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New Algebra Set!!! [#permalink]
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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#p12009482. 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#p12009503. 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#p12009564. 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#p12009625. 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#p12009706. 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#p12009737. 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#p12009758. 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#p12009809. 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#p120098210. 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#p1200987Kudos points for each correct solution!!!
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Re: New Algebra Set!!! [#permalink]
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25 Mar 2013, 12:31
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. Hi Bunuel, Query here. First of all we don't necessarily have to solve for all the three roots. Since, it is asking for a composite value i.e. the sum, we can directly obtain it. We can rearrange to get:- \(x^4-x^3-6x^2=0\) Over here, by the system of poly equations Sum of roots = - (\(coeff ofx^3/coeff of x^4\)) Similarly, Sum of roots taken two at a time = (\(coeff ofx^2/coeff of x^4\)) Product of roots = "Zero" as there is no constant term. Also, -2 satisfies the equation \(x^4-x^3-6x^2=0\) So, shouldn't the answer be -1, with roots as 0,-2,3?
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 02:07
Sinner1706 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. Hi Bunuel, Query here. First of all we don't necessarily have to solve for all the three roots. Since, it is asking for a composite value i.e. the sum, we can directly obtain it. We can rearrange to get:- \(x^4-x^3-6x^2=0\) Over here, by the system of poly equations Sum of roots = - (\(coeff ofx^3/coeff of x^4\)) Similarly, Sum of roots taken two at a time = (\(coeff ofx^2/coeff of x^4\)) Product of roots = "Zero" as there is no constant term. Also, -2 satisfies the equation \(x^4-x^3-6x^2=0\) So, shouldn't the answer be -1, with roots as 0,-2,3? It's explained in the solution above: \(x\) 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 \(x=3\).
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:03
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
Kudos points for each correct solution!!![/quote]
X^4 = X^3 + 6x^2.Solving this we get teh values of x as 0, 3 , -2.
But X cannot be negative as X is the positive fourth root of an expression(\sqrt{Expression}>= 0)
So X = 0 and 3
Sum of the values = 0+3 = 3
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:12
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
Kudos points for each correct solution!!![/quote]
Since 3 is the root of the equation x^2 + ax + 15 = 0, therefore 3 must satisfy the equation x^2 + ax + 15 = 0 .
On putting x=3 in the equation x^2 + ax + 15 = 0 , we get a= -8
On putting the value of a in the original equation x^2 + ax - b = 0, we get x^2 - 8x - b = 0. As per question x^2 - 8x - b = 0 has equal roots so Discriminant(D) = b^2 -4ac = 0 = (-8)^2 - 4.1.(-b)= 0. From this we get the value of b as -16
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:17
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. \(\frac{\sqrt{m^2+n^2}}{2}\)
Kudos points for each correct solution!!![/quote]
Adding the given two equations we get 2a^2 = (m+n)/2
Subtracting the given two equations we get 2b^2 = (m-n)/2
Multiplying above two equations we get 4a^2b^2 = (m^2 - n^2) => ab = \(\frac{\sqrt{m^2-n^2}}{2}\)
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:29
4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?
A. -39
B. -9
C. 0
D. 9
E. 39
Kudos points for each correct solution!!![/quote]
Rearranging the given equation we get
-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 when x= 2 and y = -3, then the maximum value of the whole expression is 0 + 0 -9 = -9
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:39
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
Kudos points for each correct solution!!![/quote]
On Re-arranging the given equation we get m = - x^2 - 2x +15
For m to be positive - x^2 - 2x +15 should be greater then 0. Solving we get (x + 5)(x - 3)<0. Since x is an integer we get the values of x as -4 , -3, -2, -1, 0, 1, 2 (total 7 values)
Given that X is an integer from -10 and 10, inclusive (21 values)
So, the probability is 7/21=1/3.
Answer: B.
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:44
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
Kudos points for each correct solution!!![/quote]
Rearranging we get m^2n^2 + mn - 12 = 0 => (mn + 4)(mn - 3)= 0 =>mn = -4 or mn = 3 => m= -4/n or 3/n
E. I and III only
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:55
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
Kudos points for each correct solution!!![/quote]
Rearranging the equation we get x^4 - 29x^2 + 100 = 0 => (x^2 - 25)(x^ - 4) =0 => x= +5, -5, +2, -2
From these values of x we can get -50 (-5 * +5 * +2), +50(-5 * +5 * -2) but we can't get 25
B. II only
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 06:17
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
on re-arranging we get m^3 -381m = -380 => m(m^2 - 381)= 380
On checking the options and hit and trial, we get the values of m to be 1 and -20. But since m is negative so m = -20
B. -20
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 06:29
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
\(x=(\sqrt{5}-\sqrt{7})^2\)
x= 5 + 7 - 2. \(\sqrt{35}\)
\(\sqrt{35}\) ~ 6
x = 5+7 - 2*6 = 12 - 12 = 0
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 06:34
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) = \(2.n^2 - 1\)
g(n + 12) = (n + 12)^2
f(n^2) = g(n + 12) => \(2.n^2 - 1\) = (n + 12)^2 => 2.n^2 -1 = n^2 + 144 +24.n => n^2 -24n -145 = 0
From the above equation we get n = -5 and 29
So product of values of n = -5 * 29 = -145
A. -145
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Re: New Algebra Set!!! [#permalink]
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28 Mar 2013, 01:23
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 x^4=x^3 + 6x^2 x^4 - x^3 - 6x^2 = 0 ------------> x^2 (x^2 - x - 6) = 0 -----------> x^2 (x-3)(x+2) = 0 x=0 or 3 or -2. With x = -2 original equation does not hold true so possible values for X are 3 and 0 Hence their sum is 3 Choice D. 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 x^2 + ax - b = 0 that means a^2 - 4(-b) = 0 ------> a^2 + 4b = 0 -------------Equation I x^2 + ax + 15 has 3 as a root, so 9 + 3a + 15 = 0 -------> 3a + 24 = 0 --------> a = -8 We will put the value of a in equation I -----------> a^2 + 4b = 0 --------> 64 + 4b = 0 ------> b = -16 ----> Choice B is the answer.
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Re: New Algebra Set!!! [#permalink]
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28 Mar 2013, 01:37
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 a^2 + b^2 = m a^2 - b^2 = n ------------------------ 2a^2 = m+n ----------> a^2 = (m+n)/2 --------> a = sq.root (m+n) / sq.root 2 similarly b = sq.root (m-n) / sq.root 2 So ab = (sq.root (m+n) / sq.root 2)(sq.root (m-n) / sq.root 2) -------> ab = sq.root(m^2 - n^2)/2---------------Using (a+b)(a-b) = a^2 - b^2 Choice C 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
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Re: New Algebra Set!!! [#permalink]
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28 Mar 2013, 01:49
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 there are 21 possible values of x so does m has 21 possible values from x = -4 to x = 2, m gives negative value; Total 7 values so the probability m being negative is 7/21----> 1/3 ----> choice B 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 m^2n^2 + mn = 12 ------> mn(mn+1) = 12 ----------> here mn and mn+1 are consecutive integers. So 12 is basically a product of 2 consecutive intergers those integers must be 3 and 4 or -3 and -4 so possible values for mn are mn = 3 and mn+1 = 4 --------> m = (3/n) mn = -4 and mn + 1 =-3 -------> m =-(4/n) Choice E
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Re: New Algebra Set!!! [#permalink]
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28 Mar 2013, 04:05
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 x^4 - 29x^2 + 100 = 0 --------> let X^2 be y -------> y^2 - 29y + 100 = 0 ------> y^2 - 25y - 4y + 100 = 0 --------> (y-25)(y-4)=0 y=25 -----> x^2 = 25 -----> x= 5 or -5 y=4-------> x^2 = 4 ------> x = 2 or -2 Possible values of x are 5, -5, 2, and -2 Out of options 25 can never be a product of any 3 values of x, so Choice B is the answer 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 This was a bit complex one m^3 - 381m = - 380 ------> m(m^2 - 381) = - 380 Since -380 is the product of m(m^2 - 381), any one of either m or (m^2 - 381) must be negative. Since we are given that m is a negative integer, then m^2 - 381 must be positive So m^2 - 381 > 0 -------> m^2 > 381 ---------> |m| > 19.5 ----approx. so m > 19.5-------if m is positive and m < 19.5 ------ if m is negative We know that m is negative so m must equal to -20 : choice B
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Re: New Algebra Set!!! [#permalink]
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28 Mar 2013, 04:10
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 x=(sq.root 5 - sq.root 7)^2 --------> x = 5 + 7 -2(sq.root 35) ---------using (a-b)^2 = a^2 + b^2 - 2ab 12 - 2(sq.root 35) ---------> assuming sq.root 35 = 6 ----------> 12-12 = 0 Choice A 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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29 Mar 2013, 06:54
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. Hi Bunuel, Could you please explain why -2 is not a solution? Is it because x is a non-negative number? Because if I plug in the value of -2 to the equation \(x^4=x^3+6x^2\) I get \(-2=\sqrt[4]{16}\) which can be true isn't it, as \(\sqrt[4]{16}=|2|\) Also could you explain why even root of an expression cannot be negative when we have an expression which is a square, because we know that \(\sqrt{{x^2}}=|x|\)? Thank you
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Re: New Algebra Set!!! [#permalink]
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11 Apr 2013, 15:31
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. I still cannot get how did you rearrange the given expression, bunuel . Could you please elaborate more ? Thanks in advance 
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Re: New Algebra Set!!! [#permalink]
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12 Apr 2013, 04:29
Bunuel wrote: \(m^2n^2 + mn = 12\) --> \(m^2n^2 + mn -12=0\) --> \((mn)^2 + mn - 12=0\) --> say mn=x, so we have \(x^2+x-12=0\) --> \((x+4)(x-3)=0\) --> \(x=-4\) or \(x=3\) --> \(mn=-4\) or \(mn=3\). Solving and Factoring Quadratics: http://www.purplemath.com/modules/solvquad.htmhttp://www.purplemath.com/modules/factquad.htmHope it helps. ohhhhhhhhhhhhhhhhhhhhh I thought that m is raised to the power 2n and the whole bracket raised to the power 2 ...that is what made me confused . Thanks a million , Bunuel
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Re: New Algebra Set!!!
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12 Apr 2013, 04:29
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