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How many integral values of x will satisfy |x-3|+|2x+4|+|x|

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by Jay@ManhattanReview » Sat Apr 29, 2017 4:45 am
ziyuenlau wrote:How many integral values of x will satisfy |x-3|+|2x+4|+|x| ≤ 11?

(A) 6
(B) 7
(C) 8
(D) 2
(E) 12

OA=A
We have |x-3|+|2x+4|+|x| ≤ 11

Taking value in the modulus as positive,

=> (x-3) + (2x+4) + x ≤ 11

=> 4x ≤ 10

=> x ≤ 5/2

Taking value in the modulus as negative,

=> -(x-3) - (2x+4) - x ≤ 11

=> -4x - 1 ≤ 11

=> -4x ≤ 12

=> -x ≤ 3

=> x >= -3

Thus, 5/2 >= x >= -3

=> x: {-3, -2, -1, 0, 1, 2}

There are six integer values for x.

The correct answer: A

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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by [email protected] » Sat Apr 29, 2017 9:25 am
Hi ziyuenlau,

We're told that X must be an INTEGER, so since we're adding the sums of 3 absolute values - AND we're looking for sums that are LESS than or EQUAL to 11 - there can only be a limited number of possible solutions. Based on the answer choices, we know that there are at least 2, but no more than 12, possible integer values for N. As such, we can use a bit of 'brute force' to find all of the possible values without too much trouble.

We're given the inequality: |X - 3| + |2X + 4| + |X| ≤ 11

Let's start with...
X = 0.... |-3| + |4| + |0| = 7 so X could be 0

Now let's work our way "up"...
X = 1.... |-2| + |6| + |1| = 9 so X could be 1
X = 2.... |-1| + |8| + |2| = 11 so X could be 2
X = 3.... |0| + |10| + |3| = 13 so X CANNOT be 3

Increasing X will just increase the total further, so there's no reason to test any additional positive integers...

Now let's work our way "down"....
X = -1.... |-4| + |2| + |-1| = 7 so X could be -1
X = -2.... |-5| + |0| + |-2| = 7 so X could be -2
X = -3.... |-6| + |-2| + |-3| = 11 so X could be -3
X = -4.... |-4| + |-4| + |-4| = 12 so X CANNOT be -4

Decreasing X will just increase the total further, so there's no reason to test any additional negative integers...

We have 6 possible values of X.

Final Answer: A

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