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

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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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18 Mar 2017, 19:32
shrouded1 wrote:
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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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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15 Aug 2018, 06:39
I think I have a quick way to solve this one.

1) Square root (x) > y but Can we compare x and square root (x) ? No because if 0<x<1 then square root (x) is larger than x. And if 1<x then square root (x) is smaller x. NOT SUF
2) x cube > y but can we compare x and x cube ? No because if -1<x<0 then x cube is larger than x and if 0<x<1 then x cube is smaller than x.

Together (1) (2) -> Square root (x) is larger than y, x cube is larger than y -> If we power x, x larger than y, if we root x, x still larger than y. Hence, just x must be larger than y.
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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27 Aug 2018, 23:47
shrouded1 wrote:
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).

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.

regards,
Tamal
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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28 Aug 2018, 02:36
tamal99 wrote:
shrouded1 wrote:
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).

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.

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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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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28 Aug 2018, 07:08
shrouded1 wrote:
Is $$x > y$$ ?

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

$$x\,\,\,\mathop > \limits^? \,\,\,y$$

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

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

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

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

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

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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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y &nbs [#permalink] 28 Aug 2018, 07:08

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