Is |x| < 1? means is -1 < x < 1?
(1) From |x + 1| = 2 |x - 1|, we get x = 3 and also -(x+1) = 2(x -1) implies x = 1/3
So, we have 2 values of x from statement (1). Hence, (1) is NOT SUFFICIENT to answer the question.
(2) implies that x is not equal to 3, but this is NOT SUFFICIENT to answer the question.
Combining, we are able to get a unique value for x, which is x = 1/3 as x is not equal to 3 by statement (2) and answer to the main question is "yes"
The correct answer is (C).
Tough DS inequalities
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Stmt # 1: Says |x+1| = 2|x-1|
=> x+1 = 2(x-1) or x+1 = -2(x-1)
=> x = 3 => x = 1/3
=> x = 3 or 1/3
Substitute x=3, we have |x+1| = 4 = 2|x-1| = 2|2| = 4. Hence x = 3 is valid solution.
Substitute x=1/3, we have |x+1| = 4/3 = 2|-2/3| = 4/3. Hence x = 1/3 is also a valid solution.
If x = 3 |x| > 1 and if x = 1/3 |x| < 1. Hence there is ambiguity. NOT SUFFICIENT.
Stmt # 2: Says |x-3| != 0
=> x != 3 (x is NOT EQUAL to 3). However, x can be equal to 1/3 or 2 or 5, or any number other than 3.
Hence there is ambiguity. NOT SUFFICIENT.
Combining 1 and 2, we have x = 1/3. Hence, |x| < 1. Thus SUFFICIENT. Hence C is
the answer.
=> x+1 = 2(x-1) or x+1 = -2(x-1)
=> x = 3 => x = 1/3
=> x = 3 or 1/3
Substitute x=3, we have |x+1| = 4 = 2|x-1| = 2|2| = 4. Hence x = 3 is valid solution.
Substitute x=1/3, we have |x+1| = 4/3 = 2|-2/3| = 4/3. Hence x = 1/3 is also a valid solution.
If x = 3 |x| > 1 and if x = 1/3 |x| < 1. Hence there is ambiguity. NOT SUFFICIENT.
Stmt # 2: Says |x-3| != 0
=> x != 3 (x is NOT EQUAL to 3). However, x can be equal to 1/3 or 2 or 5, or any number other than 3.
Hence there is ambiguity. NOT SUFFICIENT.
Combining 1 and 2, we have x = 1/3. Hence, |x| < 1. Thus SUFFICIENT. Hence C is
the answer.
