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Numbers - Absolute Value

For COMPETITION
Number of Total Problems: 4.
FOR PRINT ::: (Book)

Problem Num : 1

From : AMC10B

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
If $|x - 2| = p$ , where $x < 2$ , then $x - p =$
$mathrm{(A) -2 } qquad mathrm{(B) 2 } qquad mathrm{(C) 2-2p } qquad mathrm{(D) 2p-2 } qquad mathrm{(E) |2p-...$
'

Category Absolute Value

Analysis

Solution/Answer
When $x < 2,$ $x-2$ is negative so $|x - 2| = 2-x = p$ and $x = 2-p$ .
Thus $x-p = (2-p)-p = 2-2p Longrightarrow mathrm{(C)}$ .

Answer:

Problem Num : 2

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
Define $x heartsuit y$ to be $|x-y|$ for all real numbers $x$ and $y$ . Which of the following statements is not true?
$mathrm{(A) } x heartsuit y = y heartsuit x$ for all $x$ and $y$
$mathrm{(B) } 2(x heartsuit y) = (2x) heartsuit (2y)$ for all $x$ and $y$
$mathrm{(C) } x heartsuit 0 = x$ for all $x$
$mathrm{(D) } x heartsuit x = 0$ for all $x$
$mathrm{(E) } x heartsuit y > 0$ if $x eq y$
'

Category Absolute Value

Analysis

Solution/Answer
Examining statement C:
$x heartsuit 0 = |x-0| = |x|$
$|x| eq x$ when $x<0$ , but statement C says that it does for all $x$ .
Therefore the statement that is not true is $oxed{mathrm{(C)} xheartsuit 0=x ext{for all} x}$

Answer:

Problem Num : 3

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
What is the value of $x$ if $|x-1|=|x-2|$ ?
$mathrm{(A) } -frac12 qquad mathrm{(B) } frac12 qquad mathrm{(C) } 1 qquad mathrm{(D) } frac32 qquad mathr...$
'

Category Absolute Value

Analysis

Solution/Answer
$|x-1|$ is the distance between $x$ and $1$ ; $|x-2|$ is the distance between $x$ and $2$ .
Therefore, the given equation says $x$ is equidistant from $1$ and $2$ , so $x=frac{1+2}2=frac32Rightarrowmathrm{(D)}$ .
Alternatively, we can solve by casework (a method which should work for any similar problem involving absolute values of real numbers). If $x leq 1$ , then $|x - 1| = 1-x$ and $|x - 2| = 2 - x$ , so we must solve $1 - x = 2 - x$ , which has no solutions. Similarly, if $x geq 2$ , then $|x - 1| = x - 1$ and $|x - 2| = x - 2$ , so we must solve $x - 1 = x- 2$ , which also has no solutions. Finally, if $1 leq x leq 2$ , then $|x - 1| = x - 1$ and $|x - 2| = 2-x$ , so we must solve $x - 1 = 2 - x$ , which has the unique solution $x = frac32$ .

Answer:

Problem Num : 4

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
Let $a$ , $b$ , $c$ , and $d$ be real numbers with $|a-b|=2$ , $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$ ?
$mathrm{(A)} 9qquadmathrm{(B)} 12qquadmathrm{(C)} 15qquadmathrm{(D)} 18qquadmathrm{(E)} 24$
'

Category Absolute Value

Analysis

Solution/Answer
Solution 1

From $|a-b|=2$ we get that $a=bpm 2$
Similarly, $b=cpm3$ and $c=dpm4$ .
Substitution gives $a=dpm 4pm 3pm 2$ . This gives $|a-d|=|pm 4pm 3pm 2|$ . There are $2^3=8$ possibilities for the value of $pm 4pm 3pm2$ :
$4+3+2=oxed{9}$ ,
$4+3-2=oxed{5}$ ,
$4-3+2=oxed{3}$ ,
$-4+3+2=oxed{1}$ ,
$4-3-2=oxed{-1}$ ,
$-4+3-2=oxed{-3}$ ,
$-4-3+2=oxed{-5}$ ,
$-4-3-2=oxed{-9}$
Therefore, the only possible values of $|a-d|$ are 9, 5, 3, and 1. Their sum is $oxed{18}$ .

Solution 2

If we add the same constant to all of $a$ , $b$ , $c$ , and $d$ , we will not change any of the differences. Hence we can assume that $a=0$ .
From $|a-b|=2$ we get that $|b|=2$ , hence $bin{-2,2}$ .
If we multiply all four numbers by $-1$ , we will not change any of the differences. Hence we can WLOG assume that $b=2$ .
From $|b-c|=3$ we get that $cin{-1,5}$ .
From $|c-d|=4$ we get that $din{-5,1,3,9}$ .
Hence $|a-d|=|d|in{1,3,5,9}$ , and the sum of possible values is $1+3+5+9 = oxed{18}$ .

Answer:

Array ( [0] => 8215 [1] => 7772 [2] => 7790 [3] => 7895 ) 4