SRCC GBO 2009

$$0.\overline{3}$$ can be written as $$\frac{1}{3}$$. So the possible values of x, y will be as follows.

3,1 $$\Rightarrow\ (\frac{1}{x})^{\frac{1}{y}}=\frac{1}{3}$$. Here, xy=3.

9,2 $$\Rightarrow\ (\frac{1}{x})^{\frac{1}{y}}=\left(\frac{1}{9}\right)^{\frac{1}{2}}=\frac{1}{3}$$. Here, xy=18.

27,3$$\Rightarrow\ (\frac{1}{x})^{\frac{1}{y}}=\left(\frac{1}{27}\right)^{\frac{1}{3}}=\frac{1}{3}$$. Here, xy=81.

So, option B is correct.

We can do this question by looking at the options

Clearly the 2nd option is correct as $$5^2+6^2=61$$

Hence that is the correct answer

Let's look at Option 3

$$\frac{0.011\times\ 0.011}{1.21\times\ 0.01\times\ 0.01}=\frac{11\times\ 11\times\ 100}{100\times\ 121}$$

We can clearly see that on simplification that this will give 1. Hence this is the answer

When 1 is subtracted from $$10^n$$, it always results in a series of 9s.

Let the number of 9's be n. 9(n)=4707 or n=523

Hence there are 523, 9's. Thus n=523

Now we have $$\left(\frac{1}{5}\right)^{2000}=\left(\frac{2}{10}\right)^{2000}$$

As we can see the last digit of the decimal representation will be the unit digit of 2^2000. Now the cyclicity of 2 is 4, i.e powers of 2 repeats its unit digit after every 4 units. Now divide 2000 by 4, we get remainder as 0 . So we can say that 2^2000 is equivalent to 2^4. Hence the unit digit of 2^2000 is the unit digit of 2^4. Hence the unit digit is 6

 The number of digits in $$4^{50}$$ is $$\log\ 4^{50}=50\ \log\ 4$$

or 100 log 2 = 100(0.3010)=30.10

Number of digits 30+1=31

We can simplify the LHS to $$\log\ \left(\frac{a^2}{b}\times\ \frac{b}{a^2}\right)\ =\log\ 1$$ 

Now log 1 = log(a+b) Hence a+b=1 . 

$$\frac{1}{\log_ab}\times\ \frac{1}{\log_cb}\times\ \frac{1}{\log_ac}=\frac{1}{\log_ab}\times\ \frac{\log_ca}{\log_cb}=\frac{\log_ab}{\log_ab}=1$$

Let the side of the square be a cm. Hence the area shall be $$a\ cm^2$$

Hence, the diagonal of the square shall be $$a\sqrt{\ 2}$$, and a square with this side shall have an area of $$2a\ cm^2$$

Thus, the ratio of these two areas is 1:2

For the largest possible cube (of side a) in a sphere with radius r:

Diagonal of cube = diameter of the sphere, which is $$a\sqrt{\ 3}$$= 2r, or a = $$\ \frac{\ 2r}{\sqrt{\ 3}}$$

Now, the volume of this cube is $$a^3$$, or $$\ \frac{\ 8r}{3\sqrt{\ 3}}\ cm^3$$