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Bunuel
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Bunuel
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What is the median of all the values of x, which satisfy ||x–3|–5|=3 ?

A. 1
B. 2
C. 3
D. 5
E. 6

Case 1: |x–3|–5=3
|x–3| = 8
Case 1.1: x-3 = 8
x = 11
Case 1.2: x-3 = -8
x = -5

Case 2: |x–3|–5=-3
|x–3| = 2
Case 2.1: x-3 = 2
x = 5
Case 2.2: x-3 = -2
x = 1

So all the values of x = -5, 1, 5, and 11.
Therefore, the median of all the values of x = \(\frac{1+5}{2}\) = 3.
I choose C.

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Quote:
What is the median of all the values of x, which satisfy \(||x–3|–5|=3\) ?

A. 1
B. 2
C. 3
D. 5
E. 6

\(|x–3|–5=3\) , \(|x–3|–5=-3\)
\(|x–3|=8\) , \(|x–3|=2\)
\(x–3=8\) , \(x–3=-8\) , \(x–3=2\) , \(x–3=-2\)
\(x=11\) , \(x=-5\) , \(x=5\) , \(x=1\)

the median of -5,1,5,11 is 3.
ANS C.

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set1:
|x-3| -5 = 3
|x-3| = 8


set1a:
x-3= 8
x= 11


set1b:
x-3= -8
x= -5


or
set2:
|x-3| -5 = - 3
|x-3| = 2

set2a:
x-3= 2
x= 5

set2b:
x-3= -2
x= 1

values of x :


11,-5,5,1

median of (-5,1,5,11)
5+1/2= 3

hence c
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What is the median of all the values of x, which satisfy ||x–3|–5|=3||x–3|–5|=3 ?

A. 1
B. 2
C. 3
D. 5
E. 6

Let |x–3| = a

||a|–5|=3

Case 1) |a|–5=3
|a|=3+5
|a|=8

Case 2) |a|–5=-3
|a|=-3+5
|a|=2

Now insert value of a

Case 1)
a) |x-3|=8
x-3=8
x=8+3
x=11

b)|x-3|=8
x-3=-8
x=-8+3
x=-5

Case 2)
a)|x-3|=2
x-3=2
x=2+3
x=5

b)|x-3|=2
x-3=-2
x=-2+3
x=1

Values of x are = -5,1,5,11

Median = (1+5)/2 = 3

IMO Answer is C.
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Think of a number line. Absolute value means distance from 0. This equation is saying the left hand side of the equation is a distance of 3 from 0 from the right and the left.

Therefore the median which by definition is the middle 50% will be centered at 3. No algebra is needed.

Bunuel
What is the median of all the values of x, which satisfy \(||x – 3| – 5| = 3\) ?

A. 1
B. 2
C. 3
D. 5
E. 6


 

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for the Heroes of Timers Competition

 



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