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

 It is currently 14 Oct 2019, 03:46 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.  New Algebra Set!!!

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

Hide Tags

Intern  B
Joined: 29 Jul 2016
Posts: 5
Re: New Algebra Set!!!  [#permalink]

Show Tags

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 $$b=-16$$.

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!
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
Re: New Algebra Set!!!  [#permalink]

Show Tags

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 $$b=-16$$.

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.
_________________
Intern  S
Joined: 10 Jun 2016
Posts: 45
Schools: IIM-A"19
Re: New Algebra Set!!!  [#permalink]

Show Tags

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.

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 ?
_________________
Thank You Very Much,
CoolKl
Success is the Journey from Knowing to Doing

A Kudo is a gesture, to express the effort helped. Thanks for your Kudos.
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
Re: New Algebra Set!!!  [#permalink]

Show Tags

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.

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

Solving Quadratic Inequalities - Graphic Approach

Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems
_________________
Intern  B
Joined: 07 Sep 2016
Posts: 29
Re: New Algebra Set!!!  [#permalink]

Show Tags

Bunuel wrote:
SOLUTIONs:

1. If $$x=\sqrt{x^3+6x^2}$$, 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.

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 .
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
Re: New Algebra Set!!!  [#permalink]

Show Tags

gmat4varun wrote:
Bunuel wrote:
SOLUTIONs:

1. If $$x=\sqrt{x^3+6x^2}$$, 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.

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 .

$$x$$ 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 $$\sqrt{x}$$, 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.
_________________
Intern  B
Joined: 18 Jan 2017
Posts: 43
Location: India
Concentration: General Management, Entrepreneurship
GPA: 4
Re: New Algebra Set!!!  [#permalink]

Show Tags

Bunuel wrote:
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

Re-arrange and factor for x^2: $$(x^2-25)(x^2-4)=0$$.

So, we have that $$x=5$$, $$x=-5$$, $$x=2$$, or $$x=-2$$.

$$-50=5*(-5)*2$$;
$$50=5*(-5)*(-2)$$.

Only 25 is NOT a product of three possible values of x

How did you get this directly?
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
Re: New Algebra Set!!!  [#permalink]

Show Tags

srishti201996 wrote:
Bunuel wrote:
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

Re-arrange and factor for x^2: $$(x^2-25)(x^2-4)=0$$.

So, we have that $$x=5$$, $$x=-5$$, $$x=2$$, or $$x=-2$$.

$$-50=5*(-5)*2$$;
$$50=5*(-5)*(-2)$$.

Only 25 is NOT a product of three possible values of x

How did you get this directly?

For more:

7. Algebra

Other topics: Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________
Intern  B
Joined: 22 Oct 2017
Posts: 5
Re: New Algebra Set!!!  [#permalink]

Show Tags

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 $$b=-16$$.

Hi Bunuel,

hope you can help:

In the description it says "has equal roots, and one of the roots of the equation is ... 3".

I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" --> therefore i would have never thought about setting the determinant equal to zero.

Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero?

Thank you very much!

Best regards
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
Re: New Algebra Set!!!  [#permalink]

Show Tags

andr3 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 $$b=-16$$.

Hi Bunuel,

hope you can help:

In the description it says "has equal roots, and one of the roots of the equation is ... 3".

I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" --> therefore i would have never thought about setting the determinant equal to zero.

Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero?

Thank you very much!

Best regards

1. x^2 + ax - b = 0 has equal roots

2. One of the roots of the equation x^2 + ax + 15 = 0 is 3.

Those are two different equations.
_________________
Intern  B
Joined: 22 Oct 2017
Posts: 5
Re: New Algebra Set!!!  [#permalink]

Show Tags

Bunuel wrote:
andr3 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 $$b=-16$$.

Hi Bunuel,

hope you can help:

In the description it says "has equal roots, and one of the roots of the equation is ... 3".

I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" --> therefore i would have never thought about setting the determinant equal to zero.

Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero?

Thank you very much!

Best regards

1. x^2 + ax - b = 0 has equal roots

2. One of the roots of the equation x^2 + ax + 15 = 0 is 3.

Those are two different equations.

ok, than it is that language of "has equal roots", which I don't understand. could you clarify what "equal roots" means? is "equal roots" the definition of having only one solution?

thanks!
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
Re: New Algebra Set!!!  [#permalink]

Show Tags

andr3 wrote:

ok, than it is that language of "has equal roots", which I don't understand. could you clarify what "equal roots" means? is "equal roots" the definition of having only one solution?

thanks!

Yes, equal roots means that there are two roots which are equal to each other, so one distinct root.
_________________
Intern  Joined: 09 Feb 2018
Posts: 5
Re: New Algebra Set!!!  [#permalink]

Show Tags

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 $$0+0-9=-9$$.

can someone help me how the max value of the highlighted text is defined ?
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
Re: New Algebra Set!!!  [#permalink]

Show Tags

SaravanaPrabu090492 wrote:
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 $$0+0-9=-9$$.

can someone help me how the max value of the highlighted text is defined ?

$$-3(x-2)^2= (-3)*(square \ of \ a \ number)= (negative)*(non-negative) = (non-positive)$$. So, this term is negative or 0.

The same for $$-2(y+3)^2$$.
_________________
Intern  Joined: 09 Feb 2018
Posts: 5
Re: New Algebra Set!!!  [#permalink]

Show Tags

Bunuel wrote:
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

$$f(x) = 2x - 1$$, hence $$f(n^2)=2n^2-1$$.
$$g(x) = x^2$$, hence $$g(n+12)=(n+12)^2=n^2+24n+144$$
.

Since given that $$f(n^2)=g(n+12)$$, then $$2n^2-1=n^2+24n+144$$. Re-arranging gives $$n^2-24n-145=0$$.

Next, Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$:

$$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$.

Thus according to the above $$n_1*n_2=-145$$.

Having a doubt with the above highlighted text. for the first, shouldn't the whole term [2x-1] be squared like the below one ? seeking solutions.
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
Re: New Algebra Set!!!  [#permalink]

Show Tags

1
SaravanaPrabu090492 wrote:
Bunuel wrote:
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

$$f(x) = 2x - 1$$, hence $$f(n^2)=2n^2-1$$.
$$g(x) = x^2$$, hence $$g(n+12)=(n+12)^2=n^2+24n+144$$
.

Since given that $$f(n^2)=g(n+12)$$, then $$2n^2-1=n^2+24n+144$$. Re-arranging gives $$n^2-24n-145=0$$.

Next, Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$:

$$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$.

Thus according to the above $$n_1*n_2=-145$$.

Having a doubt with the above highlighted text. for the first, shouldn't the whole term [2x-1] be squared like the below one ? seeking solutions.

No.

$$f(x) = 2x - 1$$ means that we have a function which tells us what to do to get the value of the function: take a value for which you want to find the output of the function, multiply it by 2 and subtract 1.

So, to get $$f(n^2)$$ put n^2 instead of x: $$f(n^2)=2n^2-1$$.

Similarly for $$g(x) = x^2$$. To get $$g(n+12)$$ put n + 12 instead of x: $$g(n+12)=(n+12)^2=n^2+24n+144$$.
_________________
Manager  S
Status: The journey is always more beautiful than the destination
Affiliations: Computer Science
Joined: 24 Apr 2017
Posts: 50
Location: India
Concentration: Statistics, Strategy
GMAT 1: 570 Q40 V28 GPA: 3.14
Re: New Algebra Set!!!  [#permalink]

Show Tags

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 $$0+0-9=-9$$.

$$-3x^2 + 12x -2y^2 - 12y - 39$$ -->$$x(12-3x) - y(2y+12)-39$$

now x(12-3x) will be highest when x is 1.

-y(2y+12) will be higest when y = -1.
so the answer becomes 9+10-39 = -19..
where am I going wrong?
_________________
Sky is the limit. 800 is the limit.
Manager  S
Status: The journey is always more beautiful than the destination
Affiliations: Computer Science
Joined: 24 Apr 2017
Posts: 50
Location: India
Concentration: Statistics, Strategy
GMAT 1: 570 Q40 V28 GPA: 3.14
Re: New Algebra Set!!!  [#permalink]

Show Tags

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.

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

unable to understand this line. Can someone please explain?
many thanks
_________________
Sky is the limit. 800 is the limit.
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
Re: New Algebra Set!!!  [#permalink]

Show Tags

Nixondutta 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.

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

unable to understand this line. Can someone please explain?
many thanks

Check this: Solving Quadratic Inequalities: Graphic Approach

Fro more: check

Inequalities

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________
Intern  B
Joined: 27 Jan 2016
Posts: 22
Location: India
Concentration: International Business, Operations
WE: Design (Energy and Utilities)
Re: New Algebra Set!!!  [#permalink]

Show Tags

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 $$0+0-9=-9$$.

Can anybody help me with this solution. Why the maximum value of $$-3(x-2)^2$$ and $$-2(y+3)^2$$ is zero? Re: New Algebra Set!!!   [#permalink] 24 Nov 2018, 13:27

Go to page   Previous    1   2   3   4   5   6   7   8    Next  [ 145 posts ]

Display posts from previous: Sort by

New Algebra Set!!!

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

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