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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: newalgebraset14934960.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: newalgebraset14934960.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. \(\frac{\sqrt{mn}}{2}\) B. \(\frac{\sqrt{mn}}{2}\) C. \(\frac{\sqrt{m^2n^2}}{2}\) D. \(\frac{\sqrt{n^2m^2}}{2}\) E. \(\frac{\sqrt{m^2+n^2}}{2}\) Solution: newalgebraset14934960.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: newalgebraset14934960.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: newalgebraset14934960.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: newalgebraset14934960.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: newalgebraset14934960.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: newalgebraset14934960.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: newalgebraset14934960.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: newalgebraset14934980.html#p1200987Kudos points for each correct solution!!!
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 06:17
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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 rearranging 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: newalgebraset14934960.html#p1200948 x^4=x^3 + 6x^2 x^4  x^3  6x^2 = 0 > x^2 (x^2  x  6) = 0 > x^2 (x3)(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: newalgebraset14934960.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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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{mn}}{2} B. \frac{\sqrt{mn}}{2} C. \frac{\sqrt{m^2n^2}}{2} D. \frac{\sqrt{n^2m^2}}{2} E. \frac{\sqrt{m^2+n^2}}{2} Solution: newalgebraset14934960.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 (mn) / sq.root 2 So ab = (sq.root (m+n) / sq.root 2)(sq.root (mn) / sq.root 2) > ab = sq.root(m^2  n^2)/2Using (a+b)(ab) = 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: newalgebraset14934960.html#p1200962
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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: newalgebraset14934960.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: newalgebraset14934960.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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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: newalgebraset14934960.html#p1200975 x^4  29x^2 + 100 = 0 > let X^2 be y > y^2  29y + 100 = 0 > y^2  25y  4y + 100 = 0 > (y25)(y4)=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: newalgebraset14934960.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 > 19.5if 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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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: newalgebraset14934960.html#p1200982 x=(sq.root 5  sq.root 7)^2 > x = 5 + 7 2(sq.root 35) using (ab)^2 = a^2 + b^2  2ab 12  2(sq.root 35) > assuming sq.root 35 = 6 > 1212 = 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: newalgebraset14934980.html#p1200987
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Re: New Algebra Set!!! [#permalink]
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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: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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 nonnegative 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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nave81 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: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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 nonnegative 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 Note the Important Difference. On the GMAT if X^2 = 4 then x = +/ 2 or x = 2 However if x= sq.root(4) then x has to be positive i.e. 2 and it can not take negative value. when we plug 2 as the value of x in the equation we would get 2 = 4th root of 16 > 2 = 2 This is because fourth root of 16 is 2 and not 2 The rule is even root of a number can not be negative on the GMAT Regards, Abhijit.
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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
Rearrange: \((mn)^2 + mn  12=0\).
Factorize for mn: \((mn+4)(mn3)=0\). 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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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 \(x^2+x12=0\) > \((x+4)(x3)=0\) > \(x=4\) 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!!! [#permalink]
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12 Apr 2013, 05:25
Narenn wrote: nave81 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: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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 nonnegative 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 Note the Important Difference. On the GMAT if X^2 = 4 then x = +/ 2 or x = 2 However if x= sq.root(4) then x has to be positive i.e. 2 and it can not take negative value. when we plug 2 as the value of x in the equation we would get 2 = 4th root of 16 > 2 = 2 This is because fourth root of 16 is 2 and not 2 The rule is even root of a number can not be negative on the GMAT Regards, Abhijit. Hey Abhijit, This is a very interesting point that you have made here. This statement that 2 is the fourth root of 16 is admissible under normal circumstances but is not in a GMAT question. Can you mention a source which would list out such peculiar rules for GMAT quants? Thanks in advance Anshuman
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Hi Anshuman, For this you need to go thru some strategy books of GMAT Quant. In My Opinion, two important sources are, 1) Manhattan GMAT Quant strategy guides 4th or 5th Ed 2) GMAT Club Math Book  This is free and you can download it from this website free of cost (Check BB's profile) Furthermore you can check my signature. There i have attached my articles prepared on some peculiar topics. You may find them useful. Thanks, Abhijit
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22 May 2013, 22:19
Bunuel wrote: [ Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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, Request you to please let me know where I'm making a mistake. Since \(x^3+6*x^2 >= 0\) Then, \(x^2(x+6)>=0\) i.e \((x0)^2(x(6))>=0\) This implies that\( x>=6\) Hence, x=2 is a valid root, and sum of all roots should be\(x=3+(2) = 1\) Please let me know where I am going wrong. Thanks
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23 May 2013, 02:00
imhimanshu wrote: Bunuel wrote: [ Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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, Request you to please let me know where I'm making a mistake. Since \(x^3+6*x^2 >= 0\) Then, \(x^2(x+6)>=0\) i.e \((x0)^2(x(6))>=0\) This implies that\( x>=6\) Hence, x=2 is a valid root, and sum of all roots should be\(x=3+(2) = 1\) Please let me know where I am going wrong. Thanks Plug x=2 into \(x=\sqrt[4]{x^3+6x^2}\). Does the equation hold true?
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Re: New Algebra Set!!! [#permalink]
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23 May 2013, 05:03
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:
x^2 + ax + 15 = 0 equation has one root as 3. only possible values of getting product of the roots 15 , having one root 3 is 5*3. x^25x3x+15=0 take common values out => x(x5)3(x5) =0 => x=5 and x=3(what was given) , from this we get a= 8.
x^2 8x  b =0 => if roots are equal then b^24ac=0, 64+4b=0 from this b=16
Answer B



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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. \(\frac{\sqrt{mn}}{2}\) B. \(\frac{\sqrt{mn}}{2}\) C. \(\frac{\sqrt{m^2n^2}}{2}\) D. \(\frac{\sqrt{n^2m^2}}{2}\) E. \(\frac{\sqrt{m^2+n^2}}{2}\)
Solution:
a^2  b^2 = n => (a+b)(ab)=n=> (a+b)^2*(ab)^2=n^2 => (a^2 + b^2+2ab)*(a^2 + b^22ab)=n^2 => (m+2ab)*(m2ab)=n^2 => m^22ab^2=n^2 =>2ab^2=m^2n^2 => 2ab=\sqrt{\(m^2n^2\)} =>ab=\sqrt{\(m^2n^2\)}/2
Ans: C



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Re: New Algebra Set!!! [#permalink]
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23 May 2013, 05:27
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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:By using below substitution values we could reduce the powers of x. Here values (1,0,29,0,100)  taken from equation  x^4 = 29x^2  100=> (1)x^4+(0)x^329x^2+(0)x+100=0 from above explanation we get equation x^225=0 => give x=+5,5 from above calculation we get values of x =+2,2. multiplying any of the three roots we have possibility of getting product values, 50,+50. Ans:E




Re: New Algebra Set!!!
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