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Re: If 0 < x < 1, which of the following inequalities must be [#permalink]

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08 Oct 2008, 08:36

My two cents:

I beeter if you divided by x^2: x^2+x^3<x+1 -> always II better if you divided by x^2: x^2-x^3<1-x^2 always III dividing by x^2: x^2-x^3<1-x^2 if x=0.1 thus 0.01-0.001<1-0.01 always if x=0.9 thus 0.81-0.729<1-0.9 // 0.081<1 always

E

OA?

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Re: If 0 < x < 1, which of the following inequalities must be [#permalink]

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08 Oct 2008, 09:01

I also believe the answer is E since even the third equation, x^4 - x^5 < x^2 - x^3 = x^4 + x^3 < x^2 + x^5 =x^2 + x < 1+x^3 which is true since the value on the right hand side will be greater than 1.

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]

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28 May 2015, 22:43

Bunuel wrote:

scofield1521 wrote:

People, please throw some more light!!

If 0 < x < 1, which of the following inequalities must be true ?

I. \(x^5 < x^3\) II. \(x^4 + x^5 < x^3 + x^2\) III. \(x^4 - x^5 < x^2 - x^3\)

A. None B. I only C. II only D. I and II only E. I, II and III

If 0 < x < 1, then x > x^2 > x^3 > x^4 > x^5 ...

I. \(x^5 < x^3\). True.

II. \(x^4 + x^5 < x^3 + x^2\). Each term on the left hand side is less than each term on the right hand side, thus LHS < RHS.

III. \(x^4 - x^5 < x^2 - x^3\) --> \(x^4(1 - x) < x^2(1 - x)\). Since 0 < x < 1, then 1 - x > 0, so we can reduce by it: \(x^4 < x^2\). True.

Answer: E.

Hope it's clear.

Bunuel

Can we do this question algebraically without substituting fractions? By trying to prove that each of the statements given indeed have x values within the range of 0 and 1 on the number line?

If 0 < x < 1, which of the following inequalities must be true ?

I. \(x^5 < x^3\) II. \(x^4 + x^5 < x^3 + x^2\) III. \(x^4 - x^5 < x^2 - x^3\)

A. None B. I only C. II only D. I and II only E. I, II and III

If 0 < x < 1, then x > x^2 > x^3 > x^4 > x^5 ...

I. \(x^5 < x^3\). True.

II. \(x^4 + x^5 < x^3 + x^2\). Each term on the left hand side is less than each term on the right hand side, thus LHS < RHS.

III. \(x^4 - x^5 < x^2 - x^3\) --> \(x^4(1 - x) < x^2(1 - x)\). Since 0 < x < 1, then 1 - x > 0, so we can reduce by it: \(x^4 < x^2\). True.

Answer: E.

Hope it's clear.

Bunuel

Can we do this question algebraically without substituting fractions? By trying to prove that each of the statements given indeed have x values within the range of 0 and 1 on the number line?

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]

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28 May 2015, 23:52

Bunuel

Can we do this question algebraically without substituting fractions? By trying to prove that each of the statements given indeed have x values within the range of 0 and 1 on the number line?[/quote]

Using wavy line method we know that for \(0 < x < 1\) the inequality holds true

Thus all the three inequalities are true for the range \(0 < x < 1\)

Hope it's clear

Regards Harsh[/quote]

Thanks Harsh for your reply. Can you please throw some more light on the wave method? Of what I knw, signs usually alternate but what I see in your workings, signs don't always alternate. Can you explain it? Second, why are we ignoring the negative portions say in statement 1 that makes it less than -1. It shows x values dont only pertain to 0 and 1 region. So how can stmt 1 be sufficient?

Can we do this question algebraically without substituting fractions? By trying to prove that each of the statements given indeed have x values within the range of 0 and 1 on the number line?

Using wavy line method we know that for \(0 < x < 1\) the inequality holds true

Thus all the three inequalities are true for the range \(0 < x < 1\)

Hope it's clear

Regards Harsh

Quote:

Thanks Harsh for your reply. Can you please throw some more light on the wave method? Of what I knw, signs usually alternate but what I see in your workings, signs don't always alternate. Can you explain it? Second, why are we ignoring the negative portions say in statement 1 that makes it less than -1. It shows x values dont only pertain to 0 and 1 region. So how can stmt 1 be sufficient?

You are right when you say signs usually alternate around zero points of the inequality but they only alternate when the expression is sign dependent i.e. for odd powers. For example sign of \(x^3\) is dependent on the sign of \(x\) but sign of \(x^2\) is independent of sign of \(x\). Hence for expressions, signs of which are independent of the signs of their base variable, the wave would not alternate but rather bounce back to the same region it is currently in.

I would also suggest you to go through this post on Wavy Line method which will help you understand it better.

For your second query, in st-I the range of \(x\) for which the inequality holds is \(0 < x < 1\) or \(x < -1\) as seen from the wavy line diagram. However as the question asks us if the inequality is true for the range \(0 < x < 1\), we are not concerned about the other possible values of x for which the inequality is true. Once the inequality satisfies the range \(0 < x < 1\), we have our answer.

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]

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29 May 2015, 22:58

EgmatQuantExpert wrote:

Quote:

Bunuel

Can we do this question algebraically without substituting fractions? By trying to prove that each of the statements given indeed have x values within the range of 0 and 1 on the number line?

Using wavy line method we know that for \(0 < x < 1\) the inequality holds true

Thus all the three inequalities are true for the range \(0 < x < 1\)

Hope it's clear

Regards Harsh

Quote:

Thanks Harsh for your reply. Can you please throw some more light on the wave method? Of what I knw, signs usually alternate but what I see in your workings, signs don't always alternate. Can you explain it? Second, why are we ignoring the negative portions say in statement 1 that makes it less than -1. It shows x values dont only pertain to 0 and 1 region. So how can stmt 1 be sufficient?

You are right when you say signs usually alternate around zero points of the inequality but they only alternate when the expression is sign dependent i.e. for odd powers. For example sign of \(x^3\) is dependent on the sign of \(x\) but sign of \(x^2\) is independent of sign of \(x\). Hence for expressions, signs of which are independent of the signs of their base variable, the wave would not alternate but rather bounce back to the same region it is currently in.

I would also suggest you to go through this post on Wavy Line method which will help you understand it better.

For your second query, in st-I the range of \(x\) for which the inequality holds is \(0 < x < 1\) or \(x < -1\) as seen from the wavy line diagram. However as the question asks us if the inequality is true for the range \(0 < x < 1\), we are not concerned about the other possible values of x for which the inequality is true. Once the inequality satisfies the range \(0 < x < 1\), we have our answer.

Hope it's clear

Regards Harsh

Ok Harsh, but by the same logic haven't been able to crack the question below? Can you help?

If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true? I. x3 < x II. x2 < |x| III. x4 – x5 > x3 – x2 (A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III

Can we do this question algebraically without substituting fractions? By trying to prove that each of the statements given indeed have x values within the range of 0 and 1 on the number line?

Using wavy line method we know that for \(0 < x < 1\) the inequality holds true

Thus all the three inequalities are true for the range \(0 < x < 1\)

Hope it's clear

Regards Harsh

Quote:

Thanks Harsh for your reply. Can you please throw some more light on the wave method? Of what I knw, signs usually alternate but what I see in your workings, signs don't always alternate. Can you explain it? Second, why are we ignoring the negative portions say in statement 1 that makes it less than -1. It shows x values dont only pertain to 0 and 1 region. So how can stmt 1 be sufficient?

You are right when you say signs usually alternate around zero points of the inequality but they only alternate when the expression is sign dependent i.e. for odd powers. For example sign of \(x^3\) is dependent on the sign of \(x\) but sign of \(x^2\) is independent of sign of \(x\). Hence for expressions, signs of which are independent of the signs of their base variable, the wave would not alternate but rather bounce back to the same region it is currently in.

I would also suggest you to go through this post on Wavy Line method which will help you understand it better.

For your second query, in st-I the range of \(x\) for which the inequality holds is \(0 < x < 1\) or \(x < -1\) as seen from the wavy line diagram. However as the question asks us if the inequality is true for the range \(0 < x < 1\), we are not concerned about the other possible values of x for which the inequality is true. Once the inequality satisfies the range \(0 < x < 1\), we have our answer.

Hope it's clear

Regards Harsh

Ok Harsh, but by the same logic haven't been able to crack the question below? Can you help?

If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true? I. x3 < x II. x2 < |x| III. x4 – x5 > x3 – x2 (A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]

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01 Jun 2015, 08:25

Ok Harsh, but by the same logic haven't been able to crack the question below? Can you help?

If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true? I. x3 < x II. x2 < |x| III. x4 – x5 > x3 – x2 (A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III[/quote]

Please refer to this post where Japinder has presented the algebraic approach using the wavy line method.

Request you to go through it and let us know if you have trouble at any point of the solution

Regards Harsh[/quote]

Hey Harsh

Saw Japinder's solution and posted some concerns. Haven't managed a reply yet. Pls find below my comments.

Japinder two concerns: 1. why have we not flipped the inequality sign for stmt 2 when changing the sign to negative? 2. for stmt 3, we have x values less than 1 till negative infinity. how can this be sufficient as the range we want it to be is from -1 to +1

Ok Harsh, but by the same logic haven't been able to crack the question below? Can you help?

If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true? I. x3 < x II. x2 < |x| III. x4 – x5 > x3 – x2 (A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III

Please refer to this post where Japinder has presented the algebraic approach using the wavy line method.

Request you to go through it and let us know if you have trouble at any point of the solution

Regards Harsh

Hey Harsh

Saw Japinder's solution and posted some concerns. Haven't managed a reply yet. Pls find below my comments.

Japinder two concerns: 1. why have we not flipped the inequality sign for stmt 2 when changing the sign to negative? 2. for stmt 3, we have x values less than 1 till negative infinity. how can this be sufficient as the range we want it to be is from -1 to +1

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]

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24 Aug 2015, 09:23

Hi,

Please help me understand how do we derive that 0<x<1 holds true by looking at the wavy curve?

Is it because the wave crosses over in the said interval (0<x<1) ? If yes, then it doesnt matter if the area is on the positive or the negative region?

Re: If 0 < x < 1, which of the following inequalities must be [#permalink]

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23 Jan 2016, 09:41

amitdgr wrote:

If 0 < x < 1, which of the following inequalities must be true ?

I. \(x^5 < x^3\) II. \(x^4 + x^5 < x^3 + x^2\) III. \(x^4 - x^5 < x^2 - x^3\)

A. None B. I only C. II only D. I and II only E. I, II and III

I've started with I: just pick x=1/2 --> 1/32<1/8. Now look at the answer choices -which would be easier to test. As we already know "I" is correct, now, if we test III and it's correct, we don't need to test II anymore, because the only answer choice that has I and III is E. If you manipulate III as stated in many solutions here, one can see that it must be true --> Answer E (we could save some time by picking III after I)
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