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Carla
 
 

Online Word Problems #22 (last one :o) )

by Carla Wed Apr 18, 2007 8:27 pm

For this example I had tried to set up an equation like this:
8x+ 5x + 3X -4 = 16x + 5x + 3x
=> 16x - 4 = 24x ... which was wrong..
I undestood that to double the ratio you double the first part ==> 8:5 becomes 16:5
I was not clear on how to come up with this on my own for the ratio of books/clothes = 8x / (5x-4) = 16/5..
Once this is set up I am able to solve for x.. if you could help me understand how to set up this ratio with 8x / (5x - 4) = 16/5 that would be great.
Guest
 
 

by Guest Sun Apr 22, 2007 12:01 pm

The ratio by weight, measured in pounds, of books to clothes to electronics in Jorge's suitcase initially stands at 8 to 5 to 3. Jorge then removes 4 pounds of clothing from his suitcase, thereby doubling the ratio of books to clothes. Approximately how much do the electronics in the suitcase weigh, to the nearest pound?


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StaceyKoprince
ManhattanGMAT Staff
 
Posts: 9359
Joined: Wed Oct 19, 2005 9:05 am
Location: Montreal
 

ManhattanGMAT Word Translations Bank #22

by StaceyKoprince Thu Apr 26, 2007 3:16 am

You're on the right track. The problem with your original equation is that you included the third item in the ratio, electronics, but the problem says that removing the clothes doubles the ratio of books to clothes. It shouldn't actually change the ratio of books to electronics, because the weight of the books doesn't change, nor does the weight of the electronics. The "doubling" of the ratio is only because the clothing weight went down.

So when you put together your formula, you only want to represent the things that have changed relative to each other: books and clothing. Books are 8x, clothing starts at 5x. Then I remove 4# of clothing, so the clothing becomes 5x-4.

I can write ratios as fractions. So I start with a ratio of 8/5. Then I have a new ratio of 16/5 (because the ratio doubles). I can also, from the last paragraph, write the new ratio as 8x/(5x-4). So now I have two ways to write the same value (the new ratio) and I can set them equal to each other to solve for the unknown multiplier.
Stacey Koprince
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Director, Content & Curriculum
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