D01-10 : GMAT Club Tests

amitjathar

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Re: D01-10 [#permalink]

04 Dec 2014, 18:48

Hi Bunuel,
I appreciate your answer, but the method looks bit clumsy to me.
I solved it in simpler way, please guide me whether my approach is correct.

1) 1/x > -1. It means 1 > -x
That is x > -1 .
So we are not sure that if the x is greater than 1. So it is insufficient.
2) 1/x^5 > 1/x^3. It means 1/x^2 > 1 .
That is x^2 < 1.
If x^2 is less than 1, that definitely implies x is less than 1. So it is sufficient.

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Re: D01-10 [#permalink]

04 Dec 2014, 22:36

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Is x greater than 1?

(1) 1/x > -1

(2) (1/x)^5 >(1/x)^3

Alternate solution (Substituting numbers)

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amanrsingh

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Re: D01-10 [#permalink]

03 Jul 2019, 12:09

Hi Bunuel, VeritasKarishma or anyone: I tried and tried and tried but I am not able to comprehend how is B Sufficient. I have tried to solve it via Wavy Curve Method, can you please help me understand what I am doing wrong

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KarishmaB

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Re: D01-10 [#permalink]

04 Jul 2019, 22:11

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amanrsingh wrote:

Hi Bunuel, VeritasKarishma or anyone: I tried and tried and tried but I am not able to comprehend how is B Sufficient. I have tried to solve it via Wavy Curve Method, can you please help me understand what I am doing wrong

Your stmnt 2 analysis is not correct.

\(1/x^5 > 1/x^3\)

\(1/x^5 - 1/x^3 > 0\)

\((1 - x^2)/x^5 > 0\)

\((x^2 - 1)/x^5 < 0\) (Multiply both sides by -1)

\((x + 1)(x - 1)/x^5 < 0\) (Note that to use the wavy line method as discussed, we must have factors in the form (ax + b) or (ax - b), not (b - ax))

Transition points, -1, 0, 1

The expression is negative when 0 < x< 1 or x < -1

So in any case, x is less than 1.

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Re: D01-10 [#permalink]

04 Jan 2021, 06:06

Hi Bunuel,

In this question, we were not given the following info:

x is not equal to 0.

We solved the question without this. Please let me know if this was to inferred from the 2 statements given in the question. I solved the question with the approach same as yours but I had to assume that x is not equal to 0 here.

Does it make any difference whether that was given or not, as I feel this is quite an important piece of information?

Thanks.

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Bunuel

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Re: D01-10 [#permalink]

04 Jan 2021, 06:10

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theeliteguy wrote:

Hi Bunuel,

In this question, we were not given the following info:

x is not equal to 0.

We solved the question without this. Please let me know if this was to inferred from the 2 statements given in the question. I solved the question with the approach same as yours but I had to assume that x is not equal to 0 here.

Does it make any difference whether that was given or not, as I feel this is quite an important piece of information?

Thanks.

From each statement we could deduce that x cannot be 0 because if x were 0, the statements would not be correct. For example, 1/0 > -1 would not be correct because 1/0 is undefined and cannot be greater than -1.

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Re: D01-10 [#permalink]

06 Dec 2021, 23:27

Bunuel wrote:

Is \(x\) greater than 1?

(1) \(\frac{1}{x} \gt -1\)

(2) \(\frac{1}{x^5} \gt \frac{1}{x^3}\)

GMATBusters GMATinsight is 1/a < 1/b we can invert the two fractions and say a > b right? If that is true then for (1) why can't we say 1/x > -1 so x < 1/-1 i.e. x < -1 ?

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Re: D01-10 [#permalink]

07 Dec 2021, 01:11

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Hoozan wrote:

Bunuel wrote:

Is \(x\) greater than 1?

(1) \(\frac{1}{x} \gt -1\)

(2) \(\frac{1}{x^5} \gt \frac{1}{x^3}\)

GMATBusters GMATinsight is 1/a < 1/b we can invert the two fractions and say a > b right? If that is true then for (1) why can't we say 1/x > -1 so x < 1/-1 i.e. x < -1 ?

Hoozan

Rules of Cross multiplication in case of an Inequality
1) Cross multiplication of positive values doesn't affect the Inequality Sign
2) If a negative value shifts from one side to the other side then the inequality Sign FLips

Therefore,
1/a < 1/b will be a >b only if a and b are both positive (No change in sign) or both Negative (sign changes twice and remains the same)

Alternatively, You could understand it this way
If a*b is Positive then you can multiply a*b both sides of inequation without changing the inequality sign

That is why
1/x > -1 will become x < 1/-1 by multiplying -x both sides only when -x is Positive i.e. when x is Negative

I hope it clarifies your doubt.

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Re: D01-10 [#permalink]

02 Oct 2022, 00:22

Bunuel wrote:

amitjathar wrote:

Hi Bunuel,
I appreciate your answer, but the method looks bit clumsy to me.
I solved it in simpler way, please guide me whether my approach is correct.

1) 1/x > -1. It means 1 > -x
That is x > -1 .
So we are not sure that if the x is greater than 1. So it is insufficient.
2) 1/x^5 > 1/x^3. It means 1/x^2 > 1 .
That is x^2 < 1.
If x^2 is less than 1, that definitely implies x is less than 1. So it is sufficient.

No, that's not correct.

You cannot multiply 1/x > -1 by x because you don't know the sign of x. If x is positive then yes, 1 > -x but if x is negative, then when multiplying by it you should flip the sign and you get 1 < -x.

Similarly you cannot reduce 1/x^5 > 1/x^3 by 1/x^3, because you don't know its sign.

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.

Other tips on Inequalities are here: https://gmatclub.com/forum/inequalities- ... 75001.html

Hi bunuel,
Can we not simply flip the inequality sign by taking reciprocal on both sides since both sides have odd powers (ie both shall have same signs) ?
so we'll get x^5 < x^3
Is this logic correct?

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Bunuel

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Re: D01-10 [#permalink]

02 Oct 2022, 00:49

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SDW2 wrote:

Bunuel wrote:

amitjathar wrote:

Hi Bunuel,
I appreciate your answer, but the method looks bit clumsy to me.
I solved it in simpler way, please guide me whether my approach is correct.

1) 1/x > -1. It means 1 > -x
That is x > -1 .
So we are not sure that if the x is greater than 1. So it is insufficient.
2) 1/x^5 > 1/x^3. It means 1/x^2 > 1 .
That is x^2 < 1.
If x^2 is less than 1, that definitely implies x is less than 1. So it is sufficient.

No, that's not correct.

You cannot multiply 1/x > -1 by x because you don't know the sign of x. If x is positive then yes, 1 > -x but if x is negative, then when multiplying by it you should flip the sign and you get 1 < -x.

Similarly you cannot reduce 1/x^5 > 1/x^3 by 1/x^3, because you don't know its sign.

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.

Other tips on Inequalities are here: https://gmatclub.com/forum/inequalities- ... 75001.html

Hi bunuel,
Can we not simply flip the inequality sign by taking reciprocal on both sides since both sides have odd powers (ie both shall have same signs) ?
so we'll get x^5 < x^3
Is this logic correct?

You are basically multiplying 1/x^5 > 1/x^3 by x^3 and then by x^5. If x > 0, then you'd keep the sign both times and get x^3 > x^5. If < 0, then you'd flip the sign twice and still get x^3 > x^5. So, yes 1/x^5 > 1/x^3 is equivalent to x^3 > x^5: both are true when x < -1 and 0 < x< 1.

You could also multiply 1/x^5 > 1/x^3 by x^6, which would be positive, and get x > x^3, which is also equivalent to 1/x^5 > 1/x^3 and true when x < -1 and 0 < x < 1.

Hope it helps.

S

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Re: D01-10 [#permalink]

17 Dec 2022, 05:38

Bunuel wrote:

Official Solution:

Is \(x\) greater than 1?

(1) \(\frac{1}{x} \gt - 1\).

Multiply \(\frac{1}{x} \gt - 1\) by \(x^2\) (we can safely do that because a nonzero number in even power is positive) to get \(x > -x^2\);

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

\(x(1 + x) > 0\).

Two cases:

a. Both multiples are positive, (\(x > 0\) and \(x + 1 > 0\)), so \(x > 0\);

b. Both multiples are negative, (\(x < 0\) and \(x + 1 < 0\)), so \(x < -1\);

So, \(\frac{1}{x} \gt - 1\) is true when \(x < -1\) or when \(x > 0\). Thus \(x\) may or may not be greater than 1. Not sufficient.

(2) \(\frac{1}{x^5} > \frac{1}{x^3}\).

Multiply \(\frac{1}{x^5} > \frac{1}{x^3}\) by \(x^6\) (we can safely do that because a nonzero number in even power is positive) to get \(x > x^3\)

\(x - x^3 > 0\)

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

Two cases:

a. Both multiples are positive, (\(x > 0\) and \(1 - x^2 > 0\)), so \(0 < x < 1\). (Here is why: \(1 - x^2 > 0\) is the same as \(1 > x^2\), which gives \(-1 < x < 1\), which together with \(x > 0\) gives \(0 < x < 1\));

b. Both multiples are negative, (\(x < 0\) and \(1 - x^2 < 0\)), so \(x < -1\) (Here is why: \(1 - x^2 < 0\) is the same as \(1 < x^2\), which gives \(x < -1\) or \(x > 1\), which together with \(x < 0\) gives \(x < -1\)).

So, \(\frac{1}{x^5} > \frac{1}{x^3}\) is true when \(0 < x < 1\) or when \(x < -1\).

ANY \(x\) from these ranges will be less than 1. So, the answer to our original question (Is \(x\) greater than 1?), is NO. Sufficient.

Answer: B

Hi Bunuel,
we multiplied the satement 2 by x^6, but how do we know that x not zero?

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Re: D01-10 [#permalink]

17 Dec 2022, 05:51

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Stanindaw wrote:

Bunuel wrote:

Official Solution:

Is \(x\) greater than 1?

(1) \(\frac{1}{x} \gt - 1\).

Multiply \(\frac{1}{x} \gt - 1\) by \(x^2\) (we can safely do that because a nonzero number in even power is positive) to get \(x > -x^2\);

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

\(x(1 + x) > 0\).

Two cases:

a. Both multiples are positive, (\(x > 0\) and \(x + 1 > 0\)), so \(x > 0\);

b. Both multiples are negative, (\(x < 0\) and \(x + 1 < 0\)), so \(x < -1\);

So, \(\frac{1}{x} \gt - 1\) is true when \(x < -1\) or when \(x > 0\). Thus \(x\) may or may not be greater than 1. Not sufficient.

(2) \(\frac{1}{x^5} > \frac{1}{x^3}\).

Multiply \(\frac{1}{x^5} > \frac{1}{x^3}\) by \(x^6\) (we can safely do that because a nonzero number in even power is positive) to get \(x > x^3\)

\(x - x^3 > 0\)

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

Two cases:

a. Both multiples are positive, (\(x > 0\) and \(1 - x^2 > 0\)), so \(0 < x < 1\). (Here is why: \(1 - x^2 > 0\) is the same as \(1 > x^2\), which gives \(-1 < x < 1\), which together with \(x > 0\) gives \(0 < x < 1\));

b. Both multiples are negative, (\(x < 0\) and \(1 - x^2 < 0\)), so \(x < -1\) (Here is why: \(1 - x^2 < 0\) is the same as \(1 < x^2\), which gives \(x < -1\) or \(x > 1\), which together with \(x < 0\) gives \(x < -1\)).

So, \(\frac{1}{x^5} > \frac{1}{x^3}\) is true when \(0 < x < 1\) or when \(x < -1\).

ANY \(x\) from these ranges will be less than 1. So, the answer to our original question (Is \(x\) greater than 1?), is NO. Sufficient.

Answer: B

Hi Bunuel,
we multiplied the satement 2 by x^6, but how do we know that x not zero?

We are given that \(\frac{1}{x^5} > \frac{1}{x^3}\). If x were 0, this statement would not be true because division by 0 is not allowed.

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Re: D01-10 [#permalink]

01 Jun 2023, 00:11

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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

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Re D01-10 [#permalink]

12 Aug 2023, 08:35

I think this is a high-quality question and I agree with explanation.

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Re: D01-10 [#permalink]

05 Sep 2023, 12:36

Quote:

Bunuel

are there any questions like this that we can solve? I think the trick of multiplying by an even power of x is really useful.

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Bunuel

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Re: D01-10 [#permalink]

06 Sep 2023, 01:50

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aasdf1234 wrote:

Quote:

Bunuel

are there any questions like this that we can solve? I think the trick of multiplying by an even power of x is really useful.

You can find a bunch of questions where you can apply the same technique in our DS Inequality database.

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Re: D01-10 [#permalink]

08 Sep 2023, 11:08

I took such an easy route. St. 1 gives us nothing to answer x>1?
But St.2 very clear... if x>1 then x^5 should be > than x^3... and with high denominator the reciprocal of x^5 should be smaller, but that is not the case. Therefore it is suffiicient to say that whatever the case maybe, x is definitely NOT > than 1.

gmatclubot

Re: D01-10 [#permalink]

08 Sep 2023, 11:08