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How many solutions are possible for the inequality | x - 1 |

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by BTGmoderatorLU » Mon May 14, 2018 2:35 pm
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How many solutions are possible for the inequality | x - 1 | + | x - 6 | < 2?

A. 3
B. 2
C. 1
D. 0
E. 4

The OA is D.

Please, can anyone assist me with this PS question? I don't have it clear. Thanks in advance!
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by Jay@ManhattanReview » Mon May 14, 2018 11:03 pm
BTGmoderatorLU wrote:How many solutions are possible for the inequality | x - 1 | + | x - 6 | < 2?

A. 3
B. 2
C. 1
D. 0
E. 4

The OA is D.

Please, can anyone assist me with this PS question? I don't have it clear. Thanks in advance!
Looking at the inequality | x – 1 | + | x – 6 | < 2, we see that x cannot be negative since if x is negative, the value of | x – 1 | + | x – 6 | would then be greater than 7 (= |–1| + |–6|).

The value of x for which | x – 1 | and | x – 6 | would be equal is the mean of 1 and 6, i.e., 3.5.

At x = 3.5, the value of | x – 1 | + | x – 6 | = | 3.5 – 1 | + | 3.5 – 6 | = 2.5 + 2.5 = 5. This value is unacceptable since | x – 1 | + | x – 6 | < 2.

If we increase the value of x, to say x = 4, we see that at x = 4, the value of | x – 1 | + | x – 6 | = | 4 – 1 | + | 4 – 6 | = 3 + 2 = 5. This is also unacceptable since | x – 1 | + | x – 6 | < 2.

If we decrease the value of x, to say x = 3, we see that at x = 3, the value of | x – 1 | + | x – 6 | = | 3 – 1 | + | 3 – 6 | = 2 + 3 = 5. This is also unacceptable since | x – 1 | + | x – 6 | < 2.

At smaller values and larger values of x, the value of the expression | x – 1 | + | x – 6 | would rather be 5 or increase. So, there is no possible solution.

The correct answer: D

Hope this helps!

-Jay
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by GMATGuruNY » Tue May 15, 2018 1:09 am
BTGmoderatorLU wrote:How many solutions are possible for the inequality | x - 1 | + | x - 6 | < 2?

A. 3
B. 2
C. 1
D. 0
E. 4
|a-b| = the DISTANCE between a and b.

Thus:
|x-1| = the distance between x and 1.
|x-6| = the distance between x and 6.

Question stem rephrased:
For how many values of x is the sum of the two distances less than 2?

The distance between 1 and 6 is 5.

Thus, if x is BETWEEN the two endpoints in blue (or at either endpoint), then the sum of the two distances will be EQUAL TO 5:
1 <--- |x-1| ---> x <---|x-6|---> 6.
Here, |x-1| + |x-6| = the distance between 1 and 6 = 5.

By extension, if x is BEYOND either endpoint -- if x is to the left of 1 or to the right of 6 -- then the sum of the two distances will be GREATER THAN 5.

Since 5 is the least possible value for |x-1| + |x-6|, there is no value of x such that |x-1| + |x-6| < 2.

The correct answer is D.
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by Jeff@TargetTestPrep » Wed May 16, 2018 10:08 am
BTGmoderatorLU wrote:How many solutions are possible for the inequality | x - 1 | + | x - 6 | < 2?

A. 3
B. 2
C. 1
D. 0
E. 4
Notice that x - 1 = 0 means x = 1 and x - 6 = 0 means x = 6, so we have 3 cases: (1) x ≥ 6, (2) 1 ≤ x < 6, and (3) x < 1. We need to see what happen in each of these cases. Let’s do that now:

Case 1:

If x ≥ 6, then both x - 1 and x - 6 are positive, so |x - 1| = x - 1 and |x - 6| = x - 6, so |x - 1| + |x - 6| < 2 becomes:

x - 1 + x - 6 < 2

2x - 7 < 2

2x < 9

x < 9/2

However, x < 9/2 contradicts our assumption that x ≥ 6, so there are no solutions when x ≥ 6.

Case 2:

If 1 ≤ x < 6, then x - 1 is positive but x - 6 is negative, so |x - 1| = x - 1 and |x - 6| = -(x - 6) = -x + 6, so |x - 1| + |x - 6| < 2 becomes:

x - 1 + (-x + 6) < 2

5 < 2

We see that 5 < 2 is a false statement. Therefore, there are no solutions when 1 ≤ x < 6, either.

Case 3:

If x < 1, then both x - 1 and x - 6 are negative, so |x - 1| = -(x - 1) = -x + 1 and |x - 6| = -(x - 6) = -x + 6, so |x - 1| + |x - 6| < 2 becomes:

-x + 1 + (-x + 6) < 2

-2x + 7 < 2

-2x < -5

x > 5/2

However, x > 5/2 contradicts our assumption that x < 1, so there are no solutions when x < 1.

We see that there no solutions that satisfy the given inequality.

Answer: D

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