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New Algebra Set!!! [#permalink]
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18 Mar 2013, 06:56
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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: 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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19 Mar 2013, 01:25
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9. x = \((\sqrt{5}\sqrt{7})^2\) As from the given options we know that x>0. Thus, I can safely write that \(\sqrt{x} = \sqrt{7}\sqrt{5}\) or \(\sqrt{7} = \sqrt{x} +\sqrt{5}\) x can not be 1 or more than 1 as then \(\sqrt{7}\)>3 which is not true. Thus, on approximation, x=0. A.
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19 Mar 2013, 01:30
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7. \(x^4 = 29x^2  100\) or \(x^4  29x^2+100 = 0\) Let \(x^2 = t\), thus \(t^229+100 = 0\) or (t25)(t4) = 0 t = 25 > x = 5/5 or t = 4 > x = 2/2 Out of any three possible values of x, only 25 is not possible. B.
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19 Mar 2013, 06:05
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Hello Bunuel, Nice set of Questions Here goes my solutions 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) ?[/b] Sol: From the given function we have f(n^2)=g(n+12) > 2n^21= (n+12)^2 Solving this eqn we get n^2  24n  145= 0> n^2 29n+ 5n  145 =0 Possible values of n are 29 and 5. Product will be 145 and hence ans is option A
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Last edited by WoundedTiger on 19 Mar 2013, 16:10, edited 1 time in total.



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Re: New Algebra Set!!! [#permalink]
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19 Mar 2013, 07:54
my first post 1C 2B 3C 4B 5E 6D 7C 8D 9A 10A



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Re: New Algebra Set!!! [#permalink]
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19 Mar 2013, 09:57
GyanOne wrote: 4. 3x^2 + 12x 2y^2  12y  39 = 3(x^2 + 4x  13) + 2(y^26y)
Now x^2 +4x  13 has its maximum value at x = 4/2 = 2 and y^2  6y has its maximum value at y=6/2 = 3 Therefore max value of the expression = 3(4 + 8 13) + 2 (9+18) = 27 + 18 = 9
Should be B Hi GyanOne, Can you please explain how you found x/y which gives max value for the 2 cases here? Is there a formula?



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19 Mar 2013, 10:18
NeverQuit wrote: GyanOne wrote: 4. 3x^2 + 12x 2y^2  12y  39 = 3(x^2 + 4x  13) + 2(y^26y)
Now x^2 +4x  13 has its maximum value at x = 4/2 = 2 and y^2  6y has its maximum value at y=6/2 = 3 Therefore max value of the expression = 3(4 + 8 13) + 2 (9+18) = 27 + 18 = 9
Should be B Hi GyanOne, Can you please explain how you found x/y which gives max value for the 2 cases here? Is there a formula? \(3x^2 + 12x 2y^2  12y  39\) \(=3(x^24x)2(y^2+6y) 39\) \(=3(x^24x+44)2(y^2+6y+99)39\) \(=3(x2)^2 +12 2(y+3)^2 +18 39\) NOW The above expression will be minimum IFF \((x2)^2=0\) and \((y+3)^2=0\) which leaves the maximum value to be \(12+1839=9\)
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Re: New Algebra Set!!! [#permalink]
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19 Mar 2013, 16:15
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9. If x=(\sqrt{5}\sqrt{7})^2, then the best approximation of x is Ans: \sqrt{5} is aproximately 2.236 and \sqrt{7}) will be around ~ 2.6 therefore we have x= (2.236  2.6)^2 > (0.364)^2> Approximation of X is 0. Ans Option A
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19 Mar 2013, 16:18
8.If m is a negative integer and m^3 + 380 = 381m , then what is the value of m? Ans: Plug in in the answer choices and only option m=21 satisfies. So option B
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19 Mar 2013, 16:20
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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 Sol: In the given eqn let X^2= y.The eqn becomes y^2  29y +100=0, Solving for "y" we get y= 25 or y= 4 This implies X^2 = 25 or X^2= 4 Possible values of x will be 5,5 2 or 2. Only 25 is not possible as product of three factors and hence Ans should be B
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19 Mar 2013, 16:22
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 Sol: The given eqn can be written as mn (mn + 1) = 12 Now mn, mn+ will have to be consecutive Integers and possible combinations are mn=3 and mn+ 1= 4 > possible value of m will be 3/n or mn+1 = 4 and mn= 3 . possible value of m will be 3/n Ans should be III only i.e Option C
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19 Mar 2013, 16:32
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5. If x^2 + 2x 15 = m, where m 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 This was little trickier but here is my approach We need to know Probability that m is greater than 0. Integer is from 10 to 10 inclusive. Consider value of m greater than 0 only and put in the value in the given eqn For ex let us say m= 2, the eqn becomes x^2+2x 15= 2. x^2+ 2x 13=0 > Eqn will have roots of the form b+/ \sqrt{b^24ac}/ 2 We see that for m= 7, we get the eqn as x^2+ 2x 8 =0, Solving for x we get, x = 4 or 2. This seems to be the only case where the eqn gives us 2 real roots and therefore looking at option choices selected B (assuming other 2 cases will be for value of m between 10 to 0) Ans Choice B
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19 Mar 2013, 16:39
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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 Sol: Adding the 2 equations we get a= \sqrt{(m+n)/2} Subtracting Eqn 2 from 1, we get b= \sqrt{mn/2} Ans Option C
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19 Mar 2013, 16:40
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 8xb=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
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19 Mar 2013, 17:03
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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
x^4 = 29x^2  100 x^429x^2+100=0 (x^225)(x^24)=0 (x5)(x+5)(x+2)(x2)=0 roots=5, 5, 2, 2
I. 50 =(2, 5, 5) II. 25 not possible III. 50 =(2, 5, 5)
Answer is B. II only



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Re: New Algebra Set!!! [#permalink]
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19 Mar 2013, 17:14
GyanOne wrote: Edited
1. x^4 = x^3 + 6x^2 => x^2 (x^2  x  6) = 0 The roots of x^2  x  6 are 2 and 3, but 2 cannot be the value of x. So 3 and 0 are the only possible roots. => Sum of all possible solutions = 3.
Option D why 2 cannot equal to x?



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19 Mar 2013, 17:38
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Bunuel wrote: Please suggest on what category would you like the next set to be. Thank you! Hello Bunuel, I do like to see Questions on Geometry or Time Speed and Distance Thanks Mridul
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20 Mar 2013, 01:00
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1. If , then the sum of all possible solutions for x is:
A. 2 B. 0 C. 1 D. 3 E. 5
Soln: Possible solutions are x = 2 & 3, Ans(C)
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
Soln: Solving for a = 8, Ans (B)



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20 Mar 2013, 01:32
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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. B. C. D. E.
Soln: Take equation 2, square on both sides, substitute for a^2 + b^2 with m, Ans (C)
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
Soln: Used a 'gestimation' method. Since the unit digit of the resulting values on both sides has to be same, the only number from the options that will fit this condition is 20. Ans (B)



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20 Mar 2013, 01:40
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9. If , then the best approximation of x is:
A. 0 B. 1 C. 2 D. 3 E. 4
Soln: x = 12  2√35, if √35 =~ 6, then x =~ 0, Ans (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
Soln: Solving for f(x) & g(x), 2n^2  1 = (n+12)^2, Solving for n gives n = 29 or 5, Ans (A)



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Re: New Algebra Set!!! [#permalink]
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20 Mar 2013, 07:58
1. If x = (x^3 + 6x^2)^1/4, then the sum of all possible solutions for x is: A. 2 B. 0 C. 1 D. 3 E. 5 My take ==> C Approach: x = (x^3 + 6x^2)^1/4 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 So Sum of all the possible solution of x = 1 Quote: Thanks




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