NCHMCT JEE 2015 Question Paper with solutions for online practice

${\mathit{a}}^2+{\mathit{b}}^2$ = 73
a – b = 5 ...(i)
${\left(\mathit{a}\mathit{–}\mathit{b}\right)}^2$ = ${\mathit{a}}^2+{\mathit{b}}^2$ – 2ab
⇒ ${\left(5\right)}^2$ = 73 – 2ab
⇒ 2ab = 73 – 25 = 48
Now, ${\left(\mathit{a}+\mathit{b}\right)}^2$ = ${\left(\mathit{a}\mathit{–}\mathit{b}\right)}^2$ + 4ab
= ${\left(5\right)}^2$ + 2 × 2ab
= 25 + 2 × 48
= 25 + 96 = 121
⇒ a + b = 11 ...(ii)
Solving (i) and (ii)
a + b = 11
a – b = 5
––––––––
2a = 16 ⇒ a = 8
b = 11 – 8 = 3
∴ Value of a = 8.