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A. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient B. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient D. EACH statement ALONE is sufficient E. Statements (1) and (2) TOGETHER are NOT sufficient _________________

Here is my way, which worries me by its simplicity in comparison to a "750" level problem. Please let me know if there are any errors in the following approach. I am not the best with these question types and am not sure if plugging in numbers, like others did, considers an element my approach does not.

S1) 1/X > - 1 ---> 1>-X ---> (multiply by -1/turn sign)---> -1<X ..... This means X can be -.5, 0, 1, 2..insufficient

S2) 1/X^5 > 1/X^3 --->(multiply both sides by X^3)---> 1/x^2 > 1 ---> 1>X2 ---> +-1 > X ...if X is less than 1 or -1, we know its not greater than 1; sufficient.

We should probably revise the difficulty level of the question. I like your approach. +1. Can't see anything wrong with it.

dpgxxx wrote:

Here is my way, which worries me by its simplicity in comparison to a "750" level problem. Please let me know if there are any errors in the following approach. I am not the best with these question types and am not sure if plugging in numbers, like others did, considers an element my approach does not.

S1) 1/X > - 1 ---> 1>-X ---> (multiply by -1/turn sign)---> -1<X ..... This means X can be -.5, 0, 1, 2..insufficient

S2) 1/X^5 > 1/X^3 --->(multiply both sides by X^3)---> 1/x^2 > 1 ---> 1>X2 ---> +-1 > X ...if X is less than 1 or -1, we know its not greater than 1; sufficient.

Answer: B

The above approach is not right. First of all I'd suggest to state in the stem that \(x\) does not equal to zero, as \(x\) is in denominator and GMAT almost always states when providing the fraction that the variable in denominator is not zero.

Now this approach is not correct because when we multiply by variable we don't know its sign and thus don't know whether we should flip sign of inequality or not.

Algebraic solution for this question would be:

Question is \(x>1\)?

(1) \(\frac{1}{x}>- 1\) --> \(\frac{1+x}{x}>0\), two cases:

A. \(x>0\) and \(1+x>0\), \(x>-1\) --> \(x>0\);

B. \(x<0\) and \(1+x<0\), \(x<-1\) --> \(x<-1\).

We got that given inequality holds true in two ranges: \(x>0\) and \(x<-1\), thus \(x\) may or may not be greater than one. Not sufficient.

(2) \(\frac{1}{x^5}> \frac{1}{x^3}\) --> \(\frac{1-x^2}{x^5}>0\), two cases:

A. \(x>0\) (it's the same as \(x^5>0\)) and \(1-x^2>0\), \(-1<x<1\) --> \(0<x<1\);

B. \(x<0\) and \(1-x^2<0\), \(x<-1\) or \(x>1\) --> \(x<-1\);

We got that given inequality holds true in two ranges: \(0<x<1\) and \(x<-1\), ANY \(x\) from this ranges will be less than \(1\). So the answer to our original question is NO. Sufficient.

Answer: B.

As we can see the ranges for (1) and (2) obtained by this solution differ from the ranges obtained by the solution provided by dpgxxx in his/her post. _________________

We should probably revise the difficulty level of the question. I like your approach. +1. Can't see anything wrong with it.

dpgxxx wrote:

Here is my way, which worries me by its simplicity in comparison to a "750" level problem. Please let me know if there are any errors in the following approach. I am not the best with these question types and am not sure if plugging in numbers, like others did, considers an element my approach does not.

S1) 1/X > - 1 ---> 1>-X ---> (multiply by -1/turn sign)---> -1<X ..... This means X can be -.5, 0, 1, 2..insufficient

S2) 1/X^5 > 1/X^3 --->(multiply both sides by X^3)---> 1/x^2 > 1 ---> 1>X2 ---> +-1 > X ...if X is less than 1 or -1, we know its not greater than 1; sufficient.

(1) \(\frac{1}{x}>- 1\) --> \(\frac{1+x}{x}>0\), two cases:

A. \(x>0\) and \(1+x>0\), \(x>-1\) --> \(x>0\);

B. \(x<0\) and \(1+x<0\), \(x<-1\) --> \(x<-1\).

So, the given inequality holds true in two ranges: \(x>0\) and \(x<-1\), thus \(x\) may or may not be greater than 1. Not sufficient.

(2) \(\frac{1}{x^5}> \frac{1}{x^3}\) --> \(\frac{1-x^2}{x^5}>0\), two cases:

A. \(x>0\) (it's the same as \(x^5>0\)) and \(1-x^2>0\), \(-1<x<1\) --> \(0<x<1\);

B. \(x<0\) and \(1-x^2<0\), \(x<-1\) or \(x>1\) --> \(x<-1\);

We got that given inequality holds true in two ranges: \(0<x<1\) and \(x<-1\), ANY \(x\) from this ranges will be less than 1. So the answer to our original question is NO. Sufficient.

The trick to remember during solving this question is that fractions in the range \((0,1)\) taken to greater power get smaller, for example \(\left(\frac{1}{3}\right)^3 < \left(\frac{1}{3}\right)^2\), etc.

\(\frac{1}{x^5} > \frac{1}{x^3}\) is equivalent to \(x^5 < x^3\). because if two fractions have equal numerators, the fraction with a smaller denominator is the bigger one. Now we need to find the range of values of \(x\) in which \(x^5 < x^3\) holds true. This is possible in \(x \in (0,1)\) for positive \(x\) and in \(x \in (-\infty,-1)\) for negative \(x\).

We can put the values of \(x\) from these ranges to make sure it works:

You forget that \(x\) can be any number from the range \((0,\infty)\) in order to satisfy S1. \(x < -1\) is not a complete solution. Both 2 and -2 satisfy S1, as stated in the OE. S1 is not sufficient.

\(\frac{1}{x^5} > \frac{1}{x^3}\) is equivalent to \(x^5 < x^3\). because if two fractions have equal numerators, the fraction with a smaller denominator is the bigger one.

...I simply went ahead and pluggeed-in "2" and "-2" for the the second expression (just like the first). In this way, I concluded a "NO" answer for both 1/32 > 1/8 and -1/32 > -1/8

You have to be careful with the negative fractions. The highlighted expression doesn't hold true. While the absolute value of \(-\frac{1}{8}\) is greater than that of \(-\frac{1}{32}\), \(-\frac{1}{32} > -\frac{1}{8}\) because \(-\frac{1}{32}\) is closer to 0 on the number line than \(-\frac{1}{8}\) is. Justed wanted to make it clear. Hope it helps.

stilite wrote:

Since I didn't know about this rule:

\(\frac{1}{x^5} > \frac{1}{x^3}\) is equivalent to \(x^5 < x^3\). because if two fractions have equal numerators, the fraction with a smaller denominator is the bigger one.

...I simply went ahead and pluggeed-in "2" and "-2" for the the second expression (just like the first). In this way, I concluded a "NO" answer for both 1/32 > 1/8 and -1/32 > -1/8

You have to be careful with the negative fractions. The highlighted expression doesn't hold true. While the absolute value of \(-\frac{1}{8}\) is greater than that of \(-\frac{1}{32}\), \(-\frac{1}{32} > -\frac{1}{8}\) because \(-\frac{1}{32}\) is closer to 0 on the number line than \(-\frac{1}{8}\) is. Justed wanted to make it clear. Hope it helps.

stilite wrote:

Since I didn't know about this rule:

\(\frac{1}{x^5} > \frac{1}{x^3}\) is equivalent to \(x^5 < x^3\). because if two fractions have equal numerators, the fraction with a smaller denominator is the bigger one.

...I simply went ahead and pluggeed-in "2" and "-2" for the the second expression (just like the first). In this way, I concluded a "NO" answer for both 1/32 > 1/8 and -1/32 > -1/8

Let me see if I understand this question correctly - and DS problems in general since they are throwing me off left and right simply because of their demand of figuring out "yes" or "no" "yes + no" for every possibility. **So please at least tell me if I have approached this question in the correct way**

#1 The question is asking if x is greater than 1. Now I have to look at each possibility individually and plug-in numbers until I can justify whether the possibility is, in itself, enough to answer the original question stem.

#2 So, the first possibility gives us \(\frac{1}{x} > -1\)

Now I have to choose two numbers ("2" for ease) and both + and - to see if this possibility will prove the stem.

For a +2: \(\frac{1}{2} > -1\) *This holds true as +0.5 is > -1

However,

For a -2: \(\frac{1}{-2} > -1\) *This is also a confirmation of the expression itself, but creates a conflict with the stem which is only looking for the confirmation of x being greater than 1.

Therefore, the answer "A" can be ruled out and we are left with the possibilities of B, C or E.

#2 Now if I am to look at the second possibility, \(\frac{1}{x^5} > \frac{1}{x^3}\), and have not "simplified it" for whatever reason (i.e. I didn't know that rule)... I will plug in the same two numbers to see what happens.

For a +2: \(\frac{1}{2^5} > \frac{1}{2^3}\)

=\(\frac{1}{32} > \frac{1}{8}\) *This does not hold true as +3.125% is not > +12.5%

However, now...

For a -2: \(\frac{1}{-2^5} > \frac{1}{-2^3}\)

=\(\frac{1}{-32} > \frac{1}{-8}\) *This does hold true as -3.125% is > -12.5%

Now I am thoroughly confused, because I'm being tricked with the second possibility (#2) with a +# as "not true" and a -# as "true." This is the main reason I think that I am failing at these DS q's because I thought that any possibility that presents both a "Yes" and "No" answer is therefore "IN SUFFICIENT"...

Let me clarify. In the Official Explanation, I used -2 and +2 to plug into S1 because both -2 and +2 satisfy S1. So when I use different numbers that satisfy the condition (here S1) and produce a different answer to the question in the stem (if \(x\) is greater than 1), then I can conclude that a certain Statement is NOT sufficient. However, what you did with S2 is not a good way to determine if the Statement is suff or insuff. -2 satisfies S2, but +2 doesn't and this is why you can't use +2 to prove if S2 is sufficient or not. So, first you have to find the range of \(x\) that satisfies a particular Statement and only after that you can plug in different values of \(x\) from that range. What you did was using the values of \(x\) that worked for S1 but didn't quite work for S2.