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

It is currently 15 Oct 2018, 14:26

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

a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Manager
Manager
User avatar
B
Joined: 18 Dec 2012
Posts: 97
Location: India
Concentration: General Management, Strategy
GMAT 1: 660 Q49 V32
GMAT 2: 530 Q37 V25
GPA: 3.32
WE: Manufacturing and Production (Manufacturing)
GMAT ToolKit User
a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 04 Sep 2013, 00:36
3
17
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

53% (02:44) correct 47% (02:50) wrong based on 158 sessions

HideShow timer Statistics

\(a^4 + \frac{1}{a^4} = 119\) and \(a > 1\), then the value of \(a^3 - \frac{1}{a^3}\) is :

A. 27
B. 36
C. 45
D. 54
E. 63
Most Helpful Community Reply
Verbal Forum Moderator
User avatar
Joined: 10 Oct 2012
Posts: 613
Premium Member
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 04 Sep 2013, 01:48
6
1
Qoofi wrote:
\(a^4 + 1/ (a^4)\)= 119 and a > 1, then the value of \(a^3 - 1/(a^3)\)is :

A. 27
B. 36
C. 45
D. 54
E. 63


Please use the Math function to wrap the mathematical expressions.

Given that \(a^4 + \frac{1} {(a^4)}\) = 119 , adding 2 on both sides, we get : \([a^2 + \frac{1}{a^2}]^2\) \(= 121 \to a^2 + \frac{1}{a^2} = 11\)

Again, by subtracting 2 on both sides, we have \((a-\frac{1}{a})^2 = 9 \to (a-\frac{1}{a}) = 3\)

Now, \(a^3 - \frac{1}{a^3} = (a-\frac{1}{a})(a^2+\frac{1}{a^2}+1) = 12*3 = 36\)

B.
_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions

General Discussion
Intern
Intern
avatar
Joined: 31 Jan 2013
Posts: 17
Schools: ISB '15
WE: Consulting (Energy and Utilities)
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 04 Sep 2013, 01:42
Consider f(a) = a^4+(1/a)^4 = 119 and a>1

if a = 3, f(a) ~= 81
if a = 4, f(a) ~= 256

By this we can definitely say 3<a<4 and the value must definitely be closer to 3 (considering f(a) is closer to f(3)).

Substituting the value in
g(a) = a^3-(1/a)^3 ,
we could defintely say g(a) is slightly greater than g(3).
And, the most likely answer could be
36
.

/SW
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8373
Location: Pune, India
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 14 Feb 2016, 22:53
2
Qoofi wrote:
a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(a^3) is :

A. 27
B. 36
C. 45
D. 54
E. 63


If you feel that algebra is too tedious, you can approximate the value of 'a' to get to the right answer. But you must find both the minimum and maximum value to be sure of the right answer.

You know that a > 1.
\(3^4 = 81\)
\(4^4 = 256\)

To get 119, a would be more than 3 but less than 4. It will be closer to 3. Say, it will be something like 3.2 or 3.3 etc.

\(3.5^2 = 12.25\) (Note that you can easily find the square of numbers ending in 5. You write 25 at the end and multiply the first digit with the next integer and write in front so \(35^2 = .....25\) and then \(3*4 = 12\) so 1225)

But 12*12 = 144 so a is obviously less than 3.5.

\(a^3\) will be more than \(3^3 = 27\) but less than \(3.5^3\) which is approximately \(12*3.5 = 42\).

Only option 36 lies in between these values.

Answer (B)

Note that 1/a^4 and 1/a^3 have no role to play in this method because they are very small (just a few decimal points) compared to the rest of the numbers.
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Intern
avatar
Joined: 01 May 2015
Posts: 41
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 14 Feb 2016, 23:17
1
a^3 - 1/a^3 = (a - 1/a)(a^2 + 1/a^2 + 1)

So, we have to find the values of (a - 1/a) and (a^2 + 1/a^2)

a^4 + 1/a^4 = 119 will help us find this information.

a^4 + 1/a^4 = (a^2 + 1/a^2)^2 - 2

So, (a^2 + 1/a^2)^2 - 2 = 119
-> (a^2 + 1/a^2)^2 = 121
-> (a^2 + 1/a^2) = 11

Again, (a - 1/a)^2 = a^2 + 1/a^2 -2
-> (a - 1/a)^2 = 11 -2
-> (a - 1/a)^2 = 9
-> (a - 1/a) = 3

Hence, a^3 - 1/a^3 = (a - 1/a)(a^2 + 1/a^2 + 1) = 3*(11 + 1) = 36
Board of Directors
User avatar
P
Joined: 17 Jul 2014
Posts: 2654
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Premium Member Reviews Badge
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 09 Mar 2016, 20:44
spent more than 40 minutes to figure out how to solve it..
unsuccessfully...
mau5 says that he added 2 on both sides, but doesn't show that he added on the left side ...
pretty confusing one..
can anyone explain through the algebra way???
Bunuel?
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8373
Location: Pune, India
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 09 Mar 2016, 22:57
mau5 wrote:
Qoofi wrote:
\(a^4 + 1/ (a^4)\)= 119 and a > 1, then the value of \(a^3 - 1/(a^3)\)is :

A. 27
B. 36
C. 45
D. 54
E. 63


Please use the Math function to wrap the mathematical expressions.

Given that \(a^4 + \frac{1} {(a^4)}\) = 119 , adding 2 on both sides, we get : \([a^2 + \frac{1}{a^2}]^2\) \(= 121 \to a^2 + \frac{1}{a^2} = 11\)

Again, by subtracting 2 on both sides, we have \((a-\frac{1}{a})^2 = 9 \to (a-\frac{1}{a}) = 3\)

Now, \(a^3 - \frac{1}{a^3} = (a-\frac{1}{a})(a^2+\frac{1}{a^2}+1) = 12*3 = 36\)

B.


Here are more details to mau5's solution.

We need to find \(a^3 - \frac{1}{a^3}\)

\(a^3 - \frac{1}{a^3} = (a-\frac{1}{a})(a^2+\frac{1}{a^2}+1)\)

So we need the values of \(a-\frac{1}{a}\) and \((a^2+\frac{1}{a^2})\)

Given:
\(a^4 + \frac{1} {(a^4)} = 119\)

Add 2 on both sides to get:

\(a^4 + \frac{1} {(a^4)} + 2 = 119 + 2\)

which is equivalent to:
\(a^4 + \frac{1} {(a^4)} + 2*a^2*\frac{1}{a^2} = 121\)

Use \(x^2 + y^2 + 2xy = (x + y)^2\) to get

\([a^2 + \frac{1}{a^2}]^2 = 11^2\)

So, \([a^2 + \frac{1}{a^2}] = 11\)

Now, you have the value of one expression but one is still required.

Subtract 2 from both sides:

\([a^2 + \frac{1}{a^2} - 2] = 11 - 2\)

which is equivalent to:
\([a^2 + \frac{1}{a^2} - 2*a*\frac{1}{a}] = 9\)

Again, use \(x^2 + y^2 + 2xy = (x + y)^2\) to get

\((a-\frac{1}{a})^2 = 3^2\)

\((a-\frac{1}{a}) = 3\)

Now we have both the required quantities.

\(a^3 - \frac{1}{a^3} = (a-\frac{1}{a})(a^2+\frac{1}{a^2}+1) = 3*(11 + 1) = 36\)
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Intern
avatar
B
Joined: 15 Jun 2013
Posts: 47
Schools: Ivey '19 (I)
GMAT 1: 690 Q49 V35
GPA: 3.82
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 22 Jun 2017, 07:47
mau5 wrote:
Qoofi wrote:
\(a^4 + 1/ (a^4)\)= 119 and a > 1, then the value of \(a^3 - 1/(a^3)\)is :

A. 27
B. 36
C. 45
D. 54
E. 63


Please use the Math function to wrap the mathematical expressions.

Given that \(a^4 + \frac{1} {(a^4)}\) = 119 , adding 2 on both sides, we get : \([a^2 + \frac{1}{a^2}]^2\) \(= 121 \to a^2 + \frac{1}{a^2} = 11\)

Again, by subtracting 2 on both sides, we have \((a-\frac{1}{a})^2 = 9 \to (a-\frac{1}{a}) = 3\)

Now, \(a^3 - \frac{1}{a^3} = (a-\frac{1}{a})(a^2+\frac{1}{a^2}+1) = 12*3 = 36\)

B.


My solution is not as smart as the one above. But I suppose to use an approximation. if a>1 and a^4+1/(a^4)=119, so we know that 1/(a^4) is <1, so a^4~119, a^2~11.
Again, 1/a^3 is <1, so we need to calculate a^3=11*(11^0,5). What is the 11^0,5? - it is more than 3, but below 4. therefore the answer will be between 33 and 44, closer to 33. Thus, we have 36 (B)
Intern
Intern
avatar
Joined: 02 Oct 2017
Posts: 3
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 03 Mar 2018, 04:58
why a-1/a is not equal to = -3 ??? please explain.
Intern
Intern
avatar
B
Joined: 30 May 2018
Posts: 9
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

Show Tags

New post 01 Aug 2018, 02:07
sr313 wrote:
why a-1/a is not equal to = -3 ??? please explain.

Value of a cannot be -3 because a should be greater than 1(a>1) as mentioned in question

Posted from my mobile device
GMAT Club Bot
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/( &nbs [#permalink] 01 Aug 2018, 02:07
Display posts from previous: Sort by

a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(

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


Copyright

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®.