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

A is the answer.

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

-(x-11) -(x-12)<1
x>11------[2]

from 1 and 2, x can take 0 values.

To find the values of x that satisfy the inequality, let's consider the possible cases based on the sign of (x - 11) and (x - 12):

Case 1: x is between 11 and 12
If x is between 11 and 12, then both |x - 11| and |x - 12| will be positive, and their sum will be (x - 11) + (x - 12) = 2x - 23.

For this case, we want 2x - 23 < 1.
2x < 24
x < 12

Case 2: x is less than 11
If x is less than 11, then (x - 11) will be negative, and (x - 12) will also be negative.

For this case, we want -(x - 11) - (x - 12) < 1.
-2x + 23 < 1
-2x < -22
x > 11

Now, let's combine the two cases:
Since x < 12 and x > 11, the only possible value for x that satisfies the inequality is x = 11. However, we need to check this value.

For x = 11, the inequality becomes:
|11 - 11| + |11 - 12| < 1
0 + 1 < 1

This is not true, so there are no values of x that satisfy the inequality.

The correct answer is (A)

To solve this, we can imagine the number line (or draw one, on a rough sheet)
|x - 11| represents the distance of x from 11, and |x - 12| represents the distance of x from 12.

This inequality asks how many values of x are less than 1 unit away from both 11 and 12.

Because the sum of the distances from 11 and 12 must be less than 1, x must lie between 11 and 12.

Let's break down the cases:

1. If x is between 11 and 11.5, then |x - 11| < 0.5, and |x - 12| > 0.5, but < 1. So, the sum will be less than 1.
2. If x is between 11.5 and 12, then |x - 12| < 0.5, and |x - 11| > 0.5, but < 1. So, the sum will be less than 1.

So, all values of x between 11 and 12, but excluding 11.5, satisfy the inequality.

Now the thing is, we don't know whether X is an integer, or a whole number.

So, I think there are infinitely many values of x that satisfy the inequality.

I will go with E.
Waiting for OA.

Two Mods getting added will always yield a positive value, so < 1, indicates the answer lies between 0 & 1

now x could be positive, negative, zero, fraction or some combination of all these.

x cant be negative since it will "add up" inside the mod and result in a bigger number than one.


lets open the mod i.e.

(1) x-11 + x-12 < 1
= 2x-23 <1, now at x =11.5 will give us 0 and all number in 11.5 ≤ x < 12. satisfy it, since at 12 it will be = 1 and there are infinite values between 11.5 to 12 (11.6,11.7,11.999999, 11.50009 etc.)

(2) 11-x + 12-x < 1, even this equation has a limit from 11.5 ≤ x < 12. same as above
= 23 -2x <1
hence answer is Infinite

We know that a modulus of a real number x is the non-negative value of x.
So we know that the value of |x - 11| + |x - 12| will be non negative. i.e greater than or equal to zero.

We also know that the distance between 11 and 12 is 1. or the absolute difference between 11 and 12 is 1.
Now we can solve this equation |x - 11| + |x - 12| by taking three different range of x.
Case 1. for x<11
Case 2. for \(11\leq{x\leq}{12}\).
Case 3.for x>12

Since we already know that the distance between 11 and 12 is 1. Therefore for any value of x ( where x<11 or x>12) the value of \(|x - 11| + |x - 12|\geq{1}\).

For any value of x between \(11\leq{x\leq}{12}\). we will find that the sum of |x - 11| + |x - 12| is equal to 1.
We can take couple of example
For x=11.3
The value of |x - 11| will be .3 and |x - 12| will be 0.7, thus the final answer being 1.
For x=11.5
The value of |x - 11| will be 0.5 and |x - 12| will be 0.5, thus the final answer being 1.

Therefore the final answer is A.

Q: Is |x - 11| + |x - 12| < 1?

For LHS to be <1, the value of LHS must be 0 or negative.
...Since LHS has absolute value condition, it will never be equal to 0.
...Therefore, we need to that value of x which could make the LHS negative.

For x = 11, LHS = 1 ×<1
...x = -11, LHS = 46 ×<1
...x = 0, LHS = 23 ×<1
...x = 3/4, LHS = 16.5 ×<1
...x = 5/3, LHS = 19.6 ×<1

None of the values of x fit, be it - positive integer, negative integer, positive/negative proper fraction, or positive/negative improper fraction.

Therefore, option A is the correct answer.

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For the inequality |x - 11| + |x - 12| < 1 to hold true,

we can take, |x - 11| + |x - 12| = 1
now, 11<x<12
but then again, we need the value of inequality <1
which cannot happen, as for every value of x between 11 and 12 the sum will be 1.

Thus ANS. A

We can make 3 cases -

1st case, x>12,
=> x - 11 + x - 12 < 1
or 2x<24 or x<12. So x > 12 is not possible.

2nd case, 12>x>11
then, x-11 -x +12 < 1 or 1<1.
So, 12>x>11 is also not possible.

3rd case, x<11
then, 11-x+12-x<1 or 22<2x or 11<x.
This scenario is also not possible.

So no possible value of x exist for the above equation.

ANS is 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

Answer:
number of values of x satisfying the inequality
we have two values of x where function is discontinuous , x =11 and x= 12
so area of concern is x>12, 11<x<12 and another x<11 and then we test x= 11 and x= 12

Case I: x>12
x-11 +x-12 <1
which is 2x-23<1
which is 2x<24
or x<12
contradicts

Case II: 11<x<12
x-11 +12-x <1
1<1 contradicts

Case III: x<11
11-x +12-x <1
23-2x<1
2x>22
x>11
contradicts

Case IV. x= 11
0 +1 <1?
contradicts

Case V. x=12
1 +0 <1?
contradicts

So none of the values of x satisfy the equation and hence answer is 0

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

X=11
|x - 11| + |x - 12| NOT LESS THAN 1

X=12
|x - 11| + |x - 12| NOT LESS THAN 1

X=-1
|x - 11| + |x - 12| NOT LESS THAN 1

X=0
|x - 11| + |x - 12| NOT LESS THAN 1

X=1/2
|x - 11| + |x - 12| NOT LESS THAN 1

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

Absolute value are always >= 0, ie.
0 >=|x - 11| + |x - 12| < 1?

Since there is summation (+) between the two absolute values, it will be POSITIVE + POSITIVE < 1
Given that 11 & 12 are consecutive number with difference of 1, and we need to satisfy the inequality of <1, it's not really possible to find any x

x can assume infinite values between 11 and 12

Answer is A
This expression means to ask that is the distance from 11 to 12 smaller than 1. of course the distance is 1. so no such a value x.

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

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


Answer is A
Check attached solution

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three categories:
case 1 x>= 12
x-11+x-12 < 1 ==x<12 no solution
case 2 11<=x<12
x-11+12-x<1 ==no solution
case 3 x<11
11-x+12-x<1==x>11 no solution

The value satisfying this inequality is zero- Answer A

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


Solu:

Here the question says the sum of distance of X from 11 and 12 should be less than 1.. However given the distance of X from 11 to 12 itself is 1, the answer has to be Zero.