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

Imagine it this way - sum of the distance of x from 11 and 12 is less than 1

If x lies to the left of 11, the sum would always greater than 1
if x lies to the right of 12, the sum would always be greater than 1
if x lies anywhere between 11 and 12, the sum would always be equal to 1

So the equation given will never hold true for any value of x, hence answer is (A) 0

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

X^2-22X+121+x^2-24x+144<1

Solving,we get x=12 and x=11

Only two values, ans C

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There are three boundary conditions
Condition 1: x < 11,
Condition 2: \(11\leq{x}\) < 12 &
Condition 3: \(12\leq{x}\)

Condition 1 gives us,
- x + 11 - x + 12 < 1
=> - 2x < 1 - 23
=> - 2x < -22
=> 2x > 22
=> x > 11

But we assumed x <11 hence there is no solution of x in this region.

Condition 2 gives us,
x - 11 - x + 12 < 1
=> 12 - 11 < 1
=> 1 < 1
Not possible. Hence, no solution of x in this region.

Condition 3 gives us,
x - 11 + x - 12 < 1
=> 2x < 1 + 23
=> x < \(\frac{24}{2}\)
=> x < 12
But we assumed \(x\geq{12}\)
Hence, no solution in this region for x.

IMO OA should be A.

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

two positive quantities are added together here, and we were asked if there sum is less than 1.
whenever we move away from critical points of 11, and 12 the sum will always be greater than 1. however, since 11, and 12 are consecutive integers that are subtracted from x so even if we take an fractional value between 11, and 12 the difference between |x-11|, and |x-12| will always result in a value equal to one.

Thus no solution exist

IMO A

For x > 12
\((x-11)+(x-12)<1\)
we get \( x < 12 \) not possible.

For 11< x< 12
\((x-11) -(x-12)<1\)
Not possible.

For \( x< 11 \)
\(-(x-11) -(x-12)<1\)
\(-2x+23 <1\)
We get \(x>11\) not possible.

So no value of x satisfy above equation.

IMO Choice A.

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

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

checking with values...
11,12,11.9,11.5...it's not possible
Ans A

(C) 2
-Suspected to be correct.
Rationale: I set each statement equal to <1. For the first section, I wrote x - 11 <1 then added 11 to the other side; x = < 12. It would also be x - 11 > -1 to account for the absolute value. I added 11 to the other side and this is equal to x >10. I did the same thing for the second section, x - 12 < 1 then added 12 to the other side; x = < 13. It would also be x - 12 > -1 to account for the absolute value. I added 12 to the other side and this is equal to x >11. The numbers between 10 and 12 in the first section is 11, and the numbers between 11 and 13 is 12. This indicates that there are only 2 values of x to satisfy the inequality, so C is the correct answer.

any value of x it will not be less than 1

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

only values less than 1 are 0 or fractions less than 1 and numbers in negative
Given both the absolutes will always result in positive, only way is to try to see if the expression to 0

|11.5 - 11| + |11.5-12| = 0.5 + 0.5 = 1
|11.25 - 11| + |11.25-12| = 0.25 + 0.75 = 1
|11.75 - 11| + |11.75-12| = 0.75 + 0.75 = 1
|11 - 11| + |11 -12| = 0 + 1 = 1
|12 - 11| + |12 -12| = 1 + 0 = 1

There is no value of x that satisfies the inequality since both expression will balance each other.


Answer: A

The inequality is: |x - 11| + |x - 12| < 1

To find the solution, we need to consider different cases based on the absolute values:

Case 1: x < 11
In this case, both |x - 11| and |x - 12| are positive, so the inequality becomes:
-(x - 11) - (x - 12) < 1
Simplifying, we get:
-2x + 23 < 1
-2x < -22
x > 11
However, we are considering values of x less than 11 in this case, so there are no solutions in this range.

Case 2: 11 ≤ x < 12
In this case, |x - 11| is positive, and |x - 12| is negative, so the inequality becomes:
(x - 11) - (x - 12) < 1
Simplifying, we get:
1 < 1
This is not true for any x in this range.

Case 3: 12 ≤ x < 13
In this case, both |x - 11| and |x - 12| are positive, so the inequality becomes:
(x - 11) + (x - 12) < 1
Simplifying, we get:
2x - 23 < 1
2x < 24
x < 12
For this case, x must be less than 12 to satisfy the inequality. Again, no solutions

Case 4: x ≥ 13
In this case, both |x - 11| and |x - 12| are positive, so the inequality becomes:
(x - 11) + (x - 12) < 1
Simplifying, we get:
2x - 23 < 1
2x < 24
x < 12
However, we are considering values of x greater than or equal to 13 in this case, so there are no solutions in this range.

Therefore, there are no values of x within the range of 11 and 12 that satisfy the inequality.
Option A) 0

Case 2 sign needs to be reversed for second mod function.

case II: 11<x<12
x-11 - x +12 <1
1 <12 and x cannot be solved so zero values.

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|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