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Figgerin'


From pastilla: How many oz of 40% alcohol must be mixed with 10 oz of 70% alcohol to give a solution that is 45% alcohol? (fun GRE question)

After initially stumbling over the same basic subtraction as dblume (like, 7 - 4.5 = two point five, duh....) I solved the problem this way:

.4x + .7(10) = .45(10+x)
.4x + 7 = .45(10+x)
.4x + 7 = 4.5 + .45x
.4x = -2.5 + .45x
.05x = 2.5
x = 50 oz

Comments

( 5 comments — Leave a comment )
dblume
Dec. 11th, 2008 03:02 am (UTC)
Did you really make the same mistake as me? What are the odds of that? Yay for errors rudimentary math!
davidd
Dec. 11th, 2008 03:40 am (UTC)

Yup, really did.

I guess that's why they named us Duh-vid!

Edited at 2008-12-11 06:41 am (UTC)
pastilla
Dec. 11th, 2008 12:11 pm (UTC)


:: large Sanrio sticker ::



pastilla
Dec. 11th, 2008 12:12 pm (UTC)
Now, to do the field work . . .

:: assembles Pyrex cup, Everclear, lemon wedges ::
sjonsvenson
Dec. 11th, 2008 05:47 pm (UTC)
^_~
I love science experients.
Mix 10 at 70% with 10 at 40% ... nope, that's 55%, won't do so drink it.
Mix 10 at 70% with 20 at 40% ... nope, 50% is still to uch. Drink it.
Mix 10 at 70% with 30 at 40% ... (hips), 47 and a bit ... Yay, drink it.
Mhix 1o t 70& wih 40 at 40 ... hihi forty 'n six, yezzz, nother round.
hix then at ... high with 50 at lo ... (hips) ... oops dranked tho earli moust try hagain ...

Mihix 1..0 at se ve ...
*thunk*
ZZzzzzz .... snore... snore...



(so you need 150oz at 40% and 50oz at 70% to get the right mix ... and you only know the answer the morning after ^_~ )


Edited at 2008-12-11 08:48 pm (UTC)
( 5 comments — Leave a comment )

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