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

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

As the minimum distance between 11 and 12 is 1. there can be no solution that can exist which is less than 1.

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

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

Let us destroy it,

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

First case, x>11
x-11+x-12<1
2x<24
x<12
all x values could be between 11<x<12, infinity ,that is enough for us to take this answer
i can see other cases like x<11,x<12,..., this is not necessary as i don't have time in real exams

Answer:E

3 cases
x < 11
11 ≤ x < 12
x ≥ 12
for x<11 lets try with x=10
mod (10-11) + mod (10-12)<1
1+2<1 incorrect hence x<11 is not a solution range
2nd case
11 ≤ x < 12
this will give th erange of solution 1 ≤ x < 12
3rd
x ≥ 12 |x - 11| + |x - 12| = |13 - 11| + |13 - 12| = |2| + |1| = 2 + 1 = 3 incorreect 3<1

hence solution is x =[11,12)
only one value 11

Ans. (A) 0

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

We need to consider three assumption,

Assumption 1) x<=11

Equation becomes:
11-x+12-x<1
-> 23-2x<1
-> x>11, This contradicts the assumption.

Hence no value of x is possible for assumption 1.

Assumption 2) 11<x<12

Equation becomes:
x-11+12-x<1
-> 1<1 - Wrong

Hence no value of x is possible for assumption 2.

Assumption 3) x>=12

Equation becomes:
x-11+x-12<1
->x<12, This contradicts the assumption.

Hence no value of x is possible for assumption 3.

Theus we can conclude that no value of x will satisfy the given equation.

If we imagine the above, absolute expressions on a number line, we can see that the Dist from x to 11 and x to 12 is minimum and hence no solution is possible

Ans A

IF we use any integer + or negative the answer will be an integer >1
but if we use a proper fraction as x lets 1/2, 4/5 ( where numerator is < denominator) then it will always be <1
hence the answer is infinite!

IMO, value x can take would be infinite because, any fraction can take its place to get the value between 0 and 1

3 cases for which the mod values may change its sign - x> 12 , 11<=x<=12, x<11

On solving for above 3 cases - there is no solution and hence option A is correct.

If x>0
x-11+x-12<1
x<12
If x<0
-x-11-x-12<1
x>-12
-12<x<12
As it is not mentioned that is an integer, there could be infinite number between -12 and 12. Therefore (E) is the answer

For the inequality |x - 11| + |x - 12| < 1

Case 1 x<11

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

-2x<-22 , x>22

No solution for the case

Case 2. 11<x<12

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

No solution

Case 3 x>12

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

therefore 2x<24 , x<12

No solution for the case

Overall 0 solution

IMO A

|x - a | is the distance between a and x on number line.
In this case, we need sum of distances from 11, 12 to 'x' to be <1

Let's break it down into multiple cases.
1. For any x < 11, (points to left of 11)
Lets say the distance between 11 and x is y (y is positive)
The distance between 12 and x will be 1+y
So, sum of distances from 11, 12 to 'x' will be 1+2y which is >1.
Hence, no solutions in this range.

2. For any x>12,
Lets say the distance between 12 and x is y (y is positive)
The distance between 11 and x will be 1+y
So, sum of distances from 11, 12 to 'x' will be 1+2y which is >1.
Hence, no solutions in this range.

3. For any points between 11 and 12, 11=<x<=12
Lets say the distance between 12 and x is y (y is positive)
The distance between 11 and x will be 1-y
So, sum of distances from 11, 12 to 'x' will be 1.
Hence, no solutions in this range.

Since we covered all points on number line, we have no solutions for this inequality.

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

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

-(x-11)-(x-12)<1
X+11-x+12<1
X gones so can be infinite numbers E

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Let f(x) = |x - 11| + |x - 12|

If x = 10 , f(10) = 3, If x = 9, f(9) = 5. This trend will continue as long as x < 11.
If x = 13, f(13) = 3, f(14) = 5. This trend will continue as long as x > 13.
If x = 11 or x = 12, f(11) = f(12) = 1.
For any value of x between 11 and 12, sum of distance of x from 11 and 12 on number line will always be 1.

Therefore f(x) < 1 not possible.

IMO 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

BASICALLY WE NEED THE ANSWER TO BE LESS THAN ONE
SO WE CAN START THE RANGE FROM 11 OR LOWER THAN 11 UPTO INFINITY AND USE ALL THE NEGATIVE INTEGERS TO GET THE ANSWER AS LESS THAN ONE.
FOR EXAMPLE IF X = 6
[6 - 11] + [6 - 12]
-5 + -6 = - 11
IF X = -6
[-6 - 11] + [ -6 - 12]
- 17 + - 18 = - 35
THUS WE CAN USE INFINITE VALUES
THUS "E" IS THE CORRECT ANSWER.

We can make different cases and check if any value is possible.

Case 1
x>12
x-11+x-12<1
x<12
Not possible.

Case 2
x=12
1<1
Not possible

Case 3
x=11
1<1
Not possible

Case 4
11<x<12
x-11-x+12<1
1<1
Not possible

Case 5
x<11
-x+11-x+12<1
-x<-11
x>11
Not possible

Hence no values are possible.

Answer Option A

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

plugging values for x by considering boundary conditions
a) put x<11 ex: x = 10
|10-11| + |10 - 12| <1
3 < 1 --> not true

b) between 11 and 12
x = 11.5
|11.5 - 11| + |11.5 - 12| < 1
1<1
not true

c) greater than 12
x= 13
13 -11 + 13- 12 <1
3<1. --> not true

for values 11 and 12, we get 1 < 1 --> not true

There are no values of x that satisfy this condition

Option 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


For the inequality |x - 11| + |x - 12| < 1; we get

Inflection pts at x = 11, x =12

provides three ranges on the number line
x<11...................(1)
11<= x < 12........(2)
x>=12.................(3)

(1) x<11
|x - 11| + |x - 12| < 1
-x+11-x+12 < 1
-2x + 23 < 1
-2x < -22
x> 11

Inconsistent

(2) 11<= x < 12
|x - 11| + |x - 12| < 1
x-11 -x + 12 < 1
1 < 1

Inconsistent

(3) x>=12
|x - 11| + |x - 12| < 1
x-11+x-12 < 1
2x-23 < 1
2x < 24
x< 12

Inconsistent

Hence, no values of x satisfy the inequality |x - 11| + |x - 12| < 1

ie, 0 is the CORRECT answer


(A) is the CORRECT answer