gmat preparation courses

It is currently Mon Oct 15, 2018 5:48 am

All times are UTC - 5 hours [ DST ]




Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: GMAT Algebra
PostPosted: Thu Aug 05, 2010 12:49 pm 
Offline

Joined: Sun May 30, 2010 3:15 am
Posts: 424
If x / y = 3 / z, then 9y² =
A. x²/9
B. x²z
C. x²z²
D. 3x²
E. (1/9)x²z²

(C) This question is most easily solved by isolating y in the equation and substituting into the expression 9y²:
x/y = 3/z
x = 3y/z
3y = xz
y = xz/3.

If we substitute xz/3 into the expression for y, we get:
9(xz/3)² = 9(x²z²/9) = x²z².

The correct answer is choice (C).


Top
 Profile  
 
 Post subject: Re: math: algebra
PostPosted: Thu Aug 05, 2010 1:59 pm 
Offline

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Remember that we can multiply or divide each side of any equality by the same non-zero number and get the equivalent equality. That simple rule creates even simpler one, the cross-multiplication rule.

If we have
a/b = c/d we can

multiply each side by b and get
a = bc/d

or divide each side by c and get
a/(bc) = 1/d

So, basically, we can move any number from denominator of one side to the numerator of another or from numerator to the denominator.
Take a look at these animations to strengthen your understanding:
Image

Image

When we use cross-multiplication rule to solve this question we can easily find y:
Image

Then we can plug it in 9y² and simplify to get the answer.


Attachments:
abcd_3.gif [4.29 KiB]
Not downloaded yet
abcd_2.gif [4.58 KiB]
Not downloaded yet
abcd_1.gif [4.19 KiB]
Not downloaded yet
Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 posts ] 

All times are UTC - 5 hours [ DST ]


Who is online

Users browsing this forum: No registered users and 4 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group