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#1 (permalink) |
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I JUST got here.
Join Date: Jun 2006
Location: New York, New York
Posts: 20
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Please help with two GMATPrep quant questions (exponents, ratios)
I could really use some help on the solutions for these two GMATPrep questions:
1) If xy = 1, then what is the value of 2^(x+y)^2 / 2^(x-y)^2 ? 2) Kaye and Alberto have some stamps in a 5:3 ratio. Kaye gives Alberto 10 of her stamps, and the ratio becomes 7:5. How many more stamps does Kaye now have than Alberto? Answers (highlight): #1) 16 #2) 40 Thanks! |
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#2 (permalink) |
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TestMagic Guru
![]() ![]() ![]() ![]() Join Date: May 2004
Posts: 1,234
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1) 2^(x^2+y^2+2xy)/2^(x^2+y^2 - 2xy) = 2^(4xy) = 2^4 = 16
2) Initial ratio = 5:3 => After giving away 10 stamps, (5x-10)/(3x+10) = 7/5 => 25x - 50 = 21x +70 => x = 30 => Kaye has 5x-10 = 140 => Alberto has 3x+10 = 100 => Kaye has 40 more stamps than Alberto. |
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#3 (permalink) |
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I JUST got here.
Join Date: Jun 2006
Location: New York, New York
Posts: 20
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So the deal with this one:
2^(x^2+y^2+2xy)/2^(x^2+y^2 - 2xy) is that the x^2 and y^2 cancel (and then the -2xy means that the 2^-2xy can be brought up to the top of the fraction)? I guess I just haven't seen one like this before. Whenever I see addition/subtraction in a fraction, I am trained to think that you can never cancel. I guess that's not the case if the addition is in an exponent. |
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#4 (permalink) |
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Eager!
Join Date: Jun 2006
Posts: 39
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A^2-B^2 = (A-B)(A+B) (1)
A^n/A^m = A^(n-m) (2) --> 2^((x+y)^2)/2^((x-y)^2) = 2^[(x+y)^2-(x-y)^2] (according to (2)) = 2^[(x+y+x-y)(x+y-x+y)] (according to (1)) = 2^(4xy) =16 OR you can do it like GMAT-HELp and utilize only (2) to bring the denominator up and get the same result. |
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