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Is |x| + |x -1| = 1? (1) x ≥ 0 (2) x ≤ 1 11 Apr 2019, 13:57
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mendelay
Is |x| + |x -1| = 1?

(1) x ≥ 0
(2) x ≤ 1
Show more

Since there are two mods |x| & |x-1|
we will have to check values around 0 and 1
Case 1 : x<0
lets say -0.7 : |-0.7| + |-0.7-1| = 2.4

Case 2 : 0<=x<=1
lets say 0.7 : |0.7| + |0.7-1| = 1

Case 3 : x>=1
lets say 1.7 : |1.7| + |1.7-1| = 2.4

So, |x|+|x-1| = 1 if 0<=x<=1

Statement 1: NOT SUFFICIENT
Statement 2 : NOT SUFFICIENT
Combined : SUFFICIENT

Answer C
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Is |x| + |x -1| = 1? (1) x ≥ 0 (2) x ≤ 1 27 Apr 2022, 08:38
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Can we consider the stem as
lx-0l + l1-xl = 1
which means that
Distance of x from 0 + distance of 1 from x = 1
the above condition is true for 0<= x <=1
Hence Statement 1 and statement 2 are both required for the condition to be true.
Ans C.

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Is |x| + |x -1| = 1? (1) x ≥ 0 (2) x ≤ 1 27 Apr 2022, 11:13
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Is |x| + |x-1| = 1?

(1) x ≥ 0
(2) x ≤ 1
Show more

|a| = the DISTANCE between a and 0
|a-b| = the DISTANCE between a and b

Thus:
|x| = the distance between x and 0
|x-1| = the distance between x and 1

Is |x| + |x-1| = 1?
Question stem rephrased:
Is the sum of the two distances equal to 1?

Draw a number line showing 0 and 1:
0---------------------1

If x is at either endpoint or between the two endpoints, the sum of the two distances will be EQUAL TO 1:
0<------>x<------>1
Here, |x| + |x-1| = (distance between x and 0) + (distance between x and 1) = blue distance + green distance = 1

By extension, if x is BEYOND either endpoint -- if x is to the left of 0 or to the right of 1 -- then the sum of the two distances will be GREATER THAN 1.

Thus, the sum of the two distances will be equal to 1 if x is not beyond either endpoint.
Only when the two statements are combined is it guaranteed that x will not be beyond either endpoint.

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C
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Is |x| + |x -1| = 1? (1) x ≥ 0 (2) x ≤ 1 18 Jul 2023, 02:59
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|x-a| means the distance of x from 'a' on the number line.
Hence |x| represents the distance of x from 0 on the number line, and |x-1| means the distance of x from +1 on the number line. Since the sum of these 2 distances is zero, that means x is between 0 and 1 (both inclusive) on the number line and the range of x can be defined as:
x=[0,1] and now the question has become whether x lies between 0 and 1 (both inclusive) on the number line.

(1) x>=0; x can be 0,0.5,0.7,1,2,3, and so on; NOT SUFFICIENT
(2) x<=1: x can be 0, -.05, -0.7-1,-2,-3, and so on; NOT SUFFICIENT

(1)+(2): Now it is clear that x is between 0 and 1 (both inclusive); SUFFICIENT

Answer: C
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Is |x| + |x -1| = 1? (1) x ≥ 0 (2) x ≤ 1 23 Oct 2023, 23:06
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Bunuel
This one is very tricky!

Is |x| + |x -1| = 1?
(1) x ≥ 0
(2) x ≤ 1

Q is \(|x| + |x -1| = 1\). Let's check when this equation holds true. We should consider three ranges (as there are two check points \(x=0\) and \(x=1\)):

A. \(x<0\) --> \(-x-x+1=1\) --> \(x=0\), but this solution is not valid as we are checking the range \(x<0\);

B. \(0\leq{x}\leq{1}\) -->\(x-x+1=1\) --> \(1=1\), which is true. That means that for ANY value from the range \(0\leq{x}\leq{1}\), equation \(|x| + |x -1| = 1\) holds true.

C. \(x>1\) --> \(x+x-1=1\) --> \(x=1\), but this solution is not valid as we are checking the range \(x>1\).

So we get that equation \(|x| + |x -1| = 1\) holds true ONLY in the range \(0\leq{x}\leq{1}\).

Statements:
(1) \(x\geq{0}\). Not sufficient, as \(x\) must be also \(\leq{1}\);
(2) \(x\leq{1}\). Not sufficient, as \(x\) must be also \(\geq{0}\);

(1)+(2) \(0\leq{x}\leq{1}\), exactly the range we needed. Sufficient.

Answer: C.
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Bunuel, how do you know that the second range \(0\leq{x}\leq{1}\) includes both 0 and 1 and the first and second range don't include 0 and 1?
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Is |x| + |x -1| = 1? (1) x ≥ 0 (2) x ≤ 1 24 Oct 2023, 00:15
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wpt2009
Bunuel
This one is very tricky!

Is |x| + |x -1| = 1?
(1) x ≥ 0
(2) x ≤ 1

Q is \(|x| + |x -1| = 1\). Let's check when this equation holds true. We should consider three ranges (as there are two check points \(x=0\) and \(x=1\)):

A. \(x<0\) --> \(-x-x+1=1\) --> \(x=0\), but this solution is not valid as we are checking the range \(x<0\);

B. \(0\leq{x}\leq{1}\) -->\(x-x+1=1\) --> \(1=1\), which is true. That means that for ANY value from the range \(0\leq{x}\leq{1}\), equation \(|x| + |x -1| = 1\) holds true.

C. \(x>1\) --> \(x+x-1=1\) --> \(x=1\), but this solution is not valid as we are checking the range \(x>1\).

So we get that equation \(|x| + |x -1| = 1\) holds true ONLY in the range \(0\leq{x}\leq{1}\).

Statements:
(1) \(x\geq{0}\). Not sufficient, as \(x\) must be also \(\leq{1}\);
(2) \(x\leq{1}\). Not sufficient, as \(x\) must be also \(\geq{0}\);

(1)+(2) \(0\leq{x}\leq{1}\), exactly the range we needed. Sufficient.

Answer: C.

Bunuel, how do you know that the second range \(0\leq{x}\leq{1}\) includes both 0 and 1 and the first and second range don't include 0 and 1?
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Given that x can be 0 or 1, it's important to include these values in our considered ranges. Regardless of which range they fall into, the inclusion is what matters. Hence, 0 could have been included in the first range and 1 in the third; it's a matter of preference.
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