If the inequality signs were not all neatly facing the same directions, would we still be able to add?

Thanks

Hi Bunuel,

I have been trying to solve some DS questions lately and am very confused. Its probably a silly question but its been bothering me...

There are certain DS questions where one needs to manipulate the equations to decide whether the option is Sufficient or not where as some other questions like this one where we directly plug in values (usually 1,-1,0,-1/2,1/2). Is there a way you decide which approach to take before solving it ? Is it just a matter of practice ?

Cheers

Hi Bunuel,

Would it have been the same if instead of adding the inequalities we would have multiplied them?
I assume that since we know the variables are positive and the signs are always in the same direction, we can do so.

My reasoning is
- First Inequality: x < 2z < y
- Second Inequality: 2x < z < 2y

Inequalities multiplied: 2\(x^2\) < 2\(z^2\) < 2\(y^2\)
Square root each: 2x < 2z < 2y
Divide each by 2: x < z < y

I know adding inequalities is way simpler to solve this particular problem, but I just want to know if this path is also feasible.

Thank you so much!
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Hi Engr2012,
But in my previous post I was not squaring the first inequality.
- I was multiplying the first and the second inequality together --> x(2x) < z(2z) < y(2y) --> 2\(x^2\) < 2\(z^2\) < 2\(y^2\)
- I was then dividing each term by 2, yielding --> \(x^2\) < \(z^2\) < \(y^2\)
- And finally square rooting each term, yielding --> x < z < y

In my previous post, I mistakenly mixed the order of these last 2 steps.

Regarding your point of "Multiplying or dividing 2 variables in inequalities is ONLY allowed when you know the signs of the 2 variables involved" the question stem tells us that x, y, and z are positive numbers.

Is my reasoning correct then?

Thanks!
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Bunuel,
Pls give the adding and subtracting rule for inequalities.
I saw it on one of your posts.
Same sign you add and different sign you subtract...Something like that....

Given x, y, z > 0

Question: is z between x & y?

Statement 1: x < 2z < y

Visualize the number line

Case 1: 0---------x---------z-------------------2z------y......we get YES.

Case 2: 0-------z--x----2z-------------------------------y.....we get NO.

Statement 1 is Not Sufficient.


Statement 2: 2x < z < 2y

Again visualize the number line

Case 1: 0----x-----2x---------z---y----------------------2y......we get YES.

Case 2: 0----x-----2x---------y----z------------------2y.....we get NO.

Statement 2 is not Sufficient.


Combining both statements, we can add the inequalities

hence 3x < 3z < 3y, removing 3 we get

x < z < y

Combining is Sufficient.



Answer C.



Thanks,
GyM
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\(x,y,z\,\,\, > 0\,\,\,\left( * \right)\)

\(?\,\,\,\,:\,\,\,\,z\,\,{\text{between}}\,\,x\,\,{\text{and}}\,\,y\,\,\,\)

\(\left( 1 \right)\,\,\,x < 2z < y\,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,1,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,2,5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,2x < z < 2y\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,4} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\)

\(\left( {1 + 2} \right)\,\,\,\,x\,\,\mathop < \limits^{\left( * \right)} \,\,\,2x\,\,\mathop < \limits^{\left( 2 \right)} \,\,z\,\,\mathop < \limits^{\left( * \right)} \,\,\,2z\,\,\mathop < \limits^{\left( 1 \right)} \,\,\,y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x < z < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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