SRCC GBO 2012

Expression : $$\left(\sqrt[6]{27} - \sqrt{6\frac{3}{4}}\right)^2 + \frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2} + \sqrt{3}}$$

= $$[(3)^{\frac{3}{6}}-(\frac{3\sqrt3}{2})]^2$$ $$+\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2} + \sqrt{3}}$$

= $$(\sqrt3-\frac{3\sqrt3}{2})^2+$$ $$\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2} + \sqrt{3}}$$

= $$(\frac{-\sqrt3}{2})^2+$$ $$\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2} + \sqrt{3}}$$

= $$\frac{3}{4}+$$ $$\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2} + \sqrt{3}}$$

= $$\frac{9\sqrt2+3\sqrt3+8\sqrt2+8\sqrt6}{4(3\sqrt2+\sqrt3)}$$

= $$\frac{3\sqrt3(\sqrt6+1)+8\sqrt2(\sqrt3+1)}{4\sqrt3(\sqrt6+1)}$$

Expression : $$\left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{4^2}\right).....\left(1 - \frac{1}{11^2}\right)\left(1 - \frac{1}{12^2}\right)$$

= $$\left(1^2 - \frac{1}{2^2}\right)\left(1^2 - \frac{1}{3^2}\right)\left(1^2 - \frac{1}{4^2}\right).....\left(1^2 - \frac{1}{11^2}\right)\left(1^2 - \frac{1}{12^2}\right)$$

= $$\left(1 - \frac{1}{2}\right)(1+\frac{1}{2})\left(1 - \frac{1}{3}\right)(1+\frac{1}{3})\left(1 - \frac{1}{4}\right)(1+\frac{1}{4}).....\left(1 - \frac{1}{11}\right)(1+\frac{1}{11})\left(1 - \frac{1}{12}\right)(1+\frac{1}{12})$$

= $$(\frac{1}{2})(\frac{3}{2})\times(\frac{2}{3})(\frac{4}{3})\times(\frac{3}{4})(\frac{5}{4})\times(\frac{4}{5})(\frac{6}{5})\times.................\times(\frac{10}{11})(\frac{12}{11})\times(\frac{11}{12})(\frac{13}{12})$$

= $$\frac{1}{2}\times\frac{13}{12}=\frac{13}{24}$$

=> Ans - (D)

Option D has to be the smallest, since $$\sqrt{\ 26}$$ is nearly 5.1, which when multiplied with 10 shall get almost totally cancel out the positive number 51. 

Terms : $$2^{\frac{1}{2}}, 3^{\frac{1}{3}}, 8^{\frac{1}{8}}$$ and $$9^{\frac{1}{9}}$$

L.C.M. (2,3,8,9) = 72

Multiplying the exponents by 72, we get :

= $$2^{\frac{72}{2}}, 3^{\frac{72}{3}}, 8^{\frac{72}{8}}$$ and $$9^{\frac{72}{9}}$$

= $$(2)^{36},(3)^{24},(8)^9,(9)^8$$

Now, clearly, $$(2)^{36}>(8)^9$$ and $$(3)^{24}>(9)^8$$

Now, among $$(2)^{36}$$ and $$(3)^{24}$$

these can be written as : $$(8)^{12}$$ $$<$$ $$(9)^{24}$$

Thus, greatest among these are : $$3^{\frac{1}{3}}$$

=> Ans - (B)

Expression : $$3^{1001} \times 7^{1002} \times 13^{1003}$$

= $$3^{4k+1} \times 7^{4k+2} \times 13^{4k+3}$$

Unit's digit will be same as unit digit of $$(3)^1\times(7)^2\times(3)^3$$

Only unit digit considered = $$3\times9\times7=9$$

=> Ans - (D)

Expression : $$\frac{1}{1 - \sqrt{2} + \sqrt{3}} + \frac{1}{1 - \sqrt{2} - \sqrt{3}} - \frac{2}{1 + \sqrt{2} - \sqrt{3}} + \frac{3}{\sqrt{2}}$$

= $$[\frac{1}{1-\sqrt2+\sqrt3}\times\frac{(1-\sqrt2)-\sqrt3}{(1-\sqrt2)-\sqrt3}]+[\frac{1}{1-\sqrt2-\sqrt3}\times\frac{(1-\sqrt2)+\sqrt3}{(1-\sqrt2)+\sqrt3}]-2[\frac{1}{1+\sqrt2-\sqrt3}\times\frac{(1+\sqrt2)+\sqrt3}{(1+\sqrt2)+\sqrt3}]+[\frac{3}{\sqrt2}]$$

= $$[\frac{1-\sqrt2-\sqrt3}{(1-\sqrt2)^2-(\sqrt3)^2}]+[\frac{1-\sqrt2+\sqrt3}{(1-\sqrt2)^2-(\sqrt3)^2}]-2[\frac{1+\sqrt2+\sqrt3}{(1+\sqrt2)^2-(\sqrt3)^2}]+[\frac{3}{\sqrt2}]$$

= $$[\frac{1-\sqrt2-\sqrt3}{1-2\sqrt2+2-3}]+[\frac{1-\sqrt2+\sqrt3}{1-2\sqrt2+2-3}]-2[\frac{1+\sqrt2+\sqrt3}{1+2\sqrt2+2-3}]+[\frac{3}{\sqrt2}]$$

= $$[\frac{1-\sqrt2-\sqrt3}{-2\sqrt2}]+[\frac{1-\sqrt2+\sqrt3}{-2\sqrt2}]-2[\frac{1+\sqrt2+\sqrt3}{2\sqrt2}]+[\frac{6}{2\sqrt2}]$$

= $$[\frac{-1+\sqrt2+\sqrt3}{2\sqrt2}]+[\frac{-1+\sqrt2-\sqrt3}{2\sqrt2}]+[\frac{-2-2\sqrt2-2\sqrt3}{2\sqrt2}]+[\frac{6}{2\sqrt2}]$$

= $$\frac{1}{2\sqrt2}[(-1-1-2+6)+(\sqrt2+\sqrt2-2\sqrt2)+(\sqrt3-\sqrt3-2\sqrt3)]$$

= $$\frac{1}{2\sqrt2}(2-2\sqrt3)$$

= $$\frac{1-\sqrt3}{\sqrt2}$$

= $$\frac{1}{2}(\sqrt2-\sqrt6)$$

=> Ans - (D)

Let $$x=5,z=4,y=3$$

(A) : $$(x-z)^2y =1\times3=3$$ is odd

(B) : $$(x-z)^2y^2=1\times9=9$$ is odd

(C) : $$(x-z)y=1\times3=3$$ is odd

(D) : $$(x-y)^2z=4\times4=16$$ is even

=> Ans - (A)

We are given that p on being divided by d given that quotient q and remainder r, giving us the equation
p = qd + r

From the second relation, we can get the second equation to be q = d'q' + r'

Substituting the value of q from the second equation into the first equation, we get p to be (q'd' + r')d + r
p = q'd'd + r'd + r

Dividing this by dd', the first term will not give any remainder as it is a multiple of dd', and the second and third terms are not multiples of dd', giving the remainder to be r'd +r

Therefore, Option A is the correct answer. 

Expression : $$N = \sqrt{25^{64} \times 64^{25}}$$

= $$\sqrt{(5)^{128}\times(2)^{150}}$$

= $$(5)^{64}\times(2)^{75}$$

= $$(2)^{11}\times(10)^{64}$$

= $$2048\times(10)^{64}$$

Thus, there are 68 digits in the numbers and sum of digits = $$2+0+4+8=14$$

=> Ans - (C)

Let $$p=\frac{1}{q}$$

If $$p$$ increases by 20%, => $$p'=\frac{6p}{5}=\frac{6}{5q}$$

Now, $$p'=\frac{1}{q'}$$

=> $$q'=\frac{1}{p'}$$

=> $$q'=\frac{5q}{6}$$

$$\therefore$$ Percentage decrease in $$q=\frac{1}{6}\times100=16\frac{2}{3}\%$$

=> Ans - (D)