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New Algebra Set!!! [#permalink]
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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: 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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21 Mar 2013, 09:17
Answers. 1  C 2  b 3  c 4  ? 5  A 6  E 7  B 8  C ? 9  1 10 A Please do publish the OA
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Re: New Algebra Set!!! [#permalink]
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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
x^4 = 29x^2  100 x^4  29x^2 + 100 = 0 (x^2  25) (x^2  4) = 0 x^2 = 25 or x^2 = 4 x= 5, 5, 2, or 2 5 * 5 * 2 = 50 < I is possible 5 * 5 * 2 = 50 < III is possible Only II is not possible  therefore B



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10. f(n^2) = g(n+12) 2n^2  1 = (n+12)^2 2n^2  1 = n^2 + 24n + 144 n^2  24n  145 = 0 (n29)(n+5)=0 n= 29 or 5 29*(5) = 145  Answer A



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21 Mar 2013, 10:37
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9. x = (5^0.5  7^0.5)^2 x = (5^0.5)^2  2*(5^0.5)(7^0.5) + (7^0.5)^2 x = 5  2*(5^0.5)(7^0.5) + 7 x = 12  2*(5^0.5)(7^0.5)
2*(5^0.5)(7^0.5) ~= 2*(36^0.5) ~= 2*6 = 12 x ~= 1212 = 0 Answer A



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Re: New Algebra Set!!! [#permalink]
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21 Mar 2013, 11:23
1.c 2.b 3.c 4.b 5. dont understand the question 6. e 7. b 8. e (guess) 9. a 10. e



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Re: New Algebra Set!!! [#permalink]
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21 Mar 2013, 12:58
[quote="ConnectTheDots"]7. If\(x^4 = 29x^2  100\), then which of the following is NOT a product of two possible values of x?
I. 50 II. 25 III. 100
A. I only B. II only C. III only D. I and II only E. I and III only
\(x^4 = 29x^2  100\) \(=> x^429x^2+100 = 0\) \(=> x^4  25x^2  4x^2 + 100=0\) \(=> x^2(x^225)  4(x^225) = 0\) \(=>(x^2  4)(x^225) = 0\) => x = +2,2,+5,5
None of 50, 25 or 100 is possible with product of two possible values of x. what I am missing ...
> you missed the q ,it is 7. If x^4 = 29x^2  100, then which of the following is NOT a product of three possible values of x and answer is 25



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Re: New Algebra Set!!! [#permalink]
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21 Mar 2013, 13:01
surya167 wrote: Answers. 1  C 2  b 3  c 4  ? 5  A 6  E 7  B 8  C ? 9  1 10 A
Please do publish the OA : for 8 ,it is B .. Put thr value ,20 will satisfy



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Re: New Algebra Set!!! [#permalink]
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21 Mar 2013, 13:10
My answers : 1.C 2.B 3.C 4.B 5.B 6.E 7B 8B 9A 10 Aif any body has different answers ;please let me know ......



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Re: New Algebra Set!!! [#permalink]
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21 Mar 2013, 14:10
Bunuel wrote: 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
GyanOne wrote: @Bunuel, for Q5, shouldn't the original question also say that we are only looking for solutions where x is an integer?
For this to be satisfied, in x^2 + 2x 15 = m, 4  4(m15)>0 and 44(m15) must be a perfect square => 4 (16m) must be a perfect square and >0 => 16m must be a perfect square and >0 The only values that satisfy this for 10<=m<=10 are m=9,0,7 of which only 7 is positive => probability = 1/3 I agree. Proper answer to 5 as written is 10/21 If x is not explicitly constrained or if x is not constrained by the equation given, then we have to respect all real values of x, integer or not. Since x is not constrained to integers only, we have to respect the possibility that x is not an integer. Since 10 <= m <= 10 and the equation is x^2 + 2x  15 = m, we will always have x^2 + 2x + c = 0 where 25 <= c <= 5. Using quadratic formula (just to prove it, though you won't need to know it for the GMAT), we need the calculate the discriminant, which is b^2  4ac, to determine whether or not there are real values for x. Here we get 2^2  4*1*c. Since c is always negative, we will always have a positive number for the discriminant. This means that we will always have a positive number underneath the square root of the quadratic formula and, therefore, 2 real outcomes for x. There are 21 integer values of m (10 positive, 10 negative, and 0) and 10 of them are positive, so 10/21. Good question though if we constrain x to only integer values.



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Re: New Algebra Set!!! [#permalink]
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21 Mar 2013, 22:11
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7  B
x^4=29x^2100 x^429x^2+100=0 which is equal to
(x2)(x+2)(x5)(x+5) = 0
roots are 2,2,5,5
thus B is the right answer ==> 25 can't be the result of the multiplication of any three roots.



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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 Rearrange the given equation: \(x^22x+15=m\). Given that \(x\) is an integer from 10 and 10, inclusive (21 values) we need to find the probability that \(x^22x+15\) is greater than zero, so the probability that \(x^22x+15>0\). Factorize: \((x+5)(3x)>0\). 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. Answer: B.
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