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absolute value

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Source: — Data Sufficiency |

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by [email protected] » Mon Mar 03, 2014 2:09 pm
Hi buoyant,

This question is perfect for TESTing Values.

We're told that X and Y are INTEGERS and Y = |X+3| + |4-X|. We're asked "does Y = 7?" This is a YES/NO question

Fact 1: X < 4

If X=3, then Y = 6 + 1 = 7 and the answer to the question is YES
If X=-10, then Y = 7 + 14 = 21 and the answer to the question is NO
Fact 1 is INSUFFICIENT

Fact 2: X > -3

If X = 3, then Y = 7 and the answer to the question is YES
If X = 10, then Y = 13 + 6 = 19 and the answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, we have -3 < X < 4

Here we have a finite set of possibilities
If X = 3, then Y = 7 and the answer is YES
If X = 2, then Y = 5+2 = 7 and we get YES
If X = 1, then Y = 4+3 = 7 and we get YES
If X = 0, then Y = 3+4 = 7 and we get YES
If X = -1, then Y = 2+5 = 7 and we get YES
If X = -2, then Y = 1+6 = 7 and we get YES

The answer is ALWAYS YES. Thus, it's SUFFICIENT TOGETHER

Final Answer: C

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by Patrick_GMATFix » Mon Mar 03, 2014 3:53 pm
An alternative approach: rephrase the question by thinking of absolute values as measures of distance.

&#10004; Absolute value of |a-b| is the distance between a and b on the number line. for instance |5-8| = |8-5| = distance between 5 and 8 on the number line.
&#10004; Absolute value of |a+b| = |a - (-b)| and is the distance between a and -b. For instance, |2+1| = |2-(-1)| = distance from 2 to -1 on the number line.
If x and y are integers and y = |x+3| + |4-x|, does y equal 7?

1) x < 4
2) x > -3
Rephrase the question: y is the sum of the distance between x & -3 and the distance between x & 4. Is the sum of these distances equal to 7? Think about where -3 and 4 are on the number line (they are 7 apart); the only way for the two distances to add up to 7 is for x to be between -3 and 4.

Rephrase: Is x between -3 and 4 inclusive?

Only together do the statements guarantee that x falls between these values. The answer is C.
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by buoyant » Mon Mar 03, 2014 8:22 pm
Though i understand the above approaches,i don't get what was wrong with my approach.
I solved the above problem as below, but got [spoiler]ans E
[/spoiler]
for absolute value, we can take
case 1-> x is greater than or equal to 0 -> for x>0 ,y = (x+3)+(4-x)= 7
for x=0 ,y = (0+3)+(4-0)=7

case 2-> x is less than 0 -> y= -(x+3)+ [-(4-x)]= -7

so, i rephrased the qn as is y=7=? => is x greater than or equal to 0 ?

st1: Not sufficient
st2: Not sufficient

st1+ st2 : -3<x<4 => x can be -2,-1,0,1,2, and 3 => x can be +ve , 0 or -ve => not sufficient=> ans E

After i looked at the solution and solved again, i got that for each of the above values,i get y =7
but, i followed the rephrased question blindly, hence got the wrong answer.
What is wrong with my approach?

Please help.
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by Patrick_GMATFix » Mon Mar 03, 2014 9:48 pm
ase 1-> x is greater than or equal to 0 -> for x>0 ,y = (x+3)+(4-x)= 7
for x=0 ,y = (0+3)+(4-0)=7

case 2-> x is less than 0 -> y= -(x+3)+ [-(4-x)]= -7
It's incorrect to claim that if x>0, then |4-x| is the same as (4-x). For instance, if x were 5, those two expressions would not have the same value.

Likewise, it's incorrect to claim that if x<0, then |4-x| is the same as -(4-x). For instance, if x were -1, then |4-x| would be 5, but -(4-x) would be -5.

The "switch" is not whether x itself is greater than or less than 0, but rather whether the entire contents of the absolute value brackets are greater than or less than 0

What we can say correctly is that if 4-x>0, then |4-x| = +(4-x), while if 4-x<0, then |4-x| = -(4-x)
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by GMATGuruNY » Tue Mar 04, 2014 8:22 am
buoyant wrote:If x and y are integers and y = |x+3| + |4-x|, does y equal 7?

1) x < 4
2) x > -3

[spoiler]OA: C[/spoiler]
An alternate approach is to determine the CRITICAL POINTS: the values where the EXPRESSIONS inside the absolute values are equal to 0.
Here, the critical points are x=-3 (where x+3 = 0) and x=4 (where 4-x=0).

First, test the critical points themselves.
If x=-3, then y = |x+3| + |4-x| = |-3+3| + |4-(-3)| = 0+7 = 7.
If x=4, then y = |x+3| + |4-x| = |4+3| + |4-4| = 7+0 = 7.
Thus, if x=-3 or x=4, the answer to the question stem is YES.

Next, test values to EACH SIDE of the critical points.
x<-3:
If x=-4, then y = |x+3| + |4-x| = |-4+3| + |4-(-4)| = 1+8 = 9.
In this range, y>7, so the answer to the question stem is NO.

-3<x<4:
If x=0, then y = |x+3| + |4-x| = |0+3| + |4-0| = 3+4 = 7.
In this range, y=7, so the answer to the question stem is YES.

x>4:
If x=5, then y = |x+3| + |4-x| = |5+3| + |4-5| = 8+1 = 9.
In this range, y>7, so the answer to the question stem is NO.

Question stem, rephrased:
Is -3≤x≤4?

Only when the statements are combined is it guaranteed that -3≤x≤4.

The correct answer is C.
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by Matt@VeritasPrep » Tue Mar 04, 2014 5:26 pm
Maybe an easier way:

If |a| > 0, |a| = a

If |a| < 0, |a| = -a

Now suppose that (x+3) and (4-x) are both positive:

If (x+3) is positive, then |x+3| = x + 3
If (4-x) is positive, then |4-x| = 4 - x

So if (x+3) and (4-x) are BOTH positive, |x+3| + |4-x| = (x+3)+(4-x) = 7

Hence the question is basically asking us "Are both (x + 3) and (4 - x) positive?" or, more simply, "Is x greater than or equal to -3 and less than or equal to 4?" Given both statements, we can answer affirmatively!
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