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Re: New Algebra Set!!!
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22 Mar 2013, 05:16
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. \(\frac{\sqrt{m^2+n^2}}{2}\) Sum the two equations: \(2a^2=m+n\); Subtract the two equations: \(2b^2=mn\); Multiply: \(4a^2b^2=m^2n^2\); Solve for \(ab\): \(ab=\frac{\sqrt{m^2n^2}}{2}\) Answer: C.
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22 Mar 2013, 05:47
4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?A. 39 B. 9 C. 0 D. 9 E. 39 \(3x^2 + 12x 2y^2  12y  39=3x^2 + 12x122y^2  12y189=3(x^24x+4)2(y^2+6y+9)9=\) \(=3(x2)^22(y+3)^29\). So, we need to maximize the value of \(3(x2)^22(y+3)^29\). Since, the maximum value of \(3(x2)^2\) and \(2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+09=9\). Answer: B.
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22 Mar 2013, 06:08
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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22 Mar 2013, 06:13
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/nA. 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.
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22 Mar 2013, 06:20
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. 50A. I only B. II only C. III only D. I and II only E. I and III only Rearrange and factor for x^2: \((x^225)(x^24)=0\). 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 Answer: B.
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22 Mar 2013, 06:30
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 Given \(m^3 + 380 = 380m+m\). Rearrange: \(m^3m= 380m380\). \(m(m+1)(m1)=380(m1)\). Since m is a negative integer, then \(m1\neq{0}\) and we can safely reduce by \(m1\) to get \(m(m+1)=380\). So, we have that 380 is the product of two consecutive negative integers: \(380=20*(19)\), hence \(m=20\). Answer: B.
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22 Mar 2013, 06:35
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 \(x=(\sqrt{5}\sqrt{7})^2=52\sqrt{35}+7=122\sqrt{35}\). Since \(\sqrt{35}\approx{6}\), then \(122\sqrt{35}\approx{122*6}=0\). Answer: A.
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22 Mar 2013, 06:44
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 \(f(x) = 2x  1\), hence \(f(n^2)=2n^21\). \(g(x) = x^2\), hence \(g(n+12)=(n+12)^2=n^2+24n+144\). Since given that \(f(n^2)=g(n+12)\), then \(2n^21=n^2+24n+144\). Rearranging gives \(n^224n145=0\). Next, Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):
\(x_1+x_2=\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).Thus according to the above \(n_1*n_2=145\). Answer: A.
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Re: New Algebra Set!!!
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23 Mar 2013, 03:54
Hi Bunuel, Thanks for the challenging ones. Quote: Given \(m^3 + 380 = 380m+m\). I was able to backsolve this question, as I couldn't able to figure out the above step. Can you please post your thinking pattern behind this question. I mean, how did you think about breaking the term 381m into 380m + m >(A) Any particular indicators in the question that make you do so. I'm sure during exam time, I will not be able to do so. Quote: Rearrange: \(m^3m= 380m380\).
\(m(m+1)(m1)=380(m1)\). Since m is a negative integer, then \(m1\neq{0}\) and we can safely reduce by \(m1\) to get \(m(m+1)=380\).
So, we have that 380 is the product of two consecutive negative integers: \(380=20*(19)\), hence \(m=20\).
Answer: B. I have read somewhere that cancelling the like terms in GMAT is dangerous. I can sense that in your post as well( Red colored part). Request you to please enlighten me with this concept as why did you need to make extra caution while cancelling (m1) terms, instead of cancelling it without given a thought.
Thanks H



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24 Mar 2013, 03:44
imhimanshu wrote: Quote: Rearrange: \(m^3m= 380m380\).
\(m(m+1)(m1)=380(m1)\). Since m is a negative integer, then \(m1\neq{0}\) and we can safely reduce by \(m1\) to get \(m(m+1)=380\).
So, we have that 380 is the product of two consecutive negative integers: \(380=20*(19)\), hence \(m=20\).
Answer: B. I have read somewhere that cancelling the like terms in GMAT is dangerous. I can sense that in your post as well( Red colored part). Request you to please enlighten me with this concept as why did you need to make extra caution while cancelling (m1) terms, instead of cancelling it without given a thought.
Thanks H Consider equation xy=x. Here we cannot reduce by x, because x could be zero and we cannot divide by zero. What we can do is: xy=x > xyx=0 > x(y1)=0 > x=0 or y=1. Now, in the question we have \(m(m+1)(m1)=380(m1)\) and we know that m is a negative number, thus m1 does not equal to 0, which means that we can safely reduce by m1. Hope it's clear.
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Re: New Algebra Set!!!
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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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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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29 Mar 2013, 07:04
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!!!
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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, 02:22
TheNona 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
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 \(m^2n^2 + mn = 12\) > \(m^2n^2 + mn 12=0\) > \((mn)^2 + mn  12=0\) > 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.
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Re: New Algebra Set!!!
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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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