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05 Jun 2017, 14:17
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Is \(x\) between 0 and 1?
1) \(x^{2}<x\)
2) \(x\) is positive
Can someone please explain the official answer? This is from the Veritas Prep Data Sufficiency book.
1) \(x^{2}<x\)
2) \(x\) is positive
Can someone please explain the official answer? This is from the Veritas Prep Data Sufficiency book.
05 Jun 2017, 14:39
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What I don't get is why intuitively we know that x cannot be negative, since x^2 is always positive, but algebraically,
x^2<x
x^2-x<0
x(x-1)<0
x<0 or x<1
^here, x<0 or x<1. This shows x can be negative. Where am I going wrong?
x^2<x
x^2-x<0
x(x-1)<0
x<0 or x<1
^here, x<0 or x<1. This shows x can be negative. Where am I going wrong?
05 Jun 2017, 14:41
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bianalyst
Is \(x\) between 0 and 1?
1) \(x^{2}<x\)
2) \(x\) is positive
1) \(x^{2}<x\)
2) \(x\) is positive
Target question: Is x between 0 and 1?
Statement 1: x² < x
Since we know that x² is POSITIVE, we can safely divide both sides of the inequality by x²
When we do this, we get: 1 < 1/x
If 1/x is greater than 1, we know that x must be POSITIVE (i.e., x > 0).
Since x is POSITIVE, we can safely multiply both sides of the inequality by x to get: x < 1
Combine the two inequalities to get: 0 < x < 1
In other words, x IS between 0 and 1
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: x is positive
This doesn't tell us much.
It could be the case that x = 1/2, in which case, x IS between 0 and 1
Or it could be the case that x = 2, in which case, x is NOT between 0 and 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Answer:
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