If x is positive, what is the value of |x – 3| – 2|x – 4| +

I am having difficultly with this problem.

The way I see it, we start by finding the checkpoints in |x -3| - 2|x - 4| + 2|x - 6| - |x - 7|

X = 3,4,6,7

I would think we would set up ranges as follows:

x<3
3<x<4
4<x<6
6<x<7
x>7

The first problem is, my method is unlike yours (I got seemingly incorrect answers) and secondly, it takes a long time just to set up the ranges - much longer than two minutes. There must be a quicker way!

Thanks.

Your ranges are correct except for the first one that is 0<x<3, because x is positive.

If I remember correctly this was a question from the Challenge archive of MGMAT, much harder and longer to solve than anything you'll find on the real test. A quicker way (but less methodical) is to test some odd values for X, and see what you find.

(1) x is an odd integer.

If you test any odd value, you'll find that that function always equals 0. Generally speaking testing values is a good practice in the test, however you cannot be 100% sure of your answer just by testing values.

Hope it helps

Then wouldn't the last range be between just 6 and 7, not greater than 7 for the same reason (because x is positive)?

Thanks for the explanation though!

Nope. The ranges of the function taken by itself are

x<3
3<x<4
4<x<6
6<x<7
x>7

But the text says that x is positive so is wrong to consider negative values for x; this condition affects only the interval with negative values.

\(x<3\) becomes \(0<x<3\), all the others stay the same.

Ahh. I understand now. Thanks!

just found this question...https://www.manhattangmat.com/blog/index.php/2013/05/22/want-a-750-think-your-way-through-this-challenge-problem/

Just wanted to add that this is supposed to be a >750+ question.. So 600-700 is wrong. Therefore, don't be encouraged if you can't get it (I was...).

76% of the users answered this question incorrectly, so I changed the difficulty level to 700+.

If x is positive, what is the value of |x – 3| – 2|x – 4| + 2|x – 6| – |x – 7| ?

x≥1
Check points: 3, 4, 6, 7
x<3, 3<x<4, 4<x<6, 6<x<7, x>7

x<3:
|x – 3| – 2|x – 4| + 2|x – 6| – |x – 7|
-(x-3) -2 -(x-4) +2 -(x-6) - -(x-7)
-x+3 -2(-x+4) +2(-x+6) - (-x+7)
-x+3 +2x -8 -2x +12 +x-7
= 0

3<x<4:
|x – 3| – 2|x – 4| + 2|x – 6| – |x – 7|
(x-3) - 2 -(x-4) + 2 -(x-6) - -(x-7)
x-3 -2 -(x-4) +2 -(x-6) - -(x-7)
x-3 +2x -8 -2x +12 +x-7
2x-6

4<x<6
|x – 3| – 2|x – 4| + 2|x – 6| – |x – 7|
(x-3) -2(x-4) + 2 -(x-6) - -(x-7)
x-3 - 2x+8 +2 (-x+6) - (-x+7)
x-3 - 2x+8 + -2x +12 +x-7
-2x+10

6<x<7
|x – 3| – 2|x – 4| + 2|x – 6| – |x – 7|
(x-3) - 2(x-4) +2(x-6) - -(x-7)
(x-3) - 2(x-4) +2(x-6) +x-7
x-3 - 2x + 8 +2x-12 +x-7
2x-14

x>7
|x – 3| – 2|x – 4| + 2|x – 6| – |x – 7|
x-3 -2x+8 +2x-12 -x+7
=0

(1) x is an odd integer.

If x is an odd integer, then x=1, 3, 5, 7, 9, etc. Going through the five ranges:

When x=1 the result = 0
When x=5 (4<x<6) the result is -2(5)+10 = 0
When x>7 (for all odd integers) the result = 0

Therefore, as long as x is a positive, odd integer the result is always zero.
SUFFIFICIENT

(2) x>6
|x – 3| – 2|x – 4| + 2|x – 6| – |x – 7|

If x>6 then |x-6| will be positive but |x-7| may or not be positive (we are not told if x is an integer or not) therefore, the signs for |x-7| may be positive or may be negative and will affect the value of x.
INSUFFICIENT

(A)

P.S. while this problem was fairly straight forward, it took me a long time to solve. Is there any way to solve this type of problem in a quicker fashion? Thanks!

If x is positive, what is the value of V = |x – 3| – 2|x – 4| + 2|x – 6| – |x – 7| ?

Clearly:
x>7 or x<3 => V = 0
so we only find whether x>=3 and x<=7 what is V?

(1) if x is an odd integer, if x= 3, 5, 7 => V=0 => sufficient

(2) x > 6 => V = 2(x-1) with 6<x<7, V can be any of many values as x is not an integer.

Hence, A.

Hope my solution saves your time.

OFFICIAL SOLUTION FROM MANHATTAN

1) SUFFICIENT: It’s impractical to take an algebraic approach to this statement; doing so would entail a large number of cases. For instance, |x – 3| is equal to 3 – x if x < 3, but is equal to x – 3 if x > 3; similarly, the other three absolute-value expressions switch at x = 4, 6, and 7, respectively.
 
Instead, it’s more efficient to consider the first three positive odd integers (1, 3, and 5) individually and then to consider only one algebraic case, the case in which x > 7 (because when x > 7, all values are positive so we can ignore the absolute value symbols).

If x = 1, then the value is (2) – 2(3) + 2(5) – (6) = 0.
If x = 3, then the value is (0) – 2(1) + 2(3) – (4) = 0.
If x = 5, then the value is (2) – 2(1) + 2(1) – (2) = 0.
 
If x > 7, then drop the absolute value symbols and simplify:
(x – 3) – 2(x – 4) + 2(x – 6) – (x – 7) =
x – 3 – 2x + 8 + 2x – 12 – x + 7 =
(x – 2x + 2x– x) + (-3 + 8 – 12 + 7) =
0
 
Therefore, the value of the expression is also 0 for all values of x greater than or equal to 7, including the odd integers 7, 9, 11, and so on. The value of the expression is thus 0 for all positive odd integers. The statement is sufficient.
 
(2) NOT SUFFICIENT:
As determined during the discussion of statement 1, the expression is equal to 0 when x > 7 (whether odd integer, even integer, or non-integer). We still need to test the non-integer values of x between 6 and 7.
 
If x = 6.5, then the value is (3.5) – 2(2.5) + 2(0.5) – 0.5 = –1, which is not equal to 0. The expression can thus have multiple values, so the statement is insufficient.
 
The correct answer is A.

Great question. In the video below I discuss two approaches to solve this question

1. Picking values - however this is not fool-proof and may result in a mistake
2. Breaking the expression into ranges - step by step systematic way to solve these kind of questions

Hope this will help.

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