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| if x and y are positive, which of the following |
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Ron Purewal
MGMAT STAFF
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http://www.manhattangmat.com/forums/post4838.html deals with some number plugging.
there's little sense in dealing with #3 algebraically: because of the subtraction, it can clearly equal 0 (if x and y are the same number). since 1/√(x + y) is a positive number, the possibility of 0 rules out roman numeral III. (in fact, that expression can even be negative, as nothing prohibits x from being smaller than y.) -- if you want to compare two fractions, you can use the technique of cross products to perform the comparison. to use this technique, you take the two 'cross products' (one of the numerators, times the denominator of the other fraction), and associate each of the cross products with whichever fraction donated the numerator. for instance, if you're comparing 2/3 vs. 11/17, then the cross products are 2 x 17 = 34 (associated with 2/3) and 3 x 11 = 33 (associated with 11/17). because 34 is greater than 33, it follows that 2/3 is greater than 11/17. notice that this technique only applies to positive fractions... but that's all you really need: if the fractions have opposite signs, then the comparison is trivial (the positive one is bigger!), and if the fractions are both negative, then the comparison is the opposite of whatever it would be if they were positive. -- find cross products in #(i): √(x + y)/2x vs. 1/√(x + y) cross products are (x + y) vs. 2x subtract one x from both sides --> this comparison is the same as y vs. x we don't know which is bigger. find cross products in #(ii): (√x + √y)/(x + y) vs. 1/√(x + y) cross products are (√x + √y)√(x + y) vs. (x + y) divide both sides by √(x + y) to give (√x + √y) vs. √(x + y) --- remember that (quantity) divided by √(quantity) is √(quantity) -- that's the definition of what a square root is. since both of these quantities are positive, we can square them and compare the squares: (√x + √y)^2 vs. (√(x + y))^2 x + 2√xy + y vs. x + y left hand side is bigger so the original fraction is bigger than 1/√(x + y) ans = ii only there may well be a shorter and more elegant way to figure out #(ii), but i can't conjure one at the moment. a beer to anyone who can. |
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Guest
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change form of original equation by multiplying 1/sqrt(x+y) by / sqrt(x + y ) by sqrt(x+y) / sqrt (x+y) to get sqrt(x+y) / (x+y)
1) sqrt(x +y) / 2x will equal sqrt(x+y) / (x+y) if x = y, so therefore it does not have to be greater than 1/sqrt(x+y) 2) denominators are the same so it's really a matter of comparing if sqrt(x) + sqrt(y) vs sqrt(x+y), and it seems sqrt(x) + sqrt(y) is always greater. i found this by testing numbers, but i'm not sure if there is a theorem for this. 3) sqrt(x) - sqrt(y) could be negative if y is bigger than x, so therefore it does not have to be greater than 1/sqrt(x+y) because 1/sqrt(x+y) is positive |
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Ron Purewal
MGMAT STAFF
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you don't need to memorize this sort of result; you can figure it out by squaring both sides. the square of the left-hand side is x + 2√x√y + y, and the square of the right-hand side is x + y. since x and y are positive numbers, the left side is bigger. that's a proof. btw, at this point you can call it a theorem if you want. anything you've proved is a "theorem", although some theorems are of course more famous than others. |
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Guest
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It becomes very simple if you just take x and y =1.
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Ron Purewal
MGMAT STAFF
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well, yeah, but you can't solve "MUST" problems that way. you're looking for the expressions that MUST be greater than 1/√(x + y), no matter what x and y are. if the expressions work for one choice of plug-in, that doesn't tell you anything at all. (if any of the expressions doesn't work for any single plug-in, though, you can eliminate it.) -- stupid analogy: which of the following MUST be less than 10? (i) x (ii) x + 1 (iii) x + 2 obviously, the answer is "none of them" - they can all be huge, if x is huge - but your approach above would mistakenly conclude that the answer is "i, ii, and iii". -- remember that gmat problems are rarely as easy as you may think at first glance - and almost never, if users are posting them here. the problems posted on this forum tend to be drawn disproportionately from the pool of difficult problems (although there's the occasional easier one here and there on the forum). |
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| if x and y are positive, which of the following |
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