⇒ 512 + 48 - 560 = 0
⇒ 560 - 560 = 0
⇒ ${\left(565\right)}_{10}$ = ${\left(1065\right)}_{{\mathit{R}}_2}$
⇒ ${\mathit{R}}_2^3+6{\mathit{R}}_2$ + 5 = 565
⇒ ${\mathit{R}}_2^3+6{\mathit{R}}_2$ - 560 = 0
If we put ${\mathit{R}}_2$ = 8 , then
⇒ ${\left(8\right)}^3$ + 6 (8) - 560 = 0
⇒ 512 + 48 - 560 = 0
⇒ 560 - 560 = 0