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

 It is currently 16 Aug 2018, 10:55

### 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

# New Algebra Set!!!

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 28 Apr 2012
Posts: 298
Location: India
Concentration: Finance, Technology
GMAT 1: 650 Q48 V31
GMAT 2: 770 Q50 V47
WE: Information Technology (Computer Software)

### Show Tags

Updated on: 20 Mar 2013, 08:40
1
5. If x^2 + 2x -15 = -m, where m is an integer from -10 and 10, inclusive, what is the probability that m is greater than zero?

A. 2/7
B. 1/3
C. 7/20
D. 2/5
E. 3/7

$$x^2 + 2x -15 = -m$$
$$=> x^2 + 2x + 1 -16 = -m$$
$$=> (x+1)^2 = 16 -m$$
$$=> 16 -m >=0$$ and a square number

If m varies between [-10,10]

$$=> 16 - 10 <= 16 - m <= 16 - (-10)$$
$$=> 6 <= 16 - m <= 26$$

Between 6 and 16, there are 3 square numbers i.e. 9, 16, 25

For $$16 - m = 9 => m = 7$$
For$$16 - m = 16 => m = 0$$
For$$16 - m = 25 => m = -9$$

So number of possible values of m = 3 i.e. [7, 0, -9]
And for m > 0, number of possible values = 1 i.e only for m=7

So probability = 1/3
Ans = B
_________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well."
― Voltaire

Press Kudos, if I have helped.
Thanks!

Originally posted by ConnectTheDots on 18 Mar 2013, 11:25.
Last edited by ConnectTheDots on 20 Mar 2013, 08:40, edited 1 time in total.
VP
Joined: 24 Jul 2011
Posts: 1459
GMAT 1: 780 Q51 V48
GRE 1: Q800 V740

### Show Tags

18 Mar 2013, 11:55
2
@Bunuel, for Q5, shouldn't the original question also say that we are only looking for solutions where x is an integer?

For this to be satisfied, in x^2 + 2x -15 = -m, 4 - 4(m-15)>0 and 4-4(m-15) must be a perfect square
=> 4 (16-m) must be a perfect square and >0
=> 16-m must be a perfect square and >0
The only values that satisfy this for -10<=m<=10 are m=-9,0,7 of which only 7 is positive
=> probability = 1/3
_________________

GyanOne | Top MBA Rankings and MBA Admissions Blog

Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching

Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738

MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 13:00
1
1. $$x^4=x^3 + 6x^2$$

So $$x^2(x^2-x-6)=0$$
Or $$x=0, -2, 3$$
However -2 cannot be the solution as x is a positive root. So sum of the roots is 3 i.e D
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 13:05
1
2. The equation x^2 + ax - b = 0 has equal roots, and one of the roots of the equation x^2 + ax + 15 = 0 is 3. What is the value of b?

A. -64
B. -16
C. -15
D. -1/16
E. -1/64

Let the roots of $$x^2 + ax +b = 0$$ are h and h

So sum of the roots i.e $$2h = -a$$ and product of the roots i.e $$h^2 = -b$$
Now one root of $$x^2 + ax + 15=0$$ is 5
So the other root must be 5 as the product is 15.
So $$a = - (5+3) = -8$$

So, $$8 = 2h$$
or, $$h=4$$
So $$b = -4^2 = -16$$
B it is
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 13:11
1
3. If a and b are positive numbers, such that a^2 + b^2 = m and a^2 - b^2 = n, then ab in terms of m and n equals to:

2a^2 = m+n
2b^2 = m-n

4a^2b^2= m^2 - n^2
or 2ab= √(m^2-n^2)
C
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 13:17
2
4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?

A. -39
B. -9
C. 0
D. 9
E. 39

differentiating with respect to x and equating with 0

$$-6x + 12 = 0, OR x=2$$

differentiating with respect to y and equating with 0

$$-4y - 12 =0, OR y=-3$$

So max value of the expression $$-3x^2 + 12x -2y^2 - 12y - 39$$ is at (2, -3)
i.e -12+24-18+36-39 = -9
B
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 13:33
2
5. If x^2 + 2x -15 = -m, where m is an integer from -10 and 10, inclusive, what is the probability that m is greater than zero?

A. 2/7
B. 1/3
C. 7/20
D. 2/5
E. 3/7

x^2 + 2x - 15 =-m or x^2 + 2x + (m-15) = 0
or (x+5)(x-3) = -m
Now since m is an integer so x must also be an integer.
x=1 m= 12 NO
x=2 m= 7 YES
x=3 m= 0 YES
x=4 m= -9 YES
x=5 m= -20 NO
No further values will satisfy!
So m has 3 values so far!
x= 0 m= 15 NO
x= -1 m= 16 NO
x= -2 m= 15 NO
x= -3 m= 12 NO
x= -4 m= 7 YES
x= -5 m= 0 YES
x= -6 m= -9 YES
x= -7 m= -20 NO

SO m can take 3 values 7, 0 and -9 and only ONE of them is greater than 0
SO the probability is 1/3
B
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 13:42
7. If x^4 = 29x^2 - 100, then which of the following is NOT a product of three possible values of x?

I. -50
II. 25
III. 50

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

x^4 + 0x^3 - 29x^2 + 0x + 100 =0
this equation reduces to
$$(x^2-25)(x^2-4)=0$$

or x = 5, -5, 2, -2
So 50 or -50 cannot be the product of 2 of the above roots.
E
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 13:47
8. If m is a negative integer and m^3 + 380 = 381m , then what is the value of m?

A. -21
B. -20
C. -19
D. -1
E. None of the above

-1 clearly does not work
lets see for m = -19 and I have a good feeling because 380 is divisible by 19

$$-19*-19*-19 + 380$$

$$= -19(19^2-20)$$

$$=-19(361-20)$$
Not working

Lets try -20

$$-20^3 + 380$$

$$= -20(400-19)$$

$$=-20(381)$$

$$=381m$$

BINGO
B
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 13:50
9. (√5 - √7)^2 = 5+7-2√35
√35 is almost equal to √36 i.e 6

So the expression is approximated as 5+7-12 = 0
A
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 13:53
10. If f(x) = 2x - 1 and g(x) = x^2, then what is the product of all values of n for which f(n^2)=g(n+12) ?

A. -145
B. -24
C. 24
D. 145
E. None of the above

$$f(n^2) = 2n^2 - 1$$

$$g(n+12) = (n+12)^2$$

$$= n^2 + 24n +144$$

So if they are equal then

$$2n^2 - 1 = n^2 +24n+144$$

OR $$n^2-24n-145=0$$

Product of the roots of this is -145

A it is
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

18 Mar 2013, 16:15
6. If mn does not equal to zero, and m^2n^2 + mn = 12, then m could be:

I. -4/n
II. 2/n
III. 3/n

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

lets say $$mn=a$$

then $$a^2+a=12$$

or $$(a-3)(a+4)=0$$

or $$a=3, -4$$

Or $$mn=3, -4$$ or $$m=3/n$$ or $$-4/n$$

E
_________________
VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1018
Location: India
GMAT 1: 410 Q35 V11
GMAT 2: 530 Q44 V20
GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)

### Show Tags

19 Mar 2013, 01:15
souvik101990 wrote:
4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?

A. -39
B. -9
C. 0
D. 9
E. 39

differentiating with respect to x and equating with 0

$$-6x + 12 = 0, OR x=2$$

differentiating with respect to y and equating with 0

$$-4y - 12 =0, OR y=-3$$

So max value of the expression $$-3x^2 + 12x -2y^2 - 12y - 39$$ is at (2, -3)
i.e -12+24-18+36-39 = -9
B

Hi Souvik
I liked ur approach cn u pls expand on it...i mean is there any hard and fast rule that u r following.....completely missed this question....

Archit
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 615

### Show Tags

19 Mar 2013, 01:48
1
4. -3x^2 + 12x -2y^2 - 12y - 39 = -3x^2+12x-12-2y^2-12y-18-39+18+12 = -3(x-2)^2-2(y+3)^2 -9. Now a negative sign is attached to both the squares. Thus the maximum value is at x=2 and y=-3 and equals -9.

B.
_________________
MBA Section Director
Joined: 19 Mar 2012
Posts: 5124
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

19 Mar 2013, 01:57
Archit143 wrote:
souvik101990 wrote:
4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?

A. -39
B. -9
C. 0
D. 9
E. 39

differentiating with respect to x and equating with 0

$$-6x + 12 = 0, OR x=2$$

differentiating with respect to y and equating with 0

$$-4y - 12 =0, OR y=-3$$

So max value of the expression $$-3x^2 + 12x -2y^2 - 12y - 39$$ is at (2, -3)
i.e -12+24-18+36-39 = -9
B

Hi Souvik
I liked ur approach cn u pls expand on it...i mean is there any hard and fast rule that u r following.....completely missed this question....

Archit

I used calculus maxima minima. However you are free to use the "completing the square" method where you are left with 2 perfect squares and a -9 as the guy above me did.
_________________
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 615

### Show Tags

Updated on: 19 Mar 2013, 05:40
1
8.$$m^3 + 380 = 381m$$
or m^3-m = 380(m-1)
or m(m-1)(m+1) = 380(m-1)
AS m !=1, thus,

m(m+1) = 380 = 19*20

Thus, as m<0, only m=-20 satisfies.

B.
_________________

Originally posted by mau5 on 19 Mar 2013, 02:17.
Last edited by mau5 on 19 Mar 2013, 05:40, edited 1 time in total.
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 615

### Show Tags

19 Mar 2013, 02:25
1
9. x = $$(\sqrt{5}-\sqrt{7})^2$$

As from the given options we know that x>0. Thus, I can safely write that

$$\sqrt{x} = \sqrt{7}-\sqrt{5}$$

or $$\sqrt{7} = \sqrt{x} +\sqrt{5}$$

x can not be 1 or more than 1 as then $$\sqrt{7}$$>3 which is not true. Thus, on approximation, x=0.

A.
_________________
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 615

### Show Tags

19 Mar 2013, 02:30
1
7. $$x^4 = 29x^2 - 100$$

or $$x^4 - 29x^2+100 = 0$$

Let $$x^2 = t$$, thus

$$t^2-29+100 = 0$$

or (t-25)(t-4) = 0

t = 25 --> x = 5/-5

or

t = 4 --> x = 2/-2

Out of any three possible values of x, only 25 is not possible.

B.
_________________
Director
Joined: 25 Apr 2012
Posts: 701
Location: India
GPA: 3.21

### Show Tags

Updated on: 19 Mar 2013, 17:10
1
Hello Bunuel,

Nice set of Questions

Here goes my solutions

10. If f(x) = 2x - 1 and g(x) = x^2, then what is the product of all values of n for which f(n^2)=g(n+12) ?[/b]

Sol: From the given function we have

f(n^2)=g(n+12) ----> 2n^2-1= (n+12)^2
Solving this eqn we get
n^2 - 24n - 145= 0-----> n^2 -29n+ 5n - 145 =0
Possible values of n are 29 and -5.
Product will be -145 and hence ans is option A
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Originally posted by WoundedTiger on 19 Mar 2013, 07:05.
Last edited by WoundedTiger on 19 Mar 2013, 17:10, edited 1 time in total.
Intern
Joined: 29 Dec 2012
Posts: 3
WE: Information Technology (Computer Software)

### Show Tags

19 Mar 2013, 10:57
GyanOne wrote:
4. -3x^2 + 12x -2y^2 - 12y - 39 = 3(-x^2 + 4x - 13) + 2(-y^2-6y)

Now -x^2 +4x - 13 has its maximum value at x = -4/-2 = 2 and -y^2 - 6y has its maximum value at y=6/-2 = -3
Therefore max value of the expression
= 3(-4 + 8 -13) + 2 (-9+18) = -27 + 18 = -9

Should be B

Hi GyanOne,

Can you please explain how you found x/y which gives max value for the 2 cases here? Is there a formula?
Re: New Algebra Set!!! &nbs [#permalink] 19 Mar 2013, 10:57

Go to page   Previous    1   2   3   4   5   6   7    Next  [ 139 posts ]

Display posts from previous: Sort by

# Events & Promotions

 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®.