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Is |xy| > x^2*y^2 ?
(1) 0 < x^2 < 1/4
(2) 0 < y^2 < 1/9
(1) 0 < x^2 < 1/4
(2) 0 < y^2 < 1/9
Source: GMAT Prep Question Pack 1
Difficulty: Medium
--------------------
Can someone please explain what to do with |xy| > x^2y^2 before we look into the equations?
I got |xy| > (xy)^2 but I didn't know how to interpret the inequality from here. Thanks in advance
Difficulty: Medium
--------------------
Can someone please explain what to do with |xy| > x^2y^2 before we look into the equations?
I got |xy| > (xy)^2 but I didn't know how to interpret the inequality from here. Thanks in advance
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
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DelSingh
IMO C
\(|xy| > x^2y^2\)
since both sides are positive square both sides
\((xy)^2 > (xy)^4\)
\((xy)^2((xy)^2-1)<0\)
since \((xy)^2>0\) therefore \((xy)^2-1<0\)
\((xy)^2<1\)
or -1<xy<1
......so finally this is question.
finally you need both x and y to come to conclusion
STATEMENT 1==>ONLY X HENCE INSUFFICIENT.
\(0 < x^2 < 1/4\)
\(-1/2<x<1/2\)
STATEMENT 2 ==>ONLY Y HENCE INSUFFICIENT
\(0 < y^2 < 1/9\)
\(-1/3<y<1/3\)
now combining both clearly \(-1<xy<1\)
hence C
HOPE IT HELPS
Posts: 110
Here, it helps to know that numbers greater than 1, when squared, are larger. Numbers between 0 and 1, when squared, are smaller.
Once you establish that, you're just saying:
Statement 1: x^2 = smaller; y^2 = unknown (smaller or large) ==> Don't know if product is larger or smaller because you don't know magnitude
Statement 2: y^2 = smaller; x^2 = unknkown (smaller or larger) ==> Don't know if product is larger or smaller because you don't know magnitude
Statement 1 and 2: x^2 = smaller; y^2 = smaller ==> product is smaller because both numbers are smaller.
Therefore, correct answer is (C).
Once you establish that, you're just saying:
Statement 1: x^2 = smaller; y^2 = unknown (smaller or large) ==> Don't know if product is larger or smaller because you don't know magnitude
Statement 2: y^2 = smaller; x^2 = unknkown (smaller or larger) ==> Don't know if product is larger or smaller because you don't know magnitude
Statement 1 and 2: x^2 = smaller; y^2 = smaller ==> product is smaller because both numbers are smaller.
Therefore, correct answer is (C).
WoundedTiger
Two variables are confusing you.
Note that the question is just this:
Is \(|xy|> x^{2}y^{2}\)
Is \(|xy|> |xy|^2\)
Is \(z > z^2\) where \(z = |xy|\)
When is z greater than z^2? When z lies between -1 and 1 or we can say between 0 and 1 when z is positive.
1.\(0<x^{2}<1/4\)
This tells you that 0 < |x| < 1/2. Doesn't tell you anything about y so you don't know anything about z.
2. \(0<y^{2}<1/9\)
This tells you that 0 < |y| < 1/3. Doesn't tell you anything about x so you don't know anything about z.
Both together, you know that |x|*|y| is less than 1 i.e. z is less than 1. Hence z WILL BE greater than z^2.
Answer (C)
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Posts: 86
If (xy)^2 is positive, (xy)^2>(xy)4 or 1> (xy)^2
I and 2 are insufficient because in each statement other value is missing.
By combining
We know that (xy)^2 is positive. So,
(xy)^2< (1/4)*(1/9)
Clearly, 1> (xy)^2
Is my reasoning correct? Need expert help
I and 2 are insufficient because in each statement other value is missing.
By combining
We know that (xy)^2 is positive. So,
(xy)^2< (1/4)*(1/9)
Clearly, 1> (xy)^2
Is my reasoning correct? Need expert help
GMAT 1: 760 Q51 V42
Posts: 10,057
Updated on: 14 Dec 2017, 20:23
Originally posted by MathRevolution on 15 Dec 2015, 21:15.
Last edited by MathRevolution on 14 Dec 2017, 20:23, edited 1 time in total.
Last edited by MathRevolution on 14 Dec 2017, 20:23, edited 1 time in total.
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.
\(|xy| > x^2 \cdot y^2\)
\(⇔ |xy| > |xy|^2\)
\(⇔ |xy|^2 -|xy| < 0\)
\(⇔ |xy|(|xy|-1) < 0\)
\(⇔ 0 < |xy| < 1\)
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first.
Conditions 1) & 2)
\(0 < x^2 < 1/4\)
\(⇔ 0 < |x| < 1/2\)
\(0 < y^2 < 1/9\)
\(⇔ 0 < |y| < 1/3\)
Thus \(0 < |xy| < 1/6\).
Since the range of the question includes the range of the conditions, both conditions together are sufficient.
Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
Since we don't know y, the condition 1) is not sufficient.
Condition 2)
Since we don't know x, the condition 2) is not sufficient, either.
Therefore, the answer is C.
Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.
\(|xy| > x^2 \cdot y^2\)
\(⇔ |xy| > |xy|^2\)
\(⇔ |xy|^2 -|xy| < 0\)
\(⇔ |xy|(|xy|-1) < 0\)
\(⇔ 0 < |xy| < 1\)
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first.
Conditions 1) & 2)
\(0 < x^2 < 1/4\)
\(⇔ 0 < |x| < 1/2\)
\(0 < y^2 < 1/9\)
\(⇔ 0 < |y| < 1/3\)
Thus \(0 < |xy| < 1/6\).
Since the range of the question includes the range of the conditions, both conditions together are sufficient.
Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
Since we don't know y, the condition 1) is not sufficient.
Condition 2)
Since we don't know x, the condition 2) is not sufficient, either.
Therefore, the answer is C.
Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
Bunuel
9. Inequalities
- Theory
Inequalities Made Easy!
Inequalities: Tips and hints
Arithmetic with Inequalities
Wavy Line Method Application - Complex Algebraic Inequalities
Solving Quadratic Inequalities: Graphic Approach
Inequalities - Quadratic Inequalities
Graphic approach to problems with inequalities
Inequalities trick
INEQUATIONS(Inequalities) Part 1
INEQUATIONS(Inequalities) Part 2
10. Absolute Value
For more check Ultimate GMAT Quantitative Megathread
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Bunuel
both x and y should be between -1 and 1 and not equal to -1, 0 and 1.
Stmt 1: insufficient since no info about y is given
Stmt 2: insufficient since no info about x is given
Stmt 1 and 2: sufficient
IMO answer should be "C"
Posts: 1,001
Bunuel
We'll show two approaches:
The first relies on number properties and is a Logical approach.
x^2*y^2 = (xy)^2.
So we're asked when the square of a number (xy)^2 is smaller than its absolute value of a number (xy).
This happens when the square is a fraction.
Written as an equation, |xy| > (xy)^2 when 0 < (xy)^2 < 1.
Looking at our statements, (1) tells us that x^2 is a fraction. Without knowing the value of y^2, this is insufficient.
Similarly, (2) gives us only the range of y^2 and is insufficient.
Combined, we know that 0 < (xy)^2 < 1, exactly what we need.
(C) is our answer.
If this confuses you, then picking numbers is a very easy way to solve the question.
This is an Alternative approach.
(1) Say x^2 = 1/9 and y = 0. Then |xy| = 0 and x^2*y^2 = 0 so the answer is NO.
We'll try to challenge this
Say x^2 = 1/9 and y = 1. Then |xy| = 1/3 and x^2*y^2 = 1/9. Then the answer is YES.
Insufficient.
(2) is entirely symmetrical - we can choose y = 1/16 and x = 0 and then y = 1/16 and x = 1
Insufficient.
Combined:
Let's say x^2 = y^2 = 1/16. Then x and y are 1/4 or -1/4 meaning that |xy| = 1/16 which is definitely larger than (1/16)*(1/16)
So our first numbers give us a YES.
We now need to challenge this by looking for numbers that give a NO.
Trying this out for a few more pairs, we can see that x^2*y^2 is always a fraction and |xy| is always a larger fraction.
Once again, (C) is our answer.
Posts: 6,727
Bunuel
Target question: Is |xy| > x²y² ?
Statement 1: 0 < x² < 1/4
No information about y
Statement 1 is NOT SUFFICIENT
Statement 2: 0 < y² < 1/9
No information about x
Statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that -1/2 < x < 1/2, and x ≠ 0
Statement 2 tells us that -1/3 < y < 1/3, and y ≠ 0
In other words, x and y both have magnitudes that are less than 1.
Notice that, when 0 < x < 1, then x > x²
For example, if x = 1/4, then x² = 1/16
Notice that x > x²
Likewise, if y = 2/7, then y² = 4/49
Notice that y > y²
So, if -1/2 < x < 1/2, and x ≠ 0, then we know that |x| > x²
Likewise, if -1/3 < x < 1/3, and y ≠ 0, then we know that |y| > y²
When we combine both properties, we can conclude that |xy| > x²y²
So, the answer to the target question is YES, it is true that |xy| > x²y²
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
Posts: 292
Nina1987
This actually tripped me up, I ended up way overthinking this problem. Luckily, it isn't actually an issue here because statements 1) & 2) explicitly rule out 0 for x and y. In a more difficult problem I could easily see either case 1 or case 2 being viable and thus changing the answer to E.
Is |xy| > x^2*y^2 ?
The question is asking if the product of 2 numbers is more than the product of the two numbers squared.
What the question is testing is the properties of exponents around -1,0,1
There are 4 cases:
1. If either x,y = 0 then they are equal (Answer = No)
2. If both x,y are between -1, and 1, then the exponent fractions are smaller (Answer = Yes)
3. If both are less or greater than -1 and 1, then the exponent fractions are bigger (Answer = No)
4. If one is bigger and the other is between then it becomes a question of individual values, as product of fractions could cancel out to make them equal, e.g. x=1/2, y=2 so 1/2 * 2 = 1/4 * 4 = 1 (Answer = Yes/No)
(1) 0 < x^2 < 1/4
This means that x^2 is positive (obviously, since it's squared)
x^2 < 1/4
But, taking the square root gives us back the negative side of the inequality...
-1/2 < x < 1/2
We know that x≠0 because the LHS of the original statement was 0 < x^2 and NOT 0 ≤ x^2
We have no info about y so we can't make a determination, insufficient.
(2) 0 < y^2 < 1/9
The same idea as above, -1/3 < y < 1/3, y≠0
No information about x, insufficient.
(3)
y≠0, x≠0
This rules out case 1.
-1/3 < y < 1/3
-1/2 < x < 1/2
Both are between -1 and 1 which rules out case 3 and 4, and we're left with only case 2, definite Yes, C is sufficient!
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DelSingh
Asked: Is |xy| > x^2*y^2 ?
\(|xy| = \sqrt{x^2*y^2}\)
(1) 0 < x^2 < 1/4
Since there is no information about y
NOT SUFFICIENT
(2) 0 < y^2 < 1/9
Since there is no information about x
NOT SUFFICIENT
(1) + (2)
(1) 0 < x^2 < 1/4
0<|x|<1/2
(2) 0 < y^2 < 1/9
0<|y|<1/3
0<x^2*y^2<1/36<1
0<|xy|<1/6<1
Since x^2*y^2 lies between 0 & 1
\(|xy| = \sqrt{x^2*y^2} > x^2 * y^2\)
IMO C
Nina1987
VeritasKarishma Bunuel
"When is z greater than z^2? When z lies between -1 and 1 or we can say between 0 and 1 "
I think z is greater than z^2 when z lies between 0 and 1.
I am not sure why you said " ..between -1 and 1".
For any values less than 0 and greater than 1 , z^2 is greater than z.
sayan640
Check the context: z is the absolute value of xy.
When is the absolute value of a variable greater than the square of the absolute value?
When the variable is say 1/2
Also when the variable is -1/2
|-1/2| = 1/2
|-1/2|^2 = 1/4
That is why the first step in the thinking process was the range -1 to 1 (except 0) which was chopped down to 0 to 1 since all we are focussed on is |xy|, not xy.
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DelSingh
Is \(|xy| > |xy|^2\)?
Is \(|xy| - |xy|^2 > 0\)?
Is \(|xy|(1- |xy|) > 0\)?
Is \(1 > |xy| > 0\)?
(1) No information about y; INSUFFICIENT.
(2) No information about x; INSUFFICIENT
(1&2) Taken together, we can conclude 1 > |xy| > 0. SUFFICIENT.
Answer is C.
Hi Bunuel chetan2u VeritasKarishma GMATBusters nick1816
Can you please tell if my solution is correct?
This is how I solved it.
To check : |xy| > x^2*y^2
Since |xy| >= 0
Therefore, question can be simplified as : Check if x^2*y^2 <= 0
On combining A and B, we see that both x^2 and y^2 are greater than 0. Hence x^2*y^2 will be greater than 0. Hence, both together are suff to answer the question.
Thanks in advance.
Can you please tell if my solution is correct?
This is how I solved it.
To check : |xy| > x^2*y^2
Since |xy| >= 0
Therefore, question can be simplified as : Check if x^2*y^2 <= 0
On combining A and B, we see that both x^2 and y^2 are greater than 0. Hence x^2*y^2 will be greater than 0. Hence, both together are suff to answer the question.
Thanks in advance.
Laksh47
No, the solution is not correct.
Say x=3 and y=2
|x*y|=6 and x^2y^2=36.....x^2y^2 is greater
Say x=1/2 and y=1/10
|xy|=1/20 but x^2y^2=1/400..Here |xy| is greater.
So what you have to find is whether x and y are between 1 and -1. As value of fraction between 0 and 1 reduces as we increase power.
_________________
Do not be tempted to number plug on this one. Rather, use your understanding of fractions and how they are affected by exponents.
Is |xy| > x^2*y^2 ?
|xy|^2 > (x^2*y^2)^2?
x^2*y^2 > x^4*y^4?
(1) 0 < x^2 < 1/4
Nothing about y.
Insufficient.
(2) 0 < y^2 < 1/9
Nothing about x.
Insufficient.
Combo:
We know from st1 that x^2 is a fraction between 0 and 1/4 (e.g. 1/5) . This tells us that x^2 > x^4 (since any fraction raised to a progressively higher exponent will be smaller and smaller).
So now we ask:
y^2 > y^4?
St2 tells us that y^2 is a fraction b/w 0 and 1/9...so y^2 > y^4 necessarily.
x^2*y^2 > x^4*y^4 YES
Sufficient.
Answer is C.
Is |xy| > x^2*y^2 ?
|xy|^2 > (x^2*y^2)^2?
x^2*y^2 > x^4*y^4?
(1) 0 < x^2 < 1/4
Nothing about y.
Insufficient.
(2) 0 < y^2 < 1/9
Nothing about x.
Insufficient.
Combo:
We know from st1 that x^2 is a fraction between 0 and 1/4 (e.g. 1/5) . This tells us that x^2 > x^4 (since any fraction raised to a progressively higher exponent will be smaller and smaller).
So now we ask:
y^2 > y^4?
St2 tells us that y^2 is a fraction b/w 0 and 1/9...so y^2 > y^4 necessarily.
x^2*y^2 > x^4*y^4 YES
Sufficient.
Answer is C.
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