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Got this question from the October 2001 Practice Test. Section 3 - Question 14

(r+p)25 = rx + py for all values of r and p.

Column A = x + y
Column B = 50

The proposed answer is that both columns are equal.

Any ideas on how to do this?

By Chis (Chis) on Sunday, March 23, 2003 - 01:20 pm: Edit

it's true for any values of r and p, so just let r and p each = 1, and you see that 50 = x + y

By Incognito (Incognito) on Sunday, March 23, 2003 - 01:24 pm: Edit

Consider this:

(r+p)25 = rx + py for all values of r and p.

Another way to write (r+p)25 after you "defactor" it would be 25r+25p.

So what we have is:
(r+p)25
= r(25)+p(25)
= rx + py
So, by comparing these two last equations, you can see that x must equal 25, as does y....
so...
x+y = 25+25 = 50

50 = 50, so the two columns are equal.


By The Way, do you mind me asking how you got the October 2001 test?....

By Xiggi (Xiggi) on Sunday, March 23, 2003 - 02:38 pm: Edit

Not at all...

I thought I mentioned it in an earlier post about the availability of the May 2002 test. Anyhow, here it is:

http://www.collegeboard.com/prod_downloads/sat/html/may2002/sat_oct01.pdf

The May 2002 is at:
http://www.collegeboard.com/prod_downloads/sat/html/may2002/sat_may02.pdf

Here is the overall link at CB:
http://www.collegeboard.com/sat/html/may2002/test_sections.html

The answer sheets are attached to the Acrobat file. After downloading it, you can print it. I believe that the tests represent the 2 new tests that come with the 3d edition. The May 2002 was offered because of some misprints in the book. I do not have the latest book yet, so I cannot be sure.

Oh yes, thanks for the answers. It is always so much simpler when someone explains it