Is |x - 1| < 1 ? (1) (x - 1)^2 >1 (2) x < 0

Even if you aren't sure how to do the algebra, this is a good DS problem for case testing. That's because the numbers involved are pretty simple, and there's only a single variable. It's always possible, when testing cases, that you'll miss something - but it's also the best way to prove that a statement is insufficient, and it's better than just guessing or giving up because the algebra is complex.

(1)

Test extremes here. Start with a large number: x = 1000 fits the statement, since (1000-1)^2 is much greater than 1. Then, answer the question. Is |1000-1| < 1? No. It's greater.

Next, think about what you'd have to achieve to get a different answer, in this case, a yes. You'd need a much smaller value of x. The smallest positive value of x that could possibly fit the statement would be something like 2.0001. But that also gives a 'no' answer, since |2.0001-1| is still greater than 1.

Try a negative value, as well. x = -0.5 works. But again, |-0.5-1| is greater than 1, so the answer is 'no'.

If you always get a 'no', the statement is sufficient.

(2)

Same situation - test a couple of negative values of x and notice that you always get a 'no', so it's sufficient. It's nice, but not always necessary, to logically reason out why you're always getting the same answer. But if the problem is tough and you're short on time, it's okay to just notice that every case seems to give the same result and decide that the statement is sufficient.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is |x-1| < 1 ?

(1) (x-1)^2 >1
(2) x < 0


In the original condition, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. Q becomes -1<x-1<1?, 0<x<2? and for 1), x-1<-1, 1<x-1--> x<0, 2<x is derived, which is no and sufficient.
For 1), also x<0, which is no and sufficient.
Thus, the answer is D.


 For cases where we need 1 more equation, such as original conditions with“1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

I understand the question and the second statement but the interpretation of statement (1) is a bit of an annoyance. Can somebody help with that?

Is |x-1| < 1 ?

(1) (x-1)^2 >1. Since both sides of the inequality are non-negative, we can safely take the square root to get \(|x-1| > 1\) (recall that \(\sqrt{x^2}=|x|\)). Sufficient.

(2) x < 0. The question above can be rephrased as "is \(-1 < x - 1 < 1\)?". Add 1 to all three parts: "is \(0 < x < 2\)?". Thus, given statement (\(x < 0\)) gives a NO answer to the question. Sufficient.

Answer: D.

Hope it's clear.

Oh yes, didn't think of taking the square root in statement (1). That certainly makes things easier. What if I wanted to do it without taking the square root? Taking 1 one to the LHS making it (x-1)^2 - 1 > 0. How would the simplification look like in that case?

\((x-1)^2 >1\);

\(x^2 - 2x + 1 > 1\);

\(x^2 - 2x > 0\);

\(x(x - 2) > 0\);

The roots are 0 and 2. ">" sign indicates that the solution is to the left of the smaller root and to the right of the larger root. Thus x < 0 and x > 2.

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hi

For equation # 1.
(x-1)^2 >1

|x-1| > 1

so,
x is greater than 2
OR x is less than 0
please correct me if I am missing something ...

another approach, however, brings us to the scenario as under...

(x-1)^2 >1

x(x-2) > 0

so,
x is greater than 0
OR x is greater than 2
please correct me if I am missing something ...

So both approaches get us to the correct answer. Would you please, however, say if there is anything wrong with the second approach..? if yes or no, please say to me the correct approach to follow when dealing questions such as this one ..

thanks in advance ...

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