Is that the Answer?
JDesai01 wrote:If x and y are positive, which of the following must be greater than 1/(sqroot(x+y))?
I. sqroot(x+y)/2x
II. (sqroot(x) + sqroot(y))/(x+y)
III. (sqroot(x) - sqroot(y))/(x+y)
Thanks
Question (gmatprep)
Source: Beat The GMAT — Problem Solving |
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*arch*
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Hey. I think it's (I & II).
Is that the Answer?
Is that the Answer?
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alexi_laiho
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lunarpower
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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.
--
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.
Ron has been teaching various standardized tests for 20 years.
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Vignesh.4384
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Hi Ron,
I am a bit confused about this sum.
Consider i take comfortable values x = 16 and y = 9( both positeve no).
sqrt of 16 = +4 or -4
sqrt of 9 = +3 or -3
You have said that equatrion statement 2 is greater tha question.
According to question 1/SQRT(25) = +(1/5) or -(1/5).
According to statemtn 2. if you take SQRT(16) and SQRT(9) to be positve then YES statement 2 comes to 7/25 Which is greater than +(1/5) or -(1/5).
If i decide to take SQRT(16) = -4 and SQRT(9) = +3.
Statement 2 becomes -(1/25) . Sill question can be interpreted as +(1/5) or -(1/5). How can you be very sure that statement 2 is greater than question now?
The question only says X and Y are positive. It does not say weather the square root of X & Y should be positive.
Am i thinking too much ?
Regards,
Vignesh[/quote]
Thank for the idea.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.
I am a bit confused about this sum.
Consider i take comfortable values x = 16 and y = 9( both positeve no).
sqrt of 16 = +4 or -4
sqrt of 9 = +3 or -3
You have said that equatrion statement 2 is greater tha question.
According to question 1/SQRT(25) = +(1/5) or -(1/5).
According to statemtn 2. if you take SQRT(16) and SQRT(9) to be positve then YES statement 2 comes to 7/25 Which is greater than +(1/5) or -(1/5).
If i decide to take SQRT(16) = -4 and SQRT(9) = +3.
Statement 2 becomes -(1/25) . Sill question can be interpreted as +(1/5) or -(1/5). How can you be very sure that statement 2 is greater than question now?
The question only says X and Y are positive. It does not say weather the square root of X & Y should be positive.
Am i thinking too much ?
Regards,
Vignesh[/quote]
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lunarpower
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btw, you can also use good old fashioned number plugging on this problem.
because the problem tests square roots and comparisons, you should make sure to plug numbers that exhaust all combinations of the following possibilities:
* less than 1, 1, more than 1
* x < y, x = y, x > y
that's a lot of combinations, but, if you get an early enough start on the number plugging, it will work out ok within the allotted time.
because the problem tests square roots and comparisons, you should make sure to plug numbers that exhaust all combinations of the following possibilities:
* less than 1, 1, more than 1
* x < y, x = y, x > y
that's a lot of combinations, but, if you get an early enough start on the number plugging, it will work out ok within the allotted time.
Ron has been teaching various standardized tests for 20 years.
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lunarpower
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no.Vignesh.4384 wrote: sqrt of 16 = +4 or -4
sqrt of 9 = +3 or -3
the operator "√" automatically signifies the positive square root of a number. (since "√" is a function, it must yield one and only one value; that value happens to be the positive one, as in the vast majority of problems that is the more useful of the two possible values.)
you can get even really easy DS problems wrong if you don't make this realization.
Ron has been teaching various standardized tests for 20 years.
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Ian Stewart
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That's too tempting to pass up. I'll give it a shot:lunarpower wrote: 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.
If x and y are positive, which of the following must be greater than 1/(sqroot(x+y))?
I. sqroot(x+y)/2x
II. (sqroot(x) + sqroot(y))/(x+y)
III. (sqroot(x) - sqroot(y))/(x+y)
Fractions become easier to compare when we have the same numerator or the same denominator. Notice that 1/root(x+y) = root(x+y)/(x+y).
We want to know if any of I, II or III must be larger than root(x+y)/(x+y). Note that this must be positive, while III can be negative, so III does not need to be larger.
I has the same numerator, and it would be larger if its denominator, 2x, was certain to be smaller than x+y, i.e. if 2x < x+y, or x < y. We don't know whether x is smaller than y, so I does not need to be larger than root(x+y)/(x+y).
II has the same denominator as root(x+y)/(x+y). We thus want to know which has the larger numerator. That is, if we knew that
root(x+y) < root(x) + root(y)
was always true for positive x and y, we'd know that II must be larger than the quantity in the question. Well,
[root(x+y)]^2 = x+y < x + 2root(xy) + y = [root(x) + root(y)]^2
so II must be larger. That's essentially equivalent to Ron's algebraic solution above, and isn't time consuming- I think it's a good approach. Still, I find the following more 'elegant':
Of course we can make a right-triangle with legs of length root(x) and root(y). By Pythagoras, the hypotenuse will be root(x+y). But the sum of two sides of a triangle is always greater than the third side:
root(x) + root(y) > root(x+y)
which is what we were trying to prove.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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lunarpower
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you win. first shout on me if you ever come within shouting distance of earthquake country.Ian Stewart wrote:Of course we can make a right-triangle with legs of length root(x) and root(y). By Pythagoras, the hypotenuse will be root(x+y). But the sum of two sides of a triangle is always greater than the third side:
root(x) + root(y) > root(x+y)
which is what we were trying to prove.
of course, anyone capable of producing such a solution on the fly (and within the gmat's draconian time limits) would have no business trolling these forums, except as a moderator.
Ron has been teaching various standardized tests for 20 years.
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Ian Stewart
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I'm not sure what to make of the insinuation- either you think I should be a moderator, or you think I have no business here- but I don't much care as long as you get the tab next time I'm near the fault line. That said, the triangle inequality is far from a crazy thing to think up in two minutes- it's an elementary fact of mathematics, and it's the first thing that comes to my mind when I see a comparison between the sum of squares and the square of a sum, or anything similar.lunarpower wrote: you win. first shout on me if you ever come within shouting distance of earthquake country.
of course, anyone capable of producing such a solution on the fly (and within the gmat's draconian time limits) would have no business trolling these forums, except as a moderator.
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lunarpower
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holy schmackerel, i thought you were a moderator.Ian Stewart wrote:I'm not sure what to make of the insinuation- either you think I should be a moderator, or you think I have no business here- but I don't much care as long as you get the tab next time I'm near the fault line. That said, the triangle inequality is far from a crazy thing to think up in two minutes- it's an elementary fact of mathematics, and it's the first thing that comes to my mind when I see a comparison between the sum of squares and the square of a sum, or anything similar.lunarpower wrote: you win. first shout on me if you ever come within shouting distance of earthquake country.
of course, anyone capable of producing such a solution on the fly (and within the gmat's draconian time limits) would have no business trolling these forums, except as a moderator.
you're right: the pythagorean theorem and the triangle inequality are, indeed, two relatively basic facts. but it still takes exceptional insight to see that both of them are relevant to this problem - which is ostensibly altogether unrelated to geometry - AND to apply both of them successfully within the time limit.
still, i like the approach, and will shamelessly co-opt it into my own personal bag of advanced-level tricks.
Ron has been teaching various standardized tests for 20 years.
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andes1
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it is easy
only plunge numbers
ex:
x=2
y=2
nobody says x#y... am I correct?
30 seconds
the answer is only II.
RECHECK 15 SECONS
REMEMBER THE PRINCETON REVIEW BOOK
only plunge numbers
ex:
x=2
y=2
nobody says x#y... am I correct?
30 seconds
the answer is only II.
RECHECK 15 SECONS
REMEMBER THE PRINCETON REVIEW BOOK
LEARNING ENGLIS H
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kumadil2011
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Hi Ian..
I applied PT here as u advised but it seems im not getting the right answer
Let x=9, y=16, x+y=25
then √x=3 √y=4 √x+y=5
1/√(x + y)=1/5
I. [√(X+Y)]/2x = 5/18
II. [(√X) + (√Y)]/(x + y) = 7/25
III. [(√X) - (√Y)]/(x + y)= -ve
so option 1 and 2 is greater than 1/5..
please correct me
I applied PT here as u advised but it seems im not getting the right answer
Let x=9, y=16, x+y=25
then √x=3 √y=4 √x+y=5
1/√(x + y)=1/5
I. [√(X+Y)]/2x = 5/18
II. [(√X) + (√Y)]/(x + y) = 7/25
III. [(√X) - (√Y)]/(x + y)= -ve
so option 1 and 2 is greater than 1/5..
please correct me
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neelgandham
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It is a case of must be and not can be.kumadil2011 wrote:Hi Ian..
I applied PT here as u advised but it seems im not getting the right answer
Let x=9, y=16, x+y=25
then √x=3 √y=4 √x+y=5
1/√(x + y)=1/5
I. [√(X+Y)]/2x = 5/18
II. [(√X) + (√Y)]/(x + y) = 7/25
III. [(√X) - (√Y)]/(x + y)= -ve
so option 1 and 2 is greater than 1/5..
please correct me
Let us interchange the values of x and y
if y=9, x=16, x+y=25, then √y=3 √x=4 √x+y=5
1/√(x + y)=1/5 = 5/25
Only II !I. [√(x+y)]/2x = 5/32 < 5/25
II. [(√x) + (√y)]/(x + y) = 7/25 > 5/25
III. [(√x) - (√y)]/(x + y)= 1/25 < 5/25
If the question is which of the following can be greater than ...blah blah blah..blah.. then you will have to consider the I too.
Anil Gandham
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kumadil2011
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so on such questions, do you suggest to inter change numbers and choose the final answer..isnt this method risky?
