Honga

The amount of old water y is found with the equation, with n being the number of refills:

 $$y_{n}=\frac{3}{4}*y_{n-1}$$

This can be changed to a version where $$y_{n-1}$$ is irrelevant to $$y_{n}$$:

$$y_{n}=(\frac{3}{4})^{n}*y_{0}$$

With $$y_{0}$$ being the starting amount of water.

We are considering percentages hence we divide $$y_{n}$$ by $$y_{0}$$:

$$\frac{y_{n}}{y_{0}}=\frac{(\frac{3}{4})^{n}*y_{0}}{y_{0}}=(\frac{3}{4})^{n}$$

Now we can equate $$\frac{y_{n}}{y_{0}}$$ with the desired percentage and the amount of required refills:

$$\frac{y_{n}}{y_{0}}=(\frac{3}{4})^{n}=0.05$$

$$n=\frac{\log{0.05}}{\log{\frac{3}{4}}}\approx10.4$$

You'll need to refill 11 times to achieve a pollution level of 5%, you can figure the second percentage out for yourself. :)

$$\frac{2x}{\sqrt{8x}}=\frac{2x}{\sqrt{8x}}*\frac{\sqrt{2}}{\sqrt{2}}=\frac{2x\sqrt{2}}{\sqrt{16x}}=\frac{2x\sqrt{2}}{\sqrt{16}*\sqrt{x}}=\frac{2x\sqrt{2}}{4\sqrt{x}}=\frac{2\sqrt{x}\sqrt{2}}{4}=\frac{1}{2}\sqrt{2x}$$

Square meters is a different Si unit than meters, hence you can't 'fit' one in the other.

Well you can zoom in via your browser, the calculator will be enlarged while doing so (zoom 150% enlarges it to almost full screen).

The area of a square is equal to length times width.

Another known fact of a square is that length is equal to width.

Hence width*width=area.

$${\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4}} = {\frac{{\mathtt{1}}}{{\mathtt{6}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{7}}$$

$${\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{6}}}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{7}}$$

$${\frac{{\mathtt{9}}}{{\mathtt{12}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{8}}}{{\mathtt{12}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{12}}}}{\mathtt{\,\times\,}}{\mathtt{x}} = -{\mathtt{3}}$$

$${\frac{\left({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{12}}}}{\mathtt{\,\times\,}}{\mathtt{x}} = -{\mathtt{3}}$$

$${\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}{\mathtt{\,\times\,}}{\mathtt{x}}\right) = -{\mathtt{3}}$$

$${\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}}$$

$${\mathtt{x}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{12}} = {\mathtt{36}}$$

You probably gave a wrong equation or the wrong answer, 19 is a prime number and doesn't appear anywhere in the original equation (neither does a multiple of 19). This makes it highy unlikely that a fraction including 19 would be the answer.

To be honest the first part of your proof is exactly the same as the one I came up with. I think I messed up with the sine calculation though, my proof consisted of one and half page filled with sine equation reduction to get to a similar point you ended up with. Thanks for your help, +1!

D**n, multiplication seemed logical to me, nevertheless this should be able to be figured out as well.

The last number is relatively easy since it's the result of $${{\mathtt{2}}}^{{\mathtt{n}}}$$, which means the last digit regularly skips between this array of numbers: 2,4,8,6.

For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}\right)}$$ the last digit is 6.

For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}$$ the last digit is 2.

For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}$$ the last digit is 4.

For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}$$ the last digit is 8.

We know the last digit is multiplied 1122 times, $${\frac{{\mathtt{1\,122}}}{{\mathtt{4}}}} = {\frac{{\mathtt{561}}}{{\mathtt{2}}}} = {\mathtt{280.5}}$$ so it is of the form $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}$$ which means the last digit is 4.

While doing this I figured out a better way to solve this problem, we know this of the exponent:

$${\mathtt{1\,122}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{17}}$$

Another thing we know for sure is that while multiplying two numbers only the last two digits of each number affect the last two digits of the answer. Hence all digits save for the last two are irrelevant for further multiplication.

Since only the last two digits matter of the 1122 being multiplied we can drop the first two digits:

$${{\mathtt{22}}}^{{\mathtt{1\,122}}} = \left({\left({\left({\left({{\mathtt{22}}}^{{\mathtt{2}}}\right)}^{{\mathtt{3}}}\right)}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}\right)$$

$${\left({\left({{\mathtt{484}}}^{{\mathtt{3}}}\right)}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$

Only last two digits matter

$${\left({\left({{\mathtt{84}}}^{{\mathtt{3}}}\right)}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$

$${\left({{\mathtt{592\,704}}}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$

Only last two digits matter

$${\left({{\mathtt{4}}}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$

$${\left({{\mathtt{4}}}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}} = {\left({\mathtt{4\,194\,304}}\right)}^{{\mathtt{17}}}$$

Only last two digits matter

$${{\mathtt{4}}}^{{\mathtt{17}}}$$

$${{\mathtt{4}}}^{{\mathtt{17}}} = {{\mathtt{4}}}^{{\mathtt{11}}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{6}}} = {\mathtt{4\,194\,304}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{6}}}$$

Only last two digits matter

$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{6}}} = {{\mathtt{4}}}^{{\mathtt{7}}} = {\mathtt{16\,384}}$$

Last two digits of $${{\mathtt{1\,122}}}^{{\mathtt{1\,122}}}$$ are 84, case closed.

Not necessarily, the question mentions that the weight is supported by the two wires which indicates it's at rest.

The lens formula is:

$${\frac{{\mathtt{1}}}{{\mathtt{f}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{d1}}}} = {\frac{{\mathtt{1}}}{{\mathtt{d2}}}}$$

In which

f: focal length

d1:distance to source

d2:distance to image

Since the sun has an almost infinite distance to the lens d1=infinity, which means:

 $${\frac{{\mathtt{1}}}{{\mathtt{\infty}}}} = {\mathtt{0}}$$

so

 $${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{d2}}}}$$

Which means answer A is correct.