Is x > y^2? (1) x > y + 5 (2) x^2 - y^2 = 0

Is \(x>y^2\)?

(1) \(x>y+5\) --> \(x-y>5\). Clearly insufficient, for example: if \(x=1\) and \(y=-10\) then the answer is NO, but if \(x=10\) and \(y=1\) then the answer is YES. Two different answers, hence not sufficient.

(2) \(x^2-y^2=0\) --> \((x-y)(x+y)=0\) --> so either \(x-y=0\) or \(x+y=0\). Also insufficient: if \(x=1\) and \(y=1\), then answer is NO, buy if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. Two different answers, hence not sufficient.

(1)+(2) As from (1) \(x-y>5\neq{0}\), then from (2) must be true that \(x+y=0\) --> so \(x=-y\) --> substitute \(x\) in (1) --> \(-y-y>5\) --> \(y<-\frac{5}{2}<0\), as \(x=-y\), then \(x>\frac{5}{2}>0\), so \(y^2\) (or which is the same \(x^2\)) will always be more than \(x\), thus the answer to the question "Is \(x>y^2\)" is NO. Sufficient.

To elaborate more as \(x=-y>0\), the only chance for \(x>y^2\) to hold true (or which is the same for \(x>x^2\) to hold true) would be if \(x\) is fraction (\(0<x<1\)). For example if \(x=\frac{1}{2}\) and \(y=-\frac{1}{2}\) then \(x=\frac{1}{2}>y^2=\frac{1}{4}\). But the fact that \(x>\frac{5}{2}>0\) rules out this option.

Answer: C.

Hope it's clear.
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Thanks! That was a much clear explanation.

Last problem of the day:

What is the best approach to solve problems such as this:


"How many times will the digit 7 be written when listing the integers from 1 to 1000?"

Is there any combinationatorics shortcuts? The GMTClub test approached seemed very time consuming.

digits-problem-difficulty-in-gmat-club-test-97153.html

Hi, there! I'm happy to help with this.

The question: is x > y^2

Statement #1: x > y + 5

This doesn't necessarily tell us anything. If y = 1, and x = 7, then x > y^2, but if y = -6 and x = 0, then x < y^2. But itself, Statement #1 is not sufficient.

Statement #2: x^2 - y^2 = 0

This means that x^2 = y^2, which means that x = ±y. Same absolute value, but both could be positive, both could be negative, or either one could be positive and the other negative. We know that y^2 will be positive, but the x can be positive or negative, so by itself, Statement #2 is insufficient.

Combined
Now, we know that x^2 - y^2 = 0 ---> x = ±y, AND we know that x > y + 5. This leads immediate to a few conclusions
(a) x is positive and y is negative --- that's the only way they could have the same absolute value, but with x bigger than y + 5
(b) x and y must have an absolute value greater than 2.5, so that the different between positive x and negative y is more than 5

So we are comparing a positive number x, greater than 2.5, to the square of the negative number with the same absolute value. Of course, x^2 and y^2 are equal, so the question really boils down to: given that x > 2.5, is x > x^2?

For all x greater than one, the square of x is greater than x. That's because, squaring is multiplying a number by itself, and when you multiply anything by a number greater than one, it gets bigger.

Thus, if x > 2.5, when we square it, it will get bigger. Therefore, x^2 = y^2 > x for all values of x > 2.5.

Thus, combined, the statements are sufficient together. Answer = C

Does that make sense? Please let me know if you have any questions.

Mike
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Bunnel, can you please provide a graphical solution to this particular problem?

Not a good candidate for graphic approach.
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Hi i had a question.

Statement two provides us with equation x^2=y^2

Doesn't that answer our question for x>y^2 as No always ?

For example if x=-2 and y=2 then -2 is not greater than y^2..
Or x=2 and y=2; 2 is not greater than 4 ?

Am i missing something here ?

P.S: Thanks in advance for the help !

If x and y are fractions for eg. x=1/2 and y=1/2
then 1/2 > 1/4 ; i.e. x > y^2 - Hence you get a YES answer here and statement 2 becomes Insufficient.

Consider Kudos if it helped.

Hello, could someone please remove the highlighted part from the original post? (also from my post now, I suppose).
Also, I just wanted to know, if we could also write \(x^2-y^2=0\) as \(x^2=y^2\) which is the same as \(|x|=|y|\).
Just asking because I've become slightly comfortable with solving with absolute values. So is this ok?

yes, you can use \(x^2-y^2=0\) as \(x^2=y^2\) or \(|x|=|y|\)

thank you, and thank you for editing the questions I pointed out.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


Is x > y^2?

(1) x > y+5

(2) x^2-y^2 = 0

-> In the original condition, there are 2 variables(x, y), which should match with the number of equations. So, you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), it is x-y>5 for 1). In case of 2), x=y, x=-y. When x=y, it is 0>5 from x-y>5, which is impossible and becomes x=-y. Thus, x>y^2? --> -y>y^2? --> 0>y^2+y? --> 0>y(y+1)? --> -1<y<0?. In 1), -y>y+5, -5>2y, -5/2>y, -2.5>y, which is no and sufficient. Therefore, the answer is C.

During the exam, it is better to choose C since there are 2 variables.

-> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Bunuel,
the question is: IS x>y^2?
the rephrase is: is x greater or equal to 0?
So, if u combine 1 and 2, then u get x=-y, right?
Now, if y=10, then x=-10---->No
if y=-10, then x=10--->Yes.
So, how C make sense?
Thank you__
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y = 10 and x = -10 is not possible. These values does not satisfy the first statement. See, when combining we got that \(y<-\frac{5}{2}<0\).
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Thank you Bunuel. God be with you!
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Statement 1:
@y= 1 x= 7
7 > 1²
@y=20 x= 27
27 < 20²
Not Sufficient

Statement 2:
X² - y² = 0
I.e. IxI = IyI
I.e. for integer values of x and y
X < y²
For 0<x<1
x> y²
Not Sufficient

Combining
Now we can only take values of x out of the range 0 and 1

For all values of x and y now,
X is not greater than y^2
hence

Sufficient

C

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First, analyze the question: when a number (x) is greater than square of another number (y)?
condition: x must be positive [if you can find any negative/zero value for x, that's a NO to the main question], y can be positive, negative or zero.
in two situation, a number (x) can be greater than square of another number (y).
situation 1: x=y [when x and y both are positive proper fraction (0<x<1)] i.e. (1/2)>(1/2)^2.
Situation 2: x is greater than the squared value of y. [when x >1] i.e. 10>(-2)^2 or 10>(2)^2.

Statement 1: x>y+5 or x-y>5 or x-y=6 (for simplicity)
for such difference cases, six scenarios are possible.
1. both x & y are negatives. [x=-1, y=-7] thus NO to main question cause it breaks the primary condition that x must be positive.
2. x=0 , y=negative. [x=0, y=-6] thus NO to main question cause it breaks the primary condition that x must be positive.
3. x=Positive, y=negative [x=1, y=-5] thus NO to main question cause x is less than the squared value of y.
4. x=Positive, y=negative [x=5, y=-1] thus yes to main question cause x is greater than the squared value of y.
5. x=Positive, y=zero[x=6, y=0] thus yes to main question cause x is greater than the squared value of y.
6. x=Positive, y=positive [x=7, y=1] thus yes to main question cause x is greater than the squared value of y.
So statement 1 is insufficient.

Statement 2: x^2-y^2 = 0 or x^=-y^2 or IxI=IyI or x=y or x=-y
in other words, x and y are same numerical value either in same direction or in opposite direction.
1. if x is negative [i.e. -5], y can be positive [i.e. 5] or negative [-5]. but, NO to main question cause it breaks the primary condition that x must be positive.
2. if x is zero, y must be zero. but, NO to main question cause it breaks the primary condition that x must be positive.
3. if x is positive proper fraction, y can be positive or negative proper fraction. thus yes to main question cause x is greater than the squared value of y. i.e. (1/2)>(1/2)^2
4. if x is 1, y can be 1 or -1, thus NO to main question cause x equal the squared value of y.
5. if x is greater than 1, y would be greater than 1 or less than -1. thus NO to main question cause x is less than the squared value of y. i.e. (1/2)>(1/2)^2
So statement 2 is insufficient.

Combining:
statement 1 says difference between x & y is more than 5 units.
so scenarios 2,3,4 from statement 2 are cancelled out. because
2. if x is zero & y is zero, no difference between x & y.
3. if x is positive or negative proper fraction & y is positive or negative proper fraction, maximum difference between x & y is 1 unit.
4. if x is 1 & y is 1, difference between x & y is 2 unit.

The remaining scenario 1 & 5 say NO to main question cause x is less than the squared value of y.
So, My answer is C

Is \(x > y^2\)?

(1) x > y + 5

x - y > 5

We can't conclude x > y^2 from this. INSUFFICIENT.

(2) x^2 - y^2 = 0
(x+y)(x-y)

This tells us (x+y) = 0 or (x-y) = 0 or both equal 0. We can't conclude x > y^2 from this INSUFFICIENT.

(1&2) If x - y > 5, then from the second statement we can conclude that x - y = 0

x - y = 0
x = - y

-y - y > 5
-2y > 5
y < 5/2

Is - 5/2 > 25/4?

We can answer with certainty: No.

SUFFICIENT.

Answer is C.
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chetan2u Bunuel

On combining statements 1 and 2, we get x=-y

The target question is is x>y^2 or in other words, since now that we have x=-y, we can ask ourselves is y>y^2 ? This can never be hence sufficient.

Is this okay? If not, then why ?

PS: I don’t have any problem with the solutions posted above, just out of curiosity, why are we getting into x>5/2 and y<-5/2, when we can simply rephrase the target question.

Looking forward to hearing from you

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