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Is x > y ? (1) x^(1/2) > y (2) x^3 > y

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Bunuel

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08 Dec 2014, 07:48

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tarikk

gurpreetsingh

Statement 1: \(\sqrt{x} > y\) ....take x = 4, y = 1 (yes); take x = 1/4 and y = 1/3 (no) hence not sufficient.

Statement 2: \(x^3 > y\) take x = 2 and y = 3 (no) , take x = 2, y = 1 (yes) hence not sufficient.

take statement 1 and 2 together. Now the answer could be either C or E.

Either y is -ve or postive. If y is -ve then x >y always holds true.

If y is +ve then x>y^2 and x^3>y => x^6 > y^2

divide both the inequalities: x^5 >1 => x >1. Since square root of x (> 1) is greater than y => x>y.

Hence C.

The graphical approach is good, but its essential to understand that how the inequalities behaves in different domains.

Hi gurpreetsingh, could you please elaborate on:

gurpreetsingh

If y is +ve then x>y^2 and x^3>y => x^6 > y^2

. Thank you!

If y is positive, then we can square x^3 > y to get x^6 > y^2.

Check RAISING INEQUALITIES TO EVEN/ODD POWER in Tips on Inequalities.

Hope it helps.

Bunuel

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06 May 2015, 11:59

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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

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Inequality and absolute value questions from my collection
DS Questions
PS Questions

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03 Dec 2015, 05:52

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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 ?

(1) x √ >y
(2) x 3 >y

There are 2 variables (x,y) and 2 equations are given by the conditions, so there is high chance (C) will be the answer.
Looking at the conditions together,
if 0<x<1, sqrt (x)>x>x^3>y answers the question 'yes', and if x>1, x^3>x>sqrt(x)>y also answers the question 'yes'
Hence the answer becomes (C).

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.

MathRevolution

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18 Mar 2017, 20:32

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shrouded1

I thought this was a really tough question !

Is \(x > y\) ?

(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)

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. In the area of inequality, the number of equations means the number of inequalities, since the number of equations means the number of conditions.

In the original question, the number of variables is 2 and the number of equation is 0.

Hence, C is most likely the answer.

For the condition (1), we have two cases.
\(x = 1/4\) and \(y = 1/3\) satisfies the condition (1), but \(x > y\) is not true.
\(x = 4\) and \(y = 1\) satisfies the condition (1) and \(x > y\) is true.
Thus, the condition (1) is not sufficient.

For the condition (2), we also have two cases.
\(x = 2\) and \(y = 3\) satisfies the condition (2), but \(x > y\) is not true.
\(x = 3\) and \(y = 2\) satisfies the condition (2) and \(x > y\) is true.
Thus, the condition (2) is not sufficient, either.

For both the conditions (1) and (2) together, we should consider two cases.

Case 1: \(0 \leq x \leq 1\).
Since \(x^3 \leq x\), if \(x^3 > y\) by the condition (2), \(x \geq x^3 > y\), that is, \(x >y\).

Case 2: \(x > 1\)
Since \(x > \sqrt{x}\), if \(\sqrt{x} > y\) by the condition (1), \(x > \sqrt{x} >y\), that is, \(x > y\).

Hence, both the conditions (1) and (2) together are sufficient, as expected earlier.

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28 Aug 2018, 00:47

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VeritasKarishma

shrouded1

I thought this was a really tough question !

Is \(x > y\) ?

(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)

One of those gorgeous questions that seem so simple at first but surprise you later...
Best way to work on these is to fall back on your drawing skills (Yes, I love diagrams!)

Statement (1): If I can say that \(x >= \sqrt{x}\) for all values of x, then I can say that x > y. The green line shows me the region where \(x >= \sqrt{x}\) but the red line shows me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
I also understand from this statement that x >= 0.

Statement (2): If I can say that \(x >= x^3\) for all values of x, then I can say that x > y. The green lines show me the region where \(x >= x^3\) but the red lines show me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.

Attachment:

Ques.jpg

Using both together, I know x >= 0.
If 0 <= x <= 1, then we know \(x >= x^3\). Since statement (2) says that \(x^3 > y\), I can say that x > y.
If x > 1, then we know \(x > \sqrt{x}\). Since statement (1) says that \(\sqrt{x} > y\), I can deduce that x > y.
For all possible values of x, we can say x > y. Sufficient. Answer (C).

Hey VeritasKarishma,
great approach, thanks a lot.
But here I have a doubt (excuse me if silly

) -
When we are combining both graphs - its clear that X>Y when 0>x>1 AND 1>x>+.
But what about when the value of x is between "-1 to 0".....as shown with red color in second graph.

please explain.

regards,
Tamal

KarishmaB

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tamal99

VeritasKarishma

shrouded1

I thought this was a really tough question !

Is \(x > y\) ?

(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)

Attachment:

Ques.jpg

Hey VeritasKarishma,
great approach, thanks a lot.
But here I have a doubt (excuse me if silly

Stmnt 1 tells you that x >= 0 (since \(\sqrt{x}\) is not defined for negative values of x). Hence, when you use both stmnts together, negative values of x are to be ignored.

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shrouded1

Is \(x > y\) ?

(1) \(\sqrt{x} > y\)

(2) \(x^3 > y\)

(1) \(\sqrt{x} > y\)
We can say that x is positive, as we deal with square roots of positive numbers only.
\(x\geq \sqrt{x} > y\)......When \(x\geq 1\)
But when 0<x<1, \(\sqrt{x}>x\) and we can not say anything about placement of y.
\(\sqrt{x}>x>y\).......\(\sqrt{\frac{1}{4}}>\frac{1}{4}>0\)
\(\sqrt{x}>y>x\).......\(\sqrt{\frac{1}{4}}>\frac{1}{3}>\frac{1}{4}\)
Insufficient

(2) \(x^3 > y\)
\(x\geq x^3 > y\)......When \(x< 1\)
But when x>1, \(x^3>x\) and we can not say anything about placement of y.
\(x^3>x>y\).......\(2^3>2>0\)
\(x^3>y>x\).......\(2^3>5>2\)
Insufficient

Combined
x is positive and both \(x^3\) and \(\sqrt{x}\) are greater than y.
Say x>1......\(x^3>x>\sqrt{x}>y\)
But if x<1....\(\sqrt{x}>x>x^3>y\)
x=1.....\(\sqrt{x}=x=x^3>y\)
Hence, in all cases, x>y

Sufficient

C

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04 Feb 2023, 04:15

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Bunuel

shrouded1

I thought this was a really tough question !

Is \(x > y\) ?

(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)

Is x>y?

(1) \(\sqrt{x}>y\) --> if \(x=4\) and \(y=1\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{3}\) then the answer will be NO. Two different answers, hence not sufficient.

Note that from this statement we can derive that \(x\geq{0}\) because an expression under the square root cannot be negative.

(2) \(x^3>y\) --> if \(x=4\) and \(y=1\) then the answer will be YES but if \(x=2\) and \(y=3\) then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) From (1) we have that \(x\geq{0}\). Now, \(\sqrt{x}\), \(x\), \(x^3\) can be positioned on a number line only in 2 ways:

1. For \(1\leq{x}\): ------\(\sqrt{x}\)----\(x\)----\(x^3\), so \(1\leq{\sqrt{x}}\leq{x}\leq{x^3}\) (the case \(\sqrt{x}=x=x^3\) is when \(x=1\)). \(y\) is somewhere in green zone (as \(y<\sqrt{x}\) and \(y<x^3\)), so if we have this case answer is always YES: \(y<x\).

2. For \(0\leq{x}<1\): \(0\)----\(x^3\)----\(x\)----\(\sqrt{x}\)----\(1\), so \(0\leq{x^3}\leq{x}\leq{\sqrt{x}}\). \(y\) is somewhere in green zone (as \(y<\sqrt{x}\) and \(y<x^3\)), so if we have this case answer is always YES: \(y<x\).

So in both cases \(y<x\). Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel can y have negative value

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Bunuel

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04 Feb 2023, 09:06

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puneetfitness

Bunuel

shrouded1

I thought this was a really tough question !

Is \(x > y\) ?

(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)

Hi Bunuel can y have negative value

Posted from my mobile device

Good question. When we consider the two cases together, y can absolutely be some negative number, say -10. If y is negative and since we concluded that x must be positive, then in this case (y = negative) < (x = positive). So, we'd get the same YES answer to the question: x is greater than y.

Does this make sense?