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
(A) 6
(B) 7
(C) 8
(D) 2
(E) 12
OA=A
We have |x-3|+|2x+4|+|x| ≤ 11ziyuenlau 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
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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