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# New Algebra Set!!!

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18 Mar 2013, 07:56
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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

Kudos points for each correct solution!!!
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10 May 2015, 23:06
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Bunuel wrote:
Please suggest on what category would you like the next set to be. Thank you!

Hey Bunuel, I'd really like to see a Fraction/Decimal/Percent type of set.

Off topic, but just a heads up, under your "RESOURCES", you got Remainders mentioned twice (7 and 10)
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26 May 2015, 11:21
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 $$m=\frac{3}{n}$$.

Hello, how come m^2n^2 will become (mn)^2 ?
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27 May 2015, 03:28
LaxAvenger wrote:
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 $$m=\frac{3}{n}$$.

Hello, how come m^2n^2 will become (mn)^2 ?

$$(mn)^2=(mn)(mn)=m^2*n^2$$.
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27 May 2015, 11:26
1. Double square both side we will get, x^4=x^3+6x^2; rearranging we get x^2(x^2-x-6) =0
Solving above equation, x=0,-2 &3
Putting all these values in original expression we get, root under (expression); and GMAT does not support negative value in side root. So possible value of x=0 &3
Hence sum of all possible solution =3, Answer is D

9. If x=(√5−√7)2, then the best approximation of x is
X= 5+7-2*root 5*root7 =12 -2*root35 approximately =12-2*6 =12-12 =0
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27 May 2015, 13:48
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) ?
Replace n^2 instead of x in f(x) term and (n+12)^2 instead of x in g(x), we will get
2n^2-1=(n+12)^2 = n^2+144+24n, solving equation we get n^2-24n-145 =0
Resolving above equation we get n= -5 &29
Hence product of all values of n =-5*29 = -145; hence answer is A

8. If m is a negative integer and m^3 + 380 = 381m , then what is the value of m?
Re-arranging above equation we get: m^3-381m+380=0 or m(m^2 – 381) +380=0
So any value of m which satisfy “ m(m^2 -381) = -380 “ equation will be correct value of x.
Only m=-20 satisfy the given equation , hence m=-20 , answer is B

7. If x^4 = 29x^2 - 100, then which of the following is NOT a product of three possible values of x?
Assume x^2 =n ; solve equation n^2-29n+100=0 therefore n=5 &25
As n=x^2 = 25 &4 , x=+5,-5,+2 & -2.
So we can have product of possible values as -25, -50,+50 ,…..etc but x can never be +25

6. If mn does not equal to zero, and m^2n^2 + mn = 12, then m could be:
Assume mn =x; so we will have x^2+x-12=0 solving for above equation we get x= -4 & 3
So mn=-4 & 3 therefore m= -4/n & 3/n
Hence values of m is 1 and 3, answer is E

4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?
As x^2 and y^2 terms are in negative so any value (positive or negative) will give negative and minimize the value of entire term. So making x & y zero will give maximum value of the term hence -39 is the maximum value , answer is A

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:
Adding both equation we get 2 a^2 = (m+n) => a= root[ (m+n)/2]
Subtracting both equation we get 2 b ^2 = (m-n) =>b= root[ (m-n)/2]
Therefore a*b = root [(m+n)(m-n)] /2 = ½ * root (m^2 –n^2)

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?
Substitution and finding out value of a from x=3; 9+3a+15=0 => a= -8
As x^2+ax –b =0 has equal roots, so b^2 -4ac =0 therefore a^2+ 4b=0 => a^2= -4b
Substitution value of a in above equation. b= - (8*8)/4 = -16
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27 May 2015, 13:52
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.

Hi Bunuel,

I understand how you got to the three roots: 0, 3, and -2 but I missed out on the 0 root because I factored the expression like this:

x^4 = x^3 + 6x^2
x^4 = x^2(x + 6)
x^2 = x + 6
x^2 - x - 6 = 0 .... leading to only roots 3 and -2

I'm not sure why my way is incorrect since. I don't know if I broke some algebra rule. Could you please clarify my factoring method? Thanks, kindly.
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27 May 2015, 23:13
Bunuel wrote:
LaxAvenger wrote:
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 $$m=\frac{3}{n}$$.

Hello, how come m^2n^2 will become (mn)^2 ?

$$(mn)^2=(mn)(mn)=m^2*n^2$$.

Ohh.. the writing is a little bit misleading. I was thinking of:

m^2n^2 and not m^2 * n^2
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28 May 2015, 04:18
torontoclub15 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.

Hi Bunuel,

I understand how you got to the three roots: 0, 3, and -2 but I missed out on the 0 root because I factored the expression like this:

x^4 = x^3 + 6x^2
x^4 = x^2(x + 6)
x^2 = x + 6
x^2 - x - 6 = 0 .... leading to only roots 3 and -2

I'm not sure why my way is incorrect since. I don't know if I broke some algebra rule. Could you please clarify my factoring method? Thanks, kindly.

If you divide (reduce) x^4 = x^2(x + 6) by x^2, you assume, with no ground for it, that x^2 (x) does not equal to zero thus exclude a possible solution.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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11 Sep 2015, 08:17
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.

Hi Bunuel
Could you please check below is fine? Thanks

can we write ? m>0 -----> -x^2-2x+15 > 0
x^2+2x-15 < 0
(x+5)(x-3) < 0
test some big number like 100 --> (x+5)(x-3) will be positive

--- +ve--- -5 ------ -ve------- 3 -----+ve---------

range of x ==> -5<x<3
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11 Sep 2015, 22:18

Yes, that works. When you have a quadratic inequality and there are two real roots, the number line can be thought of being divided in to three regions. One to the left of x=-5, the region between -5 and 3, and the region to the right of 3. The sign of the expression will change as we go from one region to the next. This means our expression will either follow the path of +, -, + or -, +, - in the three regions. When you chose x=100 as a sample value, we found that the expression (x+5)(x-3) is positive at x=100 and we can conclude that for x>3 it is positive, this means that between -5 and 3 it will be negative, and less -5 it will be positive.

Cheers,
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08 Feb 2016, 10:07
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

Hi Bunuel,

Can you kindly help me to understand why you included the negative values of X in calculating the product of possible values of X. Can you explain that vis-a-vis question 1's solution.
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10 Dec 2016, 14:08
[quote="Bunuel"]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.

Bunuel can u please explain why it isn't C? I mean why can't x be a negative number, when in fact positive or negative x^4 both can be the given expression right? So 3 minus 2 gives us 1
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10 Dec 2016, 15:33
dinobull 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.

Bunuel can u please explain why it isn't C? I mean why can't x be a negative number, when in fact positive or negative x^4 both can be the given expression right? So 3 minus 2 gives us 1

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28 Jan 2017, 04:23
WoundedTiger 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

Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get

3^2 +3a+15=0----> Solving for a we get a=-8.

Putting the value if a in Eqn 1 we get
x^2 -8x-b=0
We know for Eqn ax^2 +bx+c=-0
Sum of roots is given by -b/a
Product of roots is c/a

Therefore from Eqn1 (after subsitution of value of a) we get
Sum of roots = -8 and roots are equal ie. -4 and -4
Product of roots will be 16. Therefore Option B

Hello
How do I know that I have to plug in 3 for X if it says that the root of the equation is 3. I dont understand. Can someone explain? Thank you very much
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28 Jan 2017, 07:52
sonnenblume92 wrote:
WoundedTiger 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

Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get

3^2 +3a+15=0----> Solving for a we get a=-8.

Putting the value if a in Eqn 1 we get
x^2 -8x-b=0
We know for Eqn ax^2 +bx+c=-0
Sum of roots is given by -b/a
Product of roots is c/a

Therefore from Eqn1 (after subsitution of value of a) we get
Sum of roots = -8 and roots are equal ie. -4 and -4
Product of roots will be 16. Therefore Option B

Hello
How do I know that I have to plug in 3 for X if it says that the root of the equation is 3. I dont understand. Can someone explain? Thank you very much

3 is a root of x^2 + ax + 15 = 0 means that when you plug 3 there the equation must hold true.
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29 Jan 2017, 13:20
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 $$0+0-9=-9$$.

Don't understand why to to maximize the value of $$-3(x-2)^2-2(y+3)^2$$ must be Zero ??
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31 Jan 2017, 10:52
coolkl 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 $$0+0-9=-9$$.

Don't understand why to to maximize the value of $$-3(x-2)^2-2(y+3)^2$$ must be Zero ??

The max value of -3(x-2)^2 is 0 and max value of -2(y+3)^2 is 0 too. You see since the square of a number is always non negative, then -2*(y+3)^2 = negative*nonnegative = nonposiitve (so -2(y+3)^2 is negative or zero).
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31 Jan 2017, 20:08
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 $$0+0-9=-9$$.

Hi Bunnel,

As per the explanation the maximum value of the expresssion is zero because the expressions in the brackets is squared (implying it will always be positive ) and the expression is multiplied by a negative number.

But what if these were the expressions:-
$$3(x-2)^2$$ and $$2(y+3)^2$$

How can we find the maximum values of such expressions?

Thanks.
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01 Feb 2017, 08:09
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 $$0+0-9=-9$$.

Hi Bunnel,

As per the explanation the maximum value of the expresssion is zero because the expressions in the brackets is squared (implying it will always be positive ) and the expression is multiplied by a negative number.

But what if these were the expressions:-
$$3(x-2)^2$$ and $$2(y+3)^2$$

How can we find the maximum values of such expressions?

Thanks.

In this case there won't be a maximum value.
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13 Mar 2017, 09:17
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!
Re: New Algebra Set!!!   [#permalink] 13 Mar 2017, 09:17

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# New Algebra Set!!!

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