Simple Algebra

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Simple Algebra

by franciskyle » Sat Mar 28, 2009 5:07 pm
Hey all - this is a simple question from a Kaplan workbook.

If (q)(34)(36)(38) = (17)(18)(19) then q =?

Am I missing something... this is a pretty easy question, but I am getting the wrong answer. Here is my method (where is my logic askew?)

(q)(34)(36)(38) = (17)(18)(19)

Factor out the two on the LHS

(2)(q)(17)(18)(19) = (17)(18)(19)

Cancel the 17,18 & 19 on both sides.

2q = 1

Therefore, q = 1/2

Where have I gone wrong in this logic?
--------------------------------------------------------------
The book gives an answer of 1/8... their logic:

(q)(34)(36)(38) = (17)(18)(19)

therefore

q = [(17)(18)(19)]/[(34)(36)(38)]

q = (17/34)(18/36)(19/38)

q = (1/2)(1/2)(1/2)

q = 1/8

What am I doing wrong?

Thanks so much! :)
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by truplayer256 » Sat Mar 28, 2009 5:35 pm
I think that the explanation that I'm going to give for this problem is going to be a little better than the book's explanation. Here it goes:

(q)(34)(36)(38)=(17)(18)(19)
q=(17)(18)(19)/(34)(36)(38)
q=(17)(18)(19)/(17)(2)(18)(2)(19)(2)
The 17's, 18's, and 19's cancel and you're left with:
q=1/(2)(2)(2) or 1/8

Let me know if you don't understand anything in my explanation.
By the way, you can't factor the 2's out from 34,36, and 38 or you'll get 2*34*36*38 and that won't really get you anywhere. Now, if you had something like 34+36+38, then you can factor a 2 out and get 2(17+18+19).

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by franciskyle » Sat Mar 28, 2009 6:02 pm
Thanks. Now I see where I have gone wrong... sometimes the simplest careless mistake can haunt you!
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by gmat740 » Sat Mar 28, 2009 8:43 pm
I solved this question verbally
may be this be of any help to you

on LHS(Left Hand Side) we see multiples of 17,18 and 19,

q[(2)(17)]*[(2)(18)]*[(2)(19)] = (17)(18)(19)

q(2)(2)(2) = 1

q =1/8