CMAT 2015 Solved Question Paper

(b) Given equations are $3\left[\frac{a}{b+1}\right]^{1/2}+2\left[\frac{b}{a+1}\right]^{1/2}$ = 14 $\;\;\;\;\;\;\;\;\;\;\;\;$.......(i)and $4\left[\frac{a}{b+1}\right]^{1/2}-2\left[\frac{b}{a+1}\right]^{1/2}$ = 7 $\;\;\;\;\;\;\;$.......(ii)Let $\left[\frac{a}{b+1}\right]^{1/2}$ = x and $\left[\frac{b}{a+1}\right]^{1/2}$ = y Now , 3x + 2y = 14 $\;\;\;\;\;\;\;\;\;\;\;\;$$\;\;\;\;\;\;\;\;\;\;\;\;$ $\;\;\;\;\;\;\;\;\;\;\;\;$........(iii)and $\;\;$ 4x - 2y = 7 $\;\;\;\;\;\;\;\;\;\;\;\;\;$$\;\;\;\;\;\;\;\;\;\;\;\;$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$........(iv)On adding Eqs. (iii) and (iv) , we get 3x + 2y = 14 $\underset{--------}{4x - 2y = 7}$ $\;\;\;\;\;\;$ 7x = 21∴ $\;\;\;\;\;\;$ x = $\frac{21}{7}$ = 3 Now, $\left[\frac{a}{b+1}\right]^{1/2}$ = 3 On squaring both sides, we get$\;\;\;\;\;\;$$\left[\frac{a}{b+1}\right] = 3^2$ ⇒ a = 9(b+1)$\;\;$ ⇒ a - 9b = 9 $\;\;\;\;\;\;\;\;\;\;\;\;$$\;\;\;\;\;\;\;\;\;\;\;\;$ .......(v)Now, putting the value of x in Eq. (iii), we get3x + 2y = 14 ⇒ 3 x 3 + 2y = 14⇒ 2y = 14 - 9∴$\;\;\;\;$ y = $\frac{5}{2}$Now , $\left[\frac{b}{a+1}\right]^{1/2} = \frac{5}{2}$On squaring both sides, we get$\;\;\;\;\;$$\left[\frac{b}{a+1}\right] = \left(\frac{5}{2}\right)^2$ ⇒ $\left[\frac{b}{a+1}\right] = \frac{25}{4}$⇒ 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 geta - 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 =$\frac{-250}{221}$