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absolute value
Source: Beat The GMAT — Problem Solving |
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fskilnik@GMATH
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$$?\,\,\,:\,\,\,\# \,\,\,{\rm{integer}}\,\,{\rm{values}}\,\,{\rm{for}}\,\,\,y = \left| {x + 5} \right| - \left| {x - 5} \right|$$harsh8686 wrote:If y=|x+5|−|x−5|, then y can take how many integer values?
A. 5
B. 10
C. 11
D. 20
E. 21
$$\left( {\rm{i}} \right)\,\,\,x \le - 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = - \left( {x + 5} \right) - \left( {5 - x} \right) = - 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,1\,\,{\rm{integer}}$$
$$\left( {{\rm{ii}}} \right)\,\,\, - 5 < x \le 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right) - \left( {5 - x} \right) = 2x\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{other}}\,\,10 \cdot 2\,\,{\rm{integers}}\,\,\,\left( * \right)\,\,\,$$
$$\left( * \right)\,\,\,x\,\,{\mathop{\rm int}} \,\,\,\left( { - 4 \le x \le 5\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)\,\,\,\,{\rm{or}}\,\,\,\,\,\left\{ \matrix{
\,x \ne {\mathop{\rm int}} \hfill \cr
\,x = {{{\mathop{\rm int}} } \over 2} \hfill \cr} \right.\,\,\,\,\,\,\,\,\left( { - 9 \le \mathop{\rm int}\,{\rm{odd}}\,\, \le 9\,\,\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)$$
$$\left( {{\rm{iii}}} \right)\,\,\,x > 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right) - \left( {x - 5} \right) = 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{already}}\,\,{\rm{obtained}}\,\,{\rm{in}}\,\,\left( {{\rm{ii}}} \right)$$
$$? = 1 + 20 = 21$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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fskilnik@GMATH
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Alternate way:harsh8686 wrote:If y=|x+5|−|x−5|, then y can take how many integer values?
A. 5
B. 10
C. 11
D. 20
E. 21
There are 21 integers between -10 and 10, both of them included (see blue interval).
It´s done!
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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GMATGuruNY
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The CRITICAL POINTS are where the expressions inside the absolute values are equal to 0.harsh8686 wrote:If y=|x+5|−|x−5|, then y can take how many integer values?
A. 5
B. 10
C. 11
D. 20
E. 21
x+5 = 0 when x=-5.
Substituting x=-5 into y = |x+5|-|x-5|, we get:
y = |-5+5| - |-5-5| = 0-10 = -10.
x-5 = 0 when x=5.
Substituting x=5 into y = |x+5|-|x-5|, we get:
y = |5+5| - |5-5| = 10-0 = 10.
The resulting values in blue indicate the range for y.
Thus, y can be equal to any integer value between -10 and 10, inclusive, implying that there are 21 integer options for y.
The correct answer is E.
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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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Student Review #3
