GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 16 Jul 2018, 19:22

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47019
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 07 Dec 2017, 21:22
1
1
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

59% (01:05) correct 41% (01:02) wrong based on 54 sessions

HideShow timer Statistics

If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 2933
Location: India
GPA: 3.12
Premium Member
Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 07 Dec 2017, 23:50
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


Since \(a^3 = 36\), a is positive.
The value of a is in the range 3 < a < 4 because \(a^3 = 36\), which is between \(27(3^3) and 64(4^3)\)

Since \(b^4 = 81\), b could be both positive or negative. The value of b is 3.

Since \(c^5 = 125\), c is positive.
The value of c is in the range 2 < c < 3 because \(c^5 = 125\), which is between \(32(2^5) and 243(3^5)\)

Hence, a>b>c(Option A) must be true.
_________________

You've got what it takes, but it will take everything you've got

SC Moderator
avatar
D
Joined: 22 May 2016
Posts: 1825
Premium Member CAT Tests
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post Updated on: 09 Dec 2017, 17:17
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b

This question is a lot harder than it looks, IMO. I'd bet many will get lucky when choosing A.

If \(n^{odd} = positive\) number, \(n\) is positive

Approximate value of \(a\)
\(a^3 = 36\), and \(36\) is positive
\(a\) must be positive
The closest integer for \(a\) is \(3\), because
\(3^3 = 27\), so \(a\) must be a little bigger than \(3\)
\(a = 3+\)

Value of \(b\)
\(b^4 = 81\)
\(b = 3\) or \(b = -3\)

Approximate value of \(c\)
\(c^5 = 125\)
\(125\) is positive. \(c\) must be positive
The two integers closest to \(c\) are \(2\) and \(3\):

\(2^5=32\) and \(3^5=243\), where

\((2^5=32) < (c^5=125) < (3^5=243)\)

Compare \(c\) to positive \(b\)

\(c\) cannot be \(3\)
\(c^5=125 \neq{243}\) (not even close)

\(c \neq{3} \neq{+b}\)
\(c = 2+\)

Is \(b\) positive or negative?
If \(b = +3\)
\(a > b > c\) (Answer A)

If \(b = -3\)
\(a > c > b\) (Answer B)

Is \(b\) positive or negative? A quandary, maybe. The principal root rule is no help.*

Negative numbers raised to even powers: brackets
I think the key is the "bracket" rule for negative numbers raised to even powers. Think of "negative four, squared."

IF \(-4\) has brackets, then:
\((-4)^2 = 16\)
That means: \((-4)(-4) = 16\)

BUT IF \(-4\) has no brackets, then:
\(-4^2 = -16\)
That means: \(- [(4)(4)] = -16\)

Expanding: \(-4^2 = ?\)
By PEMDAS order of operations, we take exponents first.
So first we raise \(4\) to the second power: \(4^2 = 16\)
Then we take the negative of the result: \(-16\)
\(-4^2 = -16\)

\(b^4 = 81\)
IF \(b\) were negative, in order to equal positive 81, \(b\) would REQUIRE brackets. Per above:
\((-3)^4 = 81\) BUT
\(-3^4 = -81\)

There are no brackets around \(b\).
So \(b\) in \(b^4 = 81\) MUST be positive

\(b = 3\), and
\(a = 3+\)
\(c = 2+\)

\(a > b > c\)

ANSWER A

Please correct me if I am mistaken. I see no other way to choose A over B. The absolute value of \(b = 3\), but I do not see how absolute value tells us the sign of \(b\).

*That is
IF we are GIVEN
\(\sqrt[4]{b^4}\) , the answer \(b\), is positive. The radical sign itself tells us to take the principal fourth root, the positive one.

But we are given an exponent, not a radical sign, such that \(b^4 = 81\) has two solutions: \(3\) and \(-3\). In other words, given \(b^4 = 81\), there is no radical sign to indicate "Take the principal (positive) fourth root." Without such direction, we still do not know if \(b\) is positive or negative.

_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"


Originally posted by generis on 09 Dec 2017, 16:33.
Last edited by generis on 09 Dec 2017, 17:17, edited 4 times in total.
Intern
Intern
avatar
B
Joined: 11 Nov 2017
Posts: 12
Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 09 Dec 2017, 16:39
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


a^3=36
Hence, a must be positive since when x is raised to an odd power, it retains it's original sign. Also, 3<a<4.

b^4=81
Hence, b could be positive or negative. b could be -3 or +3. So, b<a.

c^5 = 125
Similar to the case of a above, c has to be positive. Also, 2<c<3. Hence, c<b<a.

Answer is A.
_________________

Cheers.

Never gonna give up!!!!

SC Moderator
avatar
D
Joined: 22 May 2016
Posts: 1825
Premium Member CAT Tests
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 09 Dec 2017, 16:52
Madnov2017 wrote:
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


a^3=36
Hence, a must be positive since when x is raised to an odd power, it retains it's original sign. Also, 3<a<4.

b^4=81
Hence, b could be positive or negative. b could be -3 or +3. So, b<a.

c^5 = 125
Similar to the case of a above, c has to be positive. Also, 2<c<3. Hence, c<b<a.

Answer is A.

Madnov2017 and pushpitkc :

If a and c must be positive, and b can be positive or negative, how did you decide that b must be positive?

No disrespect, I promise: Neither of you mentions that b must be positive for Answer A to be correct. Neither do you explain why b is positive. It seems as if you kinda skipped over the issue?

Yes, no matter what, a > b.
a = 3+, and b = 3 OR -3
a > b

The question seems to me to be:
Is b > c, or is c > b?
c = 2+ (and cannot = 3)
b = 3 OR -3
If b = -3, c > b, and Answer B is correct.
If b = +3, b > c, and Answer A is correct.
How to choose the sign of b seems to me to be the only way to know which is greater, b or c

Am I missing something?
_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"

1 KUDOS received
BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 2933
Location: India
GPA: 3.12
Premium Member
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 09 Dec 2017, 23:42
1
genxer123 wrote:
Madnov2017 wrote:
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


a^3=36
Hence, a must be positive since when x is raised to an odd power, it retains it's original sign. Also, 3<a<4.

b^4=81
Hence, b could be positive or negative. b could be -3 or +3. So, b<a.

c^5 = 125
Similar to the case of a above, c has to be positive. Also, 2<c<3. Hence, c<b<a.

Answer is A.

Madnov2017 and pushpitkc :

If a and c must be positive, and b can be positive or negative, how did you decide that b must be positive?

No disrespect, I promise: Neither of you mentions that b must be positive for Answer A to be correct. Neither do you explain why b is positive. It seems as if you kinda skipped over the issue?

Yes, no matter what, a > b.
a = 3+, and b = 3 OR -3
a > b

The question seems to me to be:
Is b > c, or is c > b?
c = 2+ (and cannot = 3)
b = 3 OR -3
If b = -3, c > b, and Answer B is correct.
If b = +3, b > c, and Answer A is correct.
How to choose the sign of b seems to me to be the only way to know which is greater, b or c

Am I missing something?


Hi genxer123

No, I believe that you maybe correct.
Because in my solution, I have assumed that b is positive.
However, if b is negative then Option B(a>c>b) could also be true.

Hello Bunuel
Can you please help us with a solution for this problem.

Thanks in advance.
_________________

You've got what it takes, but it will take everything you've got

1 KUDOS received
Intern
Intern
avatar
B
Joined: 11 Nov 2017
Posts: 12
Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 10 Dec 2017, 14:46
1
genxer123 wrote:
Madnov2017 wrote:
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


a^3=36
Hence, a must be positive since when x is raised to an odd power, it retains it's original sign. Also, 3<a<4.

b^4=81
Hence, b could be positive or negative. b could be -3 or +3. So, b<a.

c^5 = 125
Similar to the case of a above, c has to be positive. Also, 2<c<3. Hence, c<b<a.

Answer is A.

Madnov2017 and pushpitkc :

If a and c must be positive, and b can be positive or negative, how did you decide that b must be positive?

No disrespect, I promise: Neither of you mentions that b must be positive for Answer A to be correct. Neither do you explain why b is positive. It seems as if you kinda skipped over the issue?

Yes, no matter what, a > b.
a = 3+, and b = 3 OR -3
a > b

The question seems to me to be:
Is b > c, or is c > b?
c = 2+ (and cannot = 3)
b = 3 OR -3
If b = -3, c > b, and Answer B is correct.
If b = +3, b > c, and Answer A is correct.
How to choose the sign of b seems to me to be the only way to know which is greater, b or c

Am I missing something?



Apologies.
It seems like I did skip over this issue. I can't think of a way to find out the sign of b.
I see that genxer123 makes use of some parenthesis rule to figure out the sign of b. However, I am not sure such a rule exists and don't think it works here. b can take a value of -3 or 3 and for b^4=81, b has to be within parenthesis, right? If we take ''-'' outside of the parenthesis, then the original equation b^4=81 fails, isn't it?

I hope I am not missing something very obvious. Don't want to sound too dumb. :D
_________________

Cheers.

Never gonna give up!!!!

1 KUDOS received
SC Moderator
avatar
D
Joined: 22 May 2016
Posts: 1825
Premium Member CAT Tests
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 10 Dec 2017, 20:36
1
Madnov2017 wrote:
Apologies.
It seems like I did skip over this issue. I can't think of a way to find out the sign of b.
I see that genxer123 makes use of some parenthesis rule to figure out the sign of b. However, I am not sure such a rule exists and don't think it works here. b can take a value of -3 or 3 and for b^4=81, b has to be within parenthesis, right? If we take ''-'' outside of the parenthesis, then the original equation b^4=81 fails, isn't it?

I hope I am not missing something very obvious. Don't want to sound too dumb. :D

Dumb you most certainly neither appear to be, sound, nor are. :-)
Quote:
I am not sure such a rule exists

It does. It is not obvious. That is exactly why Bunuel has it skulking around here.
Perhaps it is not a rule. Perhaps it is a "worldwide convention"? An "arithmetic principle"?
I don't know what to call it. I just know it exists.

It can seem counter-intuitive, but:

\(-3^4 = -81\)
\((-3)^4 = 81\)

If there is a negative number raised to an even power:
AND no parentheses? The result is negative.
Quote:
[If such a rule exists, I] don't think it works here. b can take a value of -3 or 3 and for b^4=81, b has to be within parenthesis, right?

No. For negative \(b^4\) to equal positive \(81\), the negative \(b^4\) must be in parentheses.
Positive \(b^4\) does NOT have to be in parentheses.
That is precisely why the rule DOES work.
Only positive \(b^4\) with no parentheses can = positive 81. So \(b\) must be positive.

THUS
\(3^4 = 81\)
\((3)^4 = 81\) (allowed, but not conventional)
\(-3^4 = - 81\)
\((-3)^4 = 81\)

If we had seen parentheses, \((b)^4 = 81\), we would have been stuck without a way to discern b's sign.
Both positive and negative 3 CAN go in brackets to get a positive result. Negative 3 must go in brackets to get a positive result.
Both are possible, one is required, and there would have been no way to tell the difference.

But instead we saw \(b^4=81\) Substitute \(3\) and \(-3\) for \(b\):
\(3^4 = 81\)
\(-3^4 = -81\)

So we DO have a way to discern \(b\)'s sign:
No brackets? Positive result? \(b\) must be positive
Quote:
I can't think of a way to find out the sign of b.

I can. ;) Use the "negative number raised to even power without brackets" rule: no brackets and a positive result? The number b is positive.

And that is the only reason I can find for "b must be positive."

Thanks, pushpitkc and Madnov2017 , for responding. And kudos. I appreciate it.

This question's layers are brilliant. Wish I could give its writer extra kudos.

Below are three sources who address the issue.

purplemath guy says here:

Quote:
Simplify \((–3)^2\)The square means "multiplied against itself, with two copies of the base". This means that I'll have two "minus" signs, which I can cancel:

\((–3)^2 = (–3)(–3) = (+3)(+3) = 9\)
Pay careful attention and note the difference between the above exercise and the following:

Simplify \(–3^2\)
\(–3^2 = –(3)(3) = –1(3)(3) = (–1)(9) = –9\)
In the second exercise, the square (the "to the power 2") was only on the 3; it was not on the minus sign. Those parentheses in the first exercise make all the difference in the world! Be careful with them. . .


Quote:
\(−3^2\) does not mean "the square of negative three." The exponent takes priority over the negative: it means "the negative of \(3^2\)."

See here

And finally, from the Monterrey Institute:
Quote:
That leaves us with the term \(-3^4.\) This example is a little trickier because there is a negative sign in there. One of the rules of exponential notation is that the exponent relates only to the value immediately to its left. So, -3^4 does not mean -3 • -3 • -3 • -3. It means “the opposite of 3^4,” or — (3 • 3 • 3 • 3). If we wanted the base to be -3, we’d have to use parentheses in the notation: \((-3)^4.\) Why so picky? Well, do the math:

-3^4 = – (3 • 3 • 3 • 3) = -81
(-3)^4 = -3 • -3 • -3 • -3 = 81

That’s an important difference.

My emphases. That material is HERE
_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"

Senior Manager
Senior Manager
avatar
G
Joined: 31 Jul 2017
Posts: 373
Location: Malaysia
WE: Consulting (Energy and Utilities)
Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 11 Dec 2017, 19:31
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


I can think of the following logics to establish b sign. Please let me know if I am wrong in the following logic.

1. I think as a,b,c are close to an AP - Hence a >b>c.
2. b > c.. Square on both sides - Valid.
c > b.. Square on both sides - Not valid

Hi Bunuel and Forum Experts,

Can you please help me with the below doubt.

1. In an inequality can we apply mod on both sides. For Ex -

Suppose k > m .. can we write |k| > |m|.. Please advise.
_________________

If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47019
Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 11 Dec 2017, 20:24
rahul16singh28 wrote:
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


I can think of the following logics to establish b sign. Please let me know if I am wrong in the following logic.

1. I think as a,b,c are close to an AP - Hence a >b>c.
2. b > c.. Square on both sides - Valid.
c > b.. Square on both sides - Not valid

Hi Bunuel and Forum Experts,

Can you please help me with the below doubt.

1. In an inequality can we apply mod on both sides. For Ex -

Suppose k > m .. can we write |k| > |m|.. Please advise.


-2 > -3. Is |-2| > |-3| ?
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Senior Manager
Senior Manager
avatar
G
Joined: 31 Jul 2017
Posts: 373
Location: Malaysia
WE: Consulting (Energy and Utilities)
Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 11 Dec 2017, 20:32
Bunuel wrote:
rahul16singh28 wrote:
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


I can think of the following logics to establish b sign. Please let me know if I am wrong in the following logic.

1. I think as a,b,c are close to an AP - Hence a >b>c.
2. b > c.. Square on both sides - Valid.
c > b.. Square on both sides - Not valid

Hi Bunuel and Forum Experts,

Can you please help me with the below doubt.

1. In an inequality can we apply mod on both sides. For Ex -

Suppose k > m .. can we write |k| > |m|.. Please advise.


-2 > -3. Is |-2| > |-3| ?


Thanks for clarifying the doubt Bunuel
_________________

If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!

Senior Manager
Senior Manager
avatar
G
Joined: 17 Mar 2014
Posts: 295
GMAT ToolKit User Reviews Badge CAT Tests
Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 12 Dec 2017, 15:21
pushpitkc wrote:
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


Since \(a^3 = 36\), a is positive.
The value of a is in the range 3 < a < 4 because \(a^3 = 36\), which is between \(27(3^3) and 64(4^3)\)

Since \(b^4 = 81\), b could be both positive or negative. The value of b is 3.

Since \(c^5 = 125\), c is positive.
The value of c is in the range 2 < c < 3 because \(c^5 = 125\), which is between \(32(2^5) and 243(3^5)\)

Hence, a>b>c(Option A) must be true.


How could you say value of B is 3. It could be negative. In that case Option B will be valid.
SC Moderator
avatar
D
Joined: 22 May 2016
Posts: 1825
Premium Member CAT Tests
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 12 Dec 2017, 21:41
ammuseeru wrote:
pushpitkc wrote:
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


Since \(a^3 = 36\), a is positive.
The value of a is in the range 3 < a < 4 because \(a^3 = 36\), which is between \(27(3^3) and 64(4^3)\)

Since \(b^4 = 81\), b could be both positive or negative. The value of b is 3.

Since \(c^5 = 125\), c is positive.
The value of c is in the range 2 < c < 3 because \(c^5 = 125\), which is between \(32(2^5) and 243(3^5)\)

Hence, a>b>c(Option A) must be true.


How could you say value of B is 3. It could be negative. In that case Option B will be valid.

ammuseeru , maybe you didn't read the whole topic? We have been discussing exactly that. :-)
_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"

Senior Manager
Senior Manager
avatar
G
Joined: 31 Jul 2017
Posts: 373
Location: Malaysia
WE: Consulting (Energy and Utilities)
Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

Show Tags

New post 15 Dec 2017, 07:05
Bunuel wrote:
If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?

A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b


Hi Bunuel,

Can you please provide the OE for this question. Marked this day in my calendar for your solution :-)

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________

If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!

Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be   [#permalink] 15 Dec 2017, 07:05
Display posts from previous: Sort by

If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, KarishmaB, niks18, Bunuel

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.