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# Around the World in 80 Questions (Day 4): How many values of x satisfy

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Re: Around the World in 80 Questions (Day 4): How many values of x satisfy [#permalink]
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Bunuel wrote:
How many values of x satisfy the inequality |x - 11| + |x - 12| < 1?

(A) 0
(B) 1
(C) 2
(D) 3
(E) Infinite number

 This question was provided by GMAT Club for the Around the World in 80 Questions Win over \$20,000 in prizes: Courses, Tests & more

------ REGION 1 ------- 11 --- REGION 2 --------- 12 ---------- REGION 3

REGION 1

-(x-11)-(x-12) < 1
-2x < -22
x > 11

Ignore this region

REGION 2

(x-11)-(x-12) < 1
12 - 11 < 1
1 < 1

Ignore this region

REGION 3

x - 11 + x - 12 < 1
2x - 23 < 1
x < 12

Ignore this region

No value satisfy.

IMO A
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Re: Around the World in 80 Questions (Day 4): How many values of x satisfy [#permalink]
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note that abs(x) = x if x>=0 ; -x if x<0 . We will use this definition of absolute value extensively to get to the answer

We are required to deal with abs(x-11) and abs(x-12)

Case 1 : x < 11
so the LHS of the inequality can be rewritten as (11-x)+(12-x) = 23-2x
Now 23 - 2x < 1 => 22 < 2x => x >11. This contradicts the case we are studying : x<11. Therefore no value of x<11 satisfies the inequality.

Case 2 : 11 <= x < 12
so the LHS of the inequality can be rewritten as (x-11)+(12-x) = 1
Now 1 < 1. This is wrong/false. Therefore no value of 11 <= x < 12 satisfies the inequality.

Case 3 : x >= 12
so the LHS of the inequality can be rewritten as (x-11)+(x-12) = 2x-23
Now 2x - 23 < 1 => 2x < 24 => x < 12. This contradicts the case we are studying : x >= 12. Therefore no value of x >=12 satisfies the inequality.

NO VALUE SATISFIES IN ANY OF THE CASES. SO ANSWER IS 0 i.e, A
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Re: Around the World in 80 Questions (Day 4): How many values of x satisfy [#permalink]
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Bunuel wrote:
How many values of x satisfy the inequality |x - 11| + |x - 12| < 1?

(A) 0
(B) 1
(C) 2
(D) 3
(E) Infinite number

 This question was provided by GMAT Club for the Around the World in 80 Questions Win over \$20,000 in prizes: Courses, Tests & more

|x - 11| + |x - 12| < 1

|x - a| refers to the distance between x and a on the number line
In this question, we have reference points 11 and 12

If x is to the right of 12: |x-11| will be greater than 1 => |x - 11| + |x - 12| > 1 - does not satisfy
If x = 12: |x - 11| + |x - 12| = 1 - does not satisfy
If x is between 11 and 12: |x-11| + |x-12| is the sum of distances of x from 11 and 12, which is 1 - does not satisfy
If x = 11: |x - 11| + |x - 12| = 1 - does not satisfy
If x is to the left of 11: |x-12| will be greater than 1 => |x - 11| + |x - 12| > 1 - does not satisfy

Thus, there is no possible value of x