Question 567189: Ok its been 17 years since I took algebra and I have a test for a job interview and this was one of the questions. I got the correct answer but it took me a long time with pen to paper. I know there is a better way to write it.
In printing an article of 48,000 words, a printer decides to use two sizes of type. Using the larger type, a printed page contains 1800 word. Using smaller type, a page contains 2400 words. The article is alloted 21 full pages in a magazine. How many pages must be in smaller type?
I know I need to use x+y=21 and 1800x +2400y = 48000 but I am not sure how to relate the two.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In printing an article of 48,000 words, a printer decides to use two sizes of type. Using the larger type, a printed page contains 1800 word. Using smaller type, a page contains 2400 words. The article is alloted 21 full pages in a magazine. How many pages must be in smaller type?
I know I need to use x+y=21 and 1800x +2400y = 48000 but I am not sure how to relate the two.
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Page Equation: x + y = 21
Words Equatiion: 1800x + 2400y = 48000
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Multiply thru the "Page" Eq. by 1800 to get:
1800x + 1800y = 21*1800
1800x + 2400y = 48000
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Subtract and solve for "y":
600y = 10200
y = 17 (# of pages with 1800 words)
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Solve for "x":
x + y = 21
x + 17 = 21
x = 4 (# of pages with 2400 words)
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cheers,
Stan H.
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