GoldenLeaf

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Nombre de usuarioGoldenLeaf
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avatar+1006 

GoldenLeaf  29 abr 2015
 #2
avatar+1006 
0
26 may 2014
 #1
avatar+1006 
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$${\mathtt{z}}{\mathtt{\,-\,}}{\frac{{\mathtt{8}}}{{\mathtt{z}}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{z}}{\mathtt{\,-\,}}{\frac{{\mathtt{38}}}{{\mathtt{11}}}}{\mathtt{\,\times\,}}{\mathtt{z}}$$

 

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?

25 may 2014
 #5
avatar+1006 
+13

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:

 

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

.
25 may 2014