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New Algebra Set!!!

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The next set of medium/hard PS algebra questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:

A. -2
B. 0
C. 1
D. 3
E. 5

Solution: new-algebra-set-149349-60.html#p1200948


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

Solution: new-algebra-set-149349-60.html#p1200950


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:

A. \(\frac{\sqrt{m-n}}{2}\)
B. \(\frac{\sqrt{mn}}{2}\)
C. \(\frac{\sqrt{m^2-n^2}}{2}\)
D. \(\frac{\sqrt{n^2-m^2}}{2}\)
E. \(\frac{\sqrt{m^2+n^2}}{2}\)

Solution: new-algebra-set-149349-60.html#p1200956


4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?

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

Solution: new-algebra-set-149349-60.html#p1200962


5. If x^2 + 2x -15 = -m, where x 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

Solution: new-algebra-set-149349-60.html#p1200970


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

Solution: new-algebra-set-149349-60.html#p1200973


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

Solution: new-algebra-set-149349-60.html#p1200975


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

Solution: new-algebra-set-149349-60.html#p1200980


9. If \(x=(\sqrt{5}-\sqrt{7})^2\), then the best approximation of x is:

A. 0
B. 1
C. 2
D. 3
E. 4

Solution: new-algebra-set-149349-60.html#p1200982


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

Solution: new-algebra-set-149349-80.html#p1200987


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New post 13 Mar 2017, 09:36
flexman wrote:
Bunuel wrote:
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

Since one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3, then substituting we'll get: \(3^2+3a+15=0\). Solving for \(a\) gives \(a=-8\).

Substitute \(a=-8\) in the first equation: \(x^2-8x-b=0\).

Now, we know that it has equal roots thus its discriminant must equal to zero: \(d=(-8)^2+4b=0\). Solving for \(b\) gives \(b=-16\).

Answer: B.


Hi Bunuel, how come I cannot/should not solve this problem by treating them as if they are two equations and multiple unknowns? As in, x^2 + ax + 15 = 0 ; x^2 +ax -b = 0. Subtract the two equations and you get 15+b=0 so b= -15. Thanks!


Because those are not simultaneous equations, those two are not given to be a system of equations.
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Collection of Questions:
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Re: New Algebra Set!!! [#permalink]

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New post 18 Mar 2017, 11:19
Bunuel 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

\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=-3(x^2-4x+4)-2(y^2+6y+9)-9=\)
\(=-3(x-2)^2-2(y+3)^2-9\).

So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).

Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).

Answer: B.



Hello Bunnel,

The solution is excellent. But how do we think in 2 min time, we have to split 39 as -12-18-9. It feels challenging to figure that out. Any pointers how to think ?
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New Algebra Set!!! [#permalink]

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New post 18 Mar 2017, 11:38
Bunuel wrote:
Sorry, there was a typo in the stem .

5. If x^2 + 2x -15 = -m, where x 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

Re-arrange the given equation: \(-x^2-2x+15=m\).

Given that \(x\) is an integer from -10 and 10, inclusive (21 values) we need to find the probability that \(-x^2-2x+15\) is greater than zero, so the probability that \(-x^2-2x+15>0\).

Factorize: \((x+5)(3-x)>0\). This equation holds true for \(-5<x<3\).

Since x is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

So, the probability is 7/21=1/3.

Answer: B.


Hello Bunnel,

I did the same way as above except when
-x^2-2x+15 > 0,
so took -ve sign common as
x^2+2x-15 < 0
(x+5)(x-3) < 0
x+5 < 0 or x -3 < 0
x < -5 or x < 3. So for me possible values are -6,-7-8,-9,-10. Means 5/21.
Any idea where am i wrong ?
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Re: New Algebra Set!!! [#permalink]

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New post 19 Mar 2017, 04:16
coolkl wrote:
Bunuel wrote:
Sorry, there was a typo in the stem .

5. If x^2 + 2x -15 = -m, where x 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

Re-arrange the given equation: \(-x^2-2x+15=m\).

Given that \(x\) is an integer from -10 and 10, inclusive (21 values) we need to find the probability that \(-x^2-2x+15\) is greater than zero, so the probability that \(-x^2-2x+15>0\).

Factorize: \((x+5)(3-x)>0\). This equation holds true for \(-5<x<3\).

Since x is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

So, the probability is 7/21=1/3.

Answer: B.


Hello Bunnel,

I did the same way as above except when
-x^2-2x+15 > 0,
so took -ve sign common as
x^2+2x-15 < 0
(x+5)(x-3) < 0
x+5 < 0 or x -3 < 0
x < -5 or x < 3. So for me possible values are -6,-7-8,-9,-10. Means 5/21.
Any idea where am i wrong ?


When you say that -6,-7-8,-9,-10 satisfy the inequality, did you test any of these values? Ho did you solve (x+5)(x-3) < 0 to get x < -5 or x < 3? What does x < -5 OR x < 3 even mean? It does not make sense... (x+5)(x-3) < 0 holds true for -5 < x < 3.

Check the links below for more:
Inequality tips

Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach

Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems
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New post 22 Mar 2017, 04:42
Hi Bunuel, thank you so much for posting these. Could you please share some Rate and Percent problem solving questions?

Also, if I have exhausted the question bank from GMAT prep (most questions I see are repeating now) and if I have finished Manhattan's 6 tests and the free tests posted on the forum, what resources should I use? I have used the problems listed in your signature, too. My exam is in a week from today. Thank you!

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mitko20m wrote:
Hi Bunuel, thank you so much for posting these. Could you please share some Rate and Percent problem solving questions?

Also, if I have exhausted the question bank from GMAT prep (most questions I see are repeating now) and if I have finished Manhattan's 6 tests and the free tests posted on the forum, what resources should I use? I have used the problems listed in your signature, too. My exam is in a week from today. Thank you!


Check our questions bank for all types of problems: https://gmatclub.com/forum/viewforumtags.php

Hope it helps.
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Collection of Questions:
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Re: New Algebra Set!!! [#permalink]

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New post 23 Mar 2017, 00:40
Bunuel wrote:
Sorry, there was a typo in the stem .

5. If x^2 + 2x -15 = -m, where x 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

Re-arrange the given equation: \(-x^2-2x+15=m\).

Given that \(x\) is an integer from -10 and 10, inclusive (21 values) we need to find the probability that \(-x^2-2x+15\) is greater than zero, so the probability that \(-x^2-2x+15>0\).

Factorize: \((x+5)(3-x)>0\). This equation holds true for \(-5<x<3\).

Since x is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

So, the probability is 7/21=1/3.

Answer: B.



Hi, I just solved this equation like this-

x^2+2x-15 = -m

so for m>0; -m<0

x^2+2x-15<0
(x+5)(x-3)<0

x<-5 x<3 .. So I get only 5 solutions.. (-6, -7,-8,-9, -10)

Can you please let me know whats wrong in this.. Thanks in Advance
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New post 23 Mar 2017, 01:30
RMD007 wrote:
Bunuel wrote:
Sorry, there was a typo in the stem .

5. If x^2 + 2x -15 = -m, where x 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

Re-arrange the given equation: \(-x^2-2x+15=m\).

Given that \(x\) is an integer from -10 and 10, inclusive (21 values) we need to find the probability that \(-x^2-2x+15\) is greater than zero, so the probability that \(-x^2-2x+15>0\).

Factorize: \((x+5)(3-x)>0\). This equation holds true for \(-5<x<3\).

Since x is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

So, the probability is 7/21=1/3.

Answer: B.



Hi, I just solved this equation like this-

x^2+2x-15 = -m

so for m>0; -m<0

x^2+2x-15<0
(x+5)(x-3)<0

x<-5 x<3 .. So I get only 5 solutions.. (-6, -7,-8,-9, -10)

Can you please let me know whats wrong in this.. Thanks in Advance


(x+5)(x-3)<0 holds true for -5<x<3.

This doubt is addressed two posts above yours: https://gmatclub.com/forum/new-algebra- ... l#p1822701
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Re: New Algebra Set!!! [#permalink]

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New post 23 Mar 2017, 11:59
Bunuel wrote:
flexman wrote:
Bunuel wrote:
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

Since one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3, then substituting we'll get: \(3^2+3a+15=0\). Solving for \(a\) gives \(a=-8\).

Substitute \(a=-8\) in the first equation: \(x^2-8x-b=0\).

Now, we know that it has equal roots thus its discriminant must equal to zero: \(d=(-8)^2+4b=0\). Solving for \(b\) gives \(b=-16\).

Answer: B.


Hi Bunuel, how come I cannot/should not solve this problem by treating them as if they are two equations and multiple unknowns? As in, x^2 + ax + 15 = 0 ; x^2 +ax -b = 0. Subtract the two equations and you get 15+b=0 so b= -15. Thanks!


Because those are not simultaneous equations, those two are not given to be a system of equations.


Hi Bunuel, sorry if this is a silly question, but how do we know that these are not simultaneous equations? What are the indicators here so I don't treat this as simultaneous equations next time I see a problem such as this. I guess I've been blindly treating problems such as these as a 'multiple equation multiple variable' form of equations and solving them the way you'd solve 2 or 3 equations/2 or 3 unknowns type of problems. Thanks!

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Re: New Algebra Set!!! [#permalink]

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New post 14 Jun 2017, 21:09
Bunuel wrote:
SOLUTIONs:

1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:

A. -2
B. 0
C. 1
D. 3
E. 5

Take the given expression to the 4th power: \(x^4=x^3+6x^2\);

Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);

Factorize: \(x^2(x-3)(x+2)=0\);

So, the roots are \(x=0\), \(x=3\) and \(x=-2\). But \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\).

The sum of all possible solutions for x is 0+3=3.

Answer: D.



You have been clearing doubts , one more need your help. WHy X cannot take negative value?
What I understand that If X=-3 then (-3)^4=(3)^4 .

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Re: New Algebra Set!!! [#permalink]

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New post 14 Jun 2017, 22:19
gmat4varun wrote:
Bunuel wrote:
SOLUTIONs:

1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:

A. -2
B. 0
C. 1
D. 3
E. 5

Take the given expression to the 4th power: \(x^4=x^3+6x^2\);

Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);

Factorize: \(x^2(x-3)(x+2)=0\);

So, the roots are \(x=0\), \(x=3\) and \(x=-2\). But \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\).

The sum of all possible solutions for x is 0+3=3.

Answer: D.



You have been clearing doubts , one more need your help. WHy X cannot take negative value?
What I understand that If X=-3 then (-3)^4=(3)^4 .


Please read the whole thread.

\(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)).

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. Even roots have only a positive value on the GMAT.

In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5.
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Re: New Algebra Set!!! [#permalink]

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New post 14 Jun 2017, 23:48
One of the best and trickiest collection of Algebra questions...
Kudos to you Bunuel



But they are only 10.
Do you plan on maybe adding a few more sets like these?

If they pre-exist then please do share the link...




Thanks again :)

Regards
Stone Cold.
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Re: New Algebra Set!!! [#permalink]

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New post 15 Jun 2017, 00:34
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stonecold wrote:
One of the best and trickiest collection of Algebra questions...
Kudos to you Bunuel



But they are only 10.
Do you plan on maybe adding a few more sets like these?

If they pre-exist then please do share the link...




Thanks again :)

Regards
Stone Cold.


Thank you, Stone Cold.

Those are questions written by me and specifically designed for those who aim at Q49-51. Other similar collections are in my signature.

To answer your query, yes, I'm currently working on new set, which will be published shortly. Meanwhile, please check two new and tricky questions from that set:
https://gmatclub.com/forum/if-x-and-y-a ... 42602.html
https://gmatclub.com/forum/if-x-and-y-a ... 42599.html
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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?
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New Algebra Set!!! [#permalink]

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New post 18 Jun 2017, 09:37
Bunuel wrote:
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

Re-arrange: \((mn)^2 + mn - 12=0\).

Factorize for mn: \((mn+4)(mn-3)=0\). Thus \(mn=-4\) or \(mn=3\).

So, we have that \(m=-\frac{4}{n}\) or \(m=\frac{3}{n}\).

Answer: E.


Never mind. I figured out the issue. Please ignore this comment

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Re: New Algebra Set!!! [#permalink]

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New post 14 Sep 2017, 21:01
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

m²n²+mn=12
mn(mn+1)=12
Let us use the answer choices
m=\(\frac{3}{n}\)

\(\frac{3}{n}*n\)(\(\frac{3}{n}*n\)+1)=12

3(3+1)=12
3*4=12
12=12 possible

If m=\(\frac{-4}{n}\)

\(\frac{-4}{n}\)*n(-4/n*n+1)=12

-4(-3)=12

12=12-possible

If m=\(\frac{2}{n}\)

\(\frac{2}{n}*n\)(\(\frac{2}{n}*n\)+1)=12
2(2+1)=12
6 is not equal to 12 not possible
Answer 1 and 3.

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New Algebra Set!!! [#permalink]

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New post 14 Sep 2017, 21:19
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.

Plugin answer choices start with easy nos.
Let m=-1

(-1)³+380=381*-1

-1+380≠-381.

Let m=-20

(-20)³+380=381*(-20)

-7620=-7620
Answer is B

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New Algebra Set!!!   [#permalink] 14 Sep 2017, 21:19

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