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

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Bunuel : Thanks for the great explanation. I have understood completely except for one small query I have about the second case after combining both statements. You have mentioned that Y is anywhere between \(x^3\) and 0. But cannot Y be ANYWHERE left of \(x^3\) (in other words, cannot y be negative) ?
Thanks in advance for your reply

If y is negative, then it would naturally be less than x, which from (1) is positive.
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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.
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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. 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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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

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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\(x\,\,\,\mathop > \limits^? \,\,\,y\)

\(\left( 1 \right)\,\,\,\sqrt x > y\,\,\,\,\,\left\{ \begin{gathered}
\,Take\,\,\,\left( {x;y} \right) = \left( {1\,;\,0} \right)\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,Take\,\,\,\left( {x;y} \right) = \left( {\frac{1}{4}\,;\,\frac{1}{3}} \right)\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,{x^3} > y\,\,\,\,\,\left\{ \begin{gathered}
\,(re)\,Take\,\,\,\left( {x;y} \right) = \left( {1;0} \right)\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,Take\,\,\,\left( {x;y} \right) = \left( {2\,;\,7} \right)\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\end{gathered} \right.\)


\(\left( {1 + 2} \right)\,\,\,\,\,\)

\(\left. {x = y\,\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}
\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\,\sqrt x > x\,\,\,\,\, \Rightarrow \,\,\,\,\,0 < \sqrt x < 1\,\,\,\,\,\,\,\,\left( {\sqrt x \geqslant 1\,\,\, \Rightarrow \,\,\,\sqrt x > x\,\,{\text{false,}}\,\,\,{\text{impossible}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\boxed{0 < x < 1} \hfill \\
\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\,\,{x^3} > x\,\,\,\,\, \Rightarrow \,\,\,\,\,\boxed{0 < x < 1\,\,\,{\text{false}}}\,\,\,\,\,\,\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x \ne y\)

\(x < y\left. {\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}
\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\,x < y < \sqrt x \,\,\,\,\, \Rightarrow \,\,\,\,\,\sqrt x > x\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{above}}} \,\,\,\,\,\,\boxed{0 < x < 1}\,\,\,\, \hfill \\
\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\,\,x < y < {x^3}\,\,\,\, \Rightarrow \,\,\,{x^3} > x\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{again}}} \,\,\,\,\,\,\,\boxed{0 < x < 1\,\,\,{\text{false}}}\,\,\,\,\,\,\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x < y\,\,\,{\text{is}}\,\,{\text{false}}\)

Conclusion: \(\left( {1 + 2} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


The above follows the notations and rationale taught in the GMATH method.
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Bunuel

but if i take x = 1/16 and y = 1/32 how can C be correct answer

(1/16)^3 is much less than 1/32

Not sure I follow you. x = 1/16 and y = 1/32 does not satisfy the second statement, so those values are not valid.
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my doubt is if x>0;x=2 and y=4 then also x3>y but x<y so how come C is thenswer???

nehasomani33 - Can you please help on this?
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Bunuel

Are we not considering x as negative only because\( \sqrt{-x}\) is an imaginary number?



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(1) INSUFFICIENT: If we test values here we find two sets of possible x and y values that yield

conflicting answers to the question.

x √x y Is x > y?
4 2 1 YES
1/4 1/2 1/3 NO
(2) INSUFFICIENT: If we test values here we find two sets of possible x and y values that yield

conflicting answers to the question.

X x^3 Y Is x > y?
2 8 1 YES
-1/2 -1/8 -1/4 NO
(1) AND (2) SUFFICIENT: Let’s start with statement 1 and add the constraints of statement 2. From

statement 1, we see that x has to be positive since we are taking the square root of x. There is no

point in testing negative values for y since a positive value for x against a negative y will always

yield a yes to the question. Lastly, we should consider x values between 0 and 1 and greater than 1

because proper fractions behave differently than integers with regard to exponents. When we try to

come up with x and y values that fit both conditions, we must adjust the two variables so that x is

always greater than y.

x √x x^3 y Is x > y?
2 1.4 8 1 YES
1/4 1/2 1/64 1/128 YES
Logically it also makes sense that if the cube and the square root of a number are both greater than

another number than the number itself must be greater than that other number.

The correct answer is C.

Just a quick question regarding the combination of statement 1 & 2

if we know that:

√x > y

and

x^3 > y

Can't we just combine the two statements to the following? (subtracting both statements):

√x - x^3 > 0

which can be restated as:

√x > x^3 --> (since we know from statement 1 that x must be positive) --> x > x^6

Finally simply divide by x --> 1 > x^5

Which basically says that x must be between 0 and 1, thus testing the case where x > 1 is redundant. Also once we know that x is between 0 and 1, statement 2 offers us a definite answer since
x^3 > y necessarily means that x > y (i.e. a positive fraction will always get smaller once we cube it, and x^3 remains bigger than y even after taking the cube)

Whereas √x > y could still be true or false for the question "x > y" (e.g. x=0.25 and y=0.3 --> Statement 1 true while x>y is not true / x=0.25 and y=0.1 --> Statement 1 true and x>y is true)

I saw that many really good quants here did not use the same approach, so I assume that I am missing something. I would appreciate some explanation.