Join us for MBA Spotlight – The Top 20 MBA Fair      Schedule of Events | Register

 It is currently 06 Jun 2020, 01:03 ### 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

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

Author Message
TAGS:

### Hide Tags

Manager  B
Joined: 18 Dec 2012
Posts: 91
Location: India
Concentration: General Management, Strategy
GMAT 1: 660 Q49 V32
GMAT 2: 530 Q37 V25
GPA: 3.32
WE: Manufacturing and Production (Manufacturing)
a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

### Show Tags

3
1
20 00:00

Difficulty:   75% (hard)

Question Stats: 58% (02:47) correct 42% (02:50) wrong based on 152 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
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 563
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

### Show Tags

9
1
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.
_________________
##### General Discussion
Intern  Joined: 31 Jan 2013
Posts: 15
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

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 V
Joined: 16 Oct 2010
Posts: 10506
Location: Pune, India
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

### Show Tags

3
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.

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

Intern  Joined: 01 May 2015
Posts: 35
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

### Show Tags

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 P
Joined: 17 Jul 2014
Posts: 2429
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

### Show Tags

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 V
Joined: 16 Oct 2010
Posts: 10506
Location: Pune, India
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

### Show Tags

1
1
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

Intern  B
Joined: 15 Jun 2013
Posts: 44
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

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  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

why a-1/a is not equal to = -3 ??? please explain.
Intern  B
Joined: 30 May 2018
Posts: 47
Location: India
GMAT 1: 680 Q49 V33 GMAT 2: 710 Q50 V35 GPA: 4
WE: Engineering (Energy and Utilities)
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

### Show Tags

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
Non-Human User Joined: 09 Sep 2013
Posts: 15097
Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: a^4 + 1/ (a^4) = 119 and a > 1, then the value of a^3 - 1/(   [#permalink] 26 Aug 2019, 00:48

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