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

Question

How many values of x satisfy the inequality |x - 11| + |x - 12| < 1?

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

Kinshook · Accepted Answer

Asked: How many values of x satisfy the inequality |x - 11| + |x - 12| < 1?

Case 1: x < 11
|x - 11| + |x - 12| = 11-x + 12-x< 1
23-2x < 1
2x > 22
x > 11
Not feasible since x<11

Case 2: 11<=x<12
|x - 11| + |x - 12| = x-11 + 12-x< 1
1 < 1
Not feasible

Case 3: x>=12
|x - 11| + |x - 12| = x-11 + x-12< 1
2x -23<1
2x<24
x<12
Not feasible since x>=12

IMO A

Second approach: -
|x-11|+|x-12| is the sum of distance of x from 11 and distance of x from 12
The sum of distance of x from 11 and distance of x from 12 is minimum when x lies between 11 & 12 = 1 and is NOT <1 in any case

IMO A

Sushma1215 · Answer

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

Answer : for whatever value of x, since we are taking modulus the resultant value is always going to be positive and greater than 1.

missionmba2025 · Answer

------ 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

vedha0 · Answer

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

sujoykrdatta · Answer

|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
Answer A

lewdtoucan · Answer

hello all, I think we can use the graphical way to solve "how many values can x take" problems. see Chetan's post here for a detailed explanation. here is my solution: we are given, |x-11| + |x-12| < 1 the equation can be rearranged as follows --> |x-11| < 1 - |x-12| let y = LHS and z = RHS. Plot the the functions and you'll see that curves intersect for values between 11 and 12. But the question asks specifies that LHS < RHS, therefore there are 0 points of intersection. here is a plot i generated using a graphing calc to verify the method after solving: GMAT-Club-Forum-gx17kc5y.png

shori345 · Answer

Similarly, in the 3 regions, can we plug in some values, for example in region 1 x=10, region 2 x=11.5 and region 3 x=13. For all these values, the equation does not satisfy the inequality. Is this a good approach to take?

Bunuel · Answer

A specific value might not satisfy the inequality, while another nearby one might. So plugging in alone can mislead. It’s better to follow the full analysis shown using regions.

user34745792203 · Answer

After solving for x:

x>11
x<12
⇒ 11< x < 12

So, the value of x must lie between 11 and 12.
 Since we are asked how many values of x (not just how many integers) satisfy this condition, the solution is  (E) infinite numbers.

Am I missing something?

Bunuel · Answer

Yes. 11 < x < 12 is not true, there is no x that satisfies |x - 11| + |x - 12| < 1. Please review the solutions above.

Saun2511 · Answer

can also be equal to 1

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