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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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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(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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