CMAT 2015 Solved Question Paper

(b) Given equations are
$3{\left[\frac{\mathit{a}}{\left(\mathit{b}+1\right)}\right]}^{1∕2}+2{\left[\frac{\mathit{b}}{\mathit{a}+1}\right]}^{1∕2}$ = 14 $$.......(i)
and $4{\left[\frac{\mathit{a}}{\left(\mathit{b}+1\right)}\right]}^{1∕2}-2{\left[\frac{\mathit{b}}{\mathit{a}+1}\right]}^{1∕2}$ = 7 $$.......(ii)
Let ${\left[\frac{\mathit{a}}{\left(\mathit{b}+1\right)}\right]}^{1∕2}$ = x and ${\left[\frac{\mathit{b}}{\mathit{a}+1}\right]}^{1∕2}$ = y
Now , 3x + 2y = 14 $$$$ $$........(iii)
and $$ 4x - 2y = 7 $$$$$$........(iv)
On adding Eqs. (iii) and (iv) , we get

3x + 2y = 14
$\underset{--------}{4\mathit{x}-2\mathit{y}=7}$
$$ 7x = 21
∴ $$ x = $\frac{21}{7}$ = 3
Now, ${\left[\frac{\mathit{a}}{\left(\mathit{b}+1\right)}\right]}^{1∕2}$ = 3
On squaring both sides, we get
$$$\left[\frac{\mathit{a}}{\left(\mathit{b}+1\right)}\right]=3^2$ ⇒ a = 9(b+1)
$$ ⇒ a - 9b = 9 $$$$ .......(v)
Now, putting the value of x in Eq. (iii), we get
3x + 2y = 14 ⇒ 3 x 3 + 2y = 14
⇒ 2y = 14 - 9
∴$$ y = $\frac{5}{2}$
Now , ${\left[\frac{\mathit{b}}{\mathit{a}+1}\right]}^{1∕2}=\frac{5}{2}$
On squaring both sides, we get
$$$\left[\frac{\mathit{b}}{\mathit{a}+1}\right]={\left(\frac{5}{2}\right)}^2$
⇒ $\left[\frac{\mathit{b}}{\mathit{a}+1}\right]=\frac{25}{4}$
⇒ 25a + 25 = 4b
⇒ 25a - 4b = - 25 $$........(vi)
On multiplying Eq. (v) by 25 and then, subtracting Eq.(vi) from it, we get

25a - 225b = 225
25a - 4b = - 25
$\underset{-------}{+-+}$
- 221b = 250
∴ b = $\frac{-250}{221}$
On putting the value of b in Eq. (v), we get
a - 9$\left(\frac{-250}{221}\right)$ = 9
⇒ a + $\frac{2250}{221}$ = 9
⇒ a = 9 - $\frac{2250}{221}=\frac{-261}{221}$
∴ a = $\frac{-261}{221}$ and b =$\frac{-250}{221}$