Gmat Prep 2

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by vishubn » Tue Nov 11, 2008 1:53 am
Its already been disscussed !! but i thoiguth may be people will pour ion more approaches !! its in the same post
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.

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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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by stop@800 » Tue Nov 11, 2008 2:56 am
A tricky one

I am sure, with numbers you will be able to solve this.
let me try to solve without numbers.

I
sqrt(x+y)/2x

= sqrt(x+y) / [sqrt(x+x) sqrt(x+x) ]

if I take x=y in above expression, it becomes equal to values asked
however based on relation between x and y, it will vary
so it can not be valid for MUST BE TRUE case.

II
[sqrt(x) + sqrt(y)] / (x+y)

= 1 / [ (sqrt(x) - sqrt(y) ]


III will reduce to
= 1 / [ (sqrt(x) + sqrt(y) ]


now if we look carefully
sqrt(x+y) will be less than (sqrt(x) + sqrt(y))
and
sqrt(x+y) will be greater than (sqrt(x) - sqrt(y))

hence
1/sqrt(x+y) will be greater than (sqrt(x) + sqrt(y))

and
1/sqrt(x+y) will be less than (sqrt(x) - sqrt(y))


hence ans is C [only II is true]

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by cramya » Tue Nov 11, 2008 5:45 pm
If u were to pick numbers:

Since sqrt(x+y) and sqrt(x) and sqrt(y) has to be calculated in some of the choices lets pick number where its easy to calcualte sqrt(x), sqrt(y) and sqrt(x+y)

Let x=9 y=16

1/sqrt(x+y) = 1/sqrt(9+16) = 1/5 = .20

Only stmt II will always be true



Since its a must be a true question see if u can disprove any of the choices and then that satement will not be part of the correct answer

Stmt I)

sqrt(x+y)/2x
sqrt(9+16)/2*9 = 5/18 = .27 >.20

but if we take x=16 and y=9 then this will be 5/32 = .15 < .20

Eliminate Stmt I since it must be true no matter what


Stmt II

sqrt(x)+sqrt(y)/x+y irrespective of we take x=9 y=16 or x=16 y=9 the answer will be the same(since its all additions in numerator and denominator)

i.e 3+4/25 = 7/25 = .28 > .20

Stmt II is part of the answer u should pick

Stmt III

sqrt(x)-sqrt(y) / x+y = 3-4/25 definitely less than .20

(No need to switch values for x and y here since it must be true and we have proved it may not be true in all case



Choice C) II only
Last edited by cramya on Tue Nov 11, 2008 5:51 pm, edited 1 time in total.

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by logitech » Tue Nov 11, 2008 5:49 pm
:lol:
LGTCH
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