AMC 12 Prob

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Discus: What Are My Chances?: December 2002 Archive: AMC 12 Prob
By Azn_Cowboy (Azn_Cowboy) on Sunday, December 29, 2002 - 11:26 am: Edit

If a/b + (a+10b)/(b+10a) = 2,
what does a/b equal?

By Quarky on Sunday, December 29, 2002 - 11:59 am: Edit

LOL. What in the blue hell is AMC? Is it some test for kindergarten kids or what?

One possible OBVIOUS answer is a/b = 1
Of course, if a and b are the same number, then a/b is one, and so is the ratio that is added to it, resulting in a total of 2 (and satisfying the equation).

The other possible answer is 4/5

By usa on Sunday, December 29, 2002 - 01:34 pm: Edit

amc = american math competition, also known as the first round exam to get into USA math olympiad, aka a really hard test.

By Azn_Cowboy (Azn_Cowboy) on Sunday, December 29, 2002 - 02:39 pm: Edit

Yea, they had to be two distinct integers. Forget to say that.


By caltech07 on Sunday, December 29, 2002 - 04:54 pm: Edit

Yeah, this is pretty easy... find the LCD, rewrite the condition as 10a^2 - 18ab + 8b^2 = 0, factor to (10a - 8b)(a - b) = 0, the right is exluded, so we have 10a = 8b, or a/b = 4/5

By Quarky on Sunday, December 29, 2002 - 11:52 pm: Edit

Yeah, of course, Idaho is too stupid to administer the AMC, right? BTW, I was just being sarcastic. I knew that AMC was hard. I just didn't know what it was. Thanks for the info, "usa." Where do you guys take this test? What calculators, if any, are you allowed to use?

By The (The) on Monday, December 30, 2002 - 01:19 am: Edit

Here's the answer, explained, from the official site:


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