GoldenLeaf

Find 65% of 79 c/= (Yay rudimentary cent sign!) to begin with.

$${\mathtt{79}}{\mathtt{\,\times\,}}{\mathtt{0.65}} = {\frac{{\mathtt{1\,027}}}{{\mathtt{20}}}} = {\mathtt{51.35}}$$

Add that number to the original.

$${\mathtt{79}}{\mathtt{\,\small\textbf+\,}}{\mathtt{51.35}} = {\frac{{\mathtt{2\,607}}}{{\mathtt{20}}}} = {\mathtt{130.35}}$$

The candy bar would then cost 130 c/= (Or, if you prefer, $1.30)

Alas! I have failed a math once again =(

Math has not been flowing through me recently it seems.

$${root3}{\left({\mathtt{8\,000}}\right)} = {\mathtt{20}}$$

Remember that a cube number is a number that is multiplied by itself 3 times. In this case, the answer is 20 because:

$${\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{20}} = {\mathtt{8\,000}}$$

1 + 2 = 3

You are adding one to two, which is one plus one. Thus, one plus one plus one is three.

Think of it this way, Anon.

Pretend you had four quarters of one pizza, or 1/4 for each quarter. You could divide the pizza so that it mysteriously 'vanishes', or so that when the pizza is served, well, there is no pizza to give out. Simply put, this means that there are zero quarters of pizza to give to three other people and yourself. This is why 0/4 is perfectly normal, and equals zero. Sadly, taking all four quarters for yourself will probably leave you with three angry friends and one satisfied (and probably stuffed) self.

However, if you have four slices of pizza and three friends plus yourself, you can't just not give the pizza away, even to yourself. This option is not feasible, and thus is invalid. I like to think of it as the pizza goes nasty, and thus nobody wants it, and it isn't good. This is why 4/0 is undefined, and this is simply because you can't just have something that is givable without giving it to something.

It's my weird way of thinking about it.

$${\mathtt{z}}{\mathtt{\,-\,}}{\frac{{\mathtt{8}}}{{\mathtt{z}}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{z}}{\mathtt{\,-\,}}{\frac{{\mathtt{38}}}{{\mathtt{11}}}}{\mathtt{\,\times\,}}{\mathtt{z}}$$

PEMDAS

No P

No E

M and D are present

1z - 8/z - 5z - 38/11 * 1z

z(1z - 8/z - 5z - 38/11 * 1z)

1z^2 - 8z/z - 5z^2 - 38z/11 * 1z^2

1z^2 - 8 - 5z^2 - (38z*1z^2)/11

1z^2 - 8 - 5z^2 - (1z^2*38z)/11

$${\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{z}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{z}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{z}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{38}}{\mathtt{\,\times\,}}{\mathtt{z}}\right)}{{\mathtt{11}}}}$$

11(1z^2 - 8 - 5z^2 - (1z^2*38z)/11)

11z^2 - 88 - 55z^2 - 11z^2 * 418z

(11z^2 - 88 - 55z^2 - 11z^2)/418z

$${\frac{\left({\mathtt{11}}{\mathtt{\,\times\,}}{{\mathtt{z}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{88}}{\mathtt{\,-\,}}{\mathtt{55}}{\mathtt{\,\times\,}}{{\mathtt{z}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{11}}{\mathtt{\,\times\,}}{{\mathtt{z}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{418}}{\mathtt{\,\times\,}}{\mathtt{z}}\right)}}$$

(1z/38) - (8/38) - (5z/38) - (1z/38)

(1z/38) - (5z/38) - (1z/38) - (8/38)

- (5z/38) - (8/38)

- (5z) - (8)

$${\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{z}}{\mathtt{\,-\,}}{\mathtt{8}}$$

I have no clue if this is correct. If it isn't, could somebody point me where I went wrong?

3(x+2)+4(2x+1)

3x+6+8x+4

11x+10

$${\mathtt{1\,234\,567\,890}}{\mathtt{\,\times\,}}{\mathtt{1\,234\,567\,890}} = {\mathtt{1\,524\,157\,875\,019\,052\,100}}$$

Pretty, isn't it?

Well, if she has been a widow for 3/10 of her life (her husband died after 7/10 of her life), and 3/10=24, then , to solve the puzzle, you simply have to find one tenth of her age.

$${\frac{{\mathtt{3}}}{{\mathtt{10}}}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{24}}$$

To get 1/10 from 3/10, the 3/10 must be divided by 3. Likewise, as you do to one side, you must do to both sides.

$${\frac{{\mathtt{1}}}{{\mathtt{10}}}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{8}}$$

Multiply by 10 on both sides. This will cancel out the 1/10 into 1 and find her true age.

$${\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{80}}$$

Aunt Helen is 80 years old. She started school when she was 4. She went to school for 12 years, or until she was 16. She worked for 4 years before finding her husband, Harry. She was married to Harry for 32 years, and he died when she was 52, and has spent the rest of the time since as a widow.

I think you messed up the puzzle and forgot to add 1/20 in there somewhere, because adding all of these values together gets 76 years old, though with the values given I get 80. Either way, she probably just likes to think that she is younger 

Some values:

$${\frac{{\mathtt{1}}}{{\mathtt{20}}}}{\mathtt{\,\times\,}}\left({\mathtt{80}}\right) = {\mathtt{4}}$$

$${\frac{{\mathtt{3}}}{{\mathtt{20}}}}{\mathtt{\,\times\,}}\left({\mathtt{80}}\right) = {\mathtt{12}}$$

$${\frac{{\mathtt{2}}}{{\mathtt{5}}}}{\mathtt{\,\times\,}}\left({\mathtt{80}}\right) = {\mathtt{32}}$$

$${\frac{{\mathtt{3}}}{{\mathtt{10}}}}{\mathtt{\,\times\,}}\left({\mathtt{80}}\right) = {\mathtt{24}}$$

Proof for why the puzzle isn't correct:

$$\left({\frac{{\mathtt{1}}}{{\mathtt{20}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{20}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{20}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{10}}}}\right) = {\frac{{\mathtt{19}}}{{\mathtt{20}}}} = {\mathtt{0.95}}$$