if a is less than y is less than z is less than b , is mod y-a mod less than mod y-b mod ?
1) mod z-a mod is less than mod z-b mod
2) mod y-a mod is less than mod z-b mod
mod=modulus
absolute value
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\[a < y < z < b\,\,\,\,\left( * \right)\]vaibhav101 wrote:if a < y < z < b , is |y-a| less than |y-b| ?
1) |z-a| < |z-b|
2) |y-a| < |z-b|
\[\left| {y - a} \right|\,\,\,\,\mathop < \limits^? \,\,\,\left| {y - b} \right|\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,y - a\,\,\,\,\mathop < \limits^? \,\,\,b - y\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{y\,\,\,\,\mathop < \limits^? \,\,\,\frac{{a + b}}{2}}\]
\[\left( 1 \right)\,\,\,\left. \begin{gathered}
{\text{dist}}\left( {z,a} \right) < {\text{dist}}\left( {z,b} \right)\,\,\, \hfill \\
a < y < z < b\,\,\,\left( * \right) \hfill \\
\end{gathered} \right\}\,\,\,\,\, \Rightarrow \,\,\,\left( {y < } \right)\,\,z < \frac{{a + b}}{2}\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \]
\[\left( 2 \right)\,\,\left. \begin{gathered}
{\text{dist}}\left( {y,a} \right) < {\text{dist}}\left( {z,b} \right) \hfill \\
a < y < z < b\,\,\,\left( * \right) \hfill \\
\end{gathered} \right\}\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,y < \frac{{a + b}}{2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \]
\[\left( {**} \right)\,\,\,y \geqslant \frac{{a + b}}{2}\,\,\,\, \Rightarrow \,\,\,\,{\text{dist}}\left( {y,a} \right) \geqslant {\text{dist}}\left( {y,b} \right) > {\text{dist}}\left( {z,b} \right)\,\,\, \Rightarrow \,\,\,\,{\text{impossible}}\]
The answer is __D__
The above follows the notations and rationale taught in the GMATH method.
Regards,
fskilnik.
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if a < y < z < b , is |y-a| less than |y-b| ?
1) |z-a| < |z-b|
2) |y-a| < |z-b|
|a-b| = the distance between a and b.
Since a < y < z < b, we get the following number line:
a..........y..........z..........b
Statement 1: |z-a| < |z-b|
Since the distance between y and a is less than the distance between z and a, we get:
|y-a| < |z-a|
Since the distance between z and b is less than the distance between y and b, we get:
|z-b| < |y-b|
Linking the inequalities above to the inequality in Statement 1, we get:
|y-a| < |z-a| < |z-b| < |y-b|
|y-a| < |y-b|
Thus, the answer to the question stem is YES.
SUFFICIENT.
Statement 2: |y-a| < |z-b|
Since the distance between z and b is less than the distance between y and b, we get:
|z-b| < |y-b|
Linking the inequality above to the inequality in Statement 2, we get:
|y-a| < |z-b| < |y-b|
|y-a| < |y-b|
Thus, the answer to the question stem is YES.
SUFFICIENT.
The correct answer is D.
1) |z-a| < |z-b|
2) |y-a| < |z-b|
|a-b| = the distance between a and b.
Since a < y < z < b, we get the following number line:
a..........y..........z..........b
Statement 1: |z-a| < |z-b|
Since the distance between y and a is less than the distance between z and a, we get:
|y-a| < |z-a|
Since the distance between z and b is less than the distance between y and b, we get:
|z-b| < |y-b|
Linking the inequalities above to the inequality in Statement 1, we get:
|y-a| < |z-a| < |z-b| < |y-b|
|y-a| < |y-b|
Thus, the answer to the question stem is YES.
SUFFICIENT.
Statement 2: |y-a| < |z-b|
Since the distance between z and b is less than the distance between y and b, we get:
|z-b| < |y-b|
Linking the inequality above to the inequality in Statement 2, we get:
|y-a| < |z-b| < |y-b|
|y-a| < |y-b|
Thus, the answer to the question stem is YES.
SUFFICIENT.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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