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Re: New Algebra Set!!!
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Updated on: 20 Mar 2013, 08:40
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
\(x^2 + 2x 15 = m\) \(=> x^2 + 2x + 1 16 = m\) \(=> (x+1)^2 = 16 m\) \(=> 16 m >=0\) and a square number
If m varies between [10,10]
\(=> 16  10 <= 16  m <= 16  (10)\) \(=> 6 <= 16  m <= 26\)
Between 6 and 16, there are 3 square numbers i.e. 9, 16, 25
For \(16  m = 9 => m = 7\) For\(16  m = 16 => m = 0\) For\(16  m = 25 => m = 9\)
So number of possible values of m = 3 i.e. [7, 0, 9] And for m > 0, number of possible values = 1 i.e only for m=7
So probability = 1/3 Ans = B



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Re: New Algebra Set!!!
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18 Mar 2013, 11:55
@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
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Re: New Algebra Set!!!
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18 Mar 2013, 13:00
1. \(x^4=x^3 + 6x^2\) So \(x^2(x^2x6)=0\) Or \(x=0, 2, 3\) However 2 cannot be the solution as x is a positive root. So sum of the roots is 3 i.e D
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Re: New Algebra Set!!!
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18 Mar 2013, 13:05
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 Let the roots of \(x^2 + ax +b = 0\) are h and h So sum of the roots i.e \(2h = a\) and product of the roots i.e \(h^2 = b\) Now one root of \(x^2 + ax + 15=0\) is 5 So the other root must be 5 as the product is 15. So \(a =  (5+3) = 8\) So, \(8 = 2h\) or, \(h=4\) So \(b = 4^2 = 16\) B it is
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18 Mar 2013, 13:11
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: 2a^2 = m+n 2b^2 = mn 4a^2b^2= m^2  n^2 or 2ab= √(m^2n^2) C
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Re: New Algebra Set!!!
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18 Mar 2013, 13:17
4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ? A. 39 B. 9 C. 0 D. 9 E. 39 differentiating with respect to x and equating with 0 \(6x + 12 = 0, OR x=2\) differentiating with respect to y and equating with 0 \(4y  12 =0, OR y=3\) So max value of the expression \(3x^2 + 12x 2y^2  12y  39\) is at (2, 3) i.e 12+2418+3639 = 9 B
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Re: New Algebra Set!!!
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18 Mar 2013, 13:33
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 x^2 + 2x  15 =m or x^2 + 2x + (m15) = 0 or (x+5)(x3) = m Now since m is an integer so x must also be an integer. x=1 m= 12 NO x=2 m= 7 YES x=3 m= 0 YES x=4 m= 9 YES x=5 m= 20 NO No further values will satisfy! So m has 3 values so far! x= 0 m= 15 NO x= 1 m= 16 NO x= 2 m= 15 NO x= 3 m= 12 NO x= 4 m= 7 YES x= 5 m= 0 YES x= 6 m= 9 YES x= 7 m= 20 NO SO m can take 3 values 7, 0 and 9 and only ONE of them is greater than 0 SO the probability is 1/3 B
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Re: New Algebra Set!!!
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18 Mar 2013, 13:42
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 + 0x^3  29x^2 + 0x + 100 =0 this equation reduces to \((x^225)(x^24)=0\) or x = 5, 5, 2, 2 So 50 or 50 cannot be the product of 2 of the above roots. E
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Re: New Algebra Set!!!
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18 Mar 2013, 13:47
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 Lets just start with the bottom and solve 1 clearly does not work lets see for m = 19 and I have a good feeling because 380 is divisible by 19 \(19*19*19 + 380\) \(= 19(19^220)\) \(=19(36120)\) Not working Lets try 20 \(20^3 + 380\) \(= 20(40019)\) \(=20(381)\) \(=381m\) BINGO B
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Re: New Algebra Set!!!
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18 Mar 2013, 13:50
9. (√5  √7)^2 = 5+72√35 √35 is almost equal to √36 i.e 6 So the expression is approximated as 5+712 = 0 A
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18 Mar 2013, 13:53
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 \(f(n^2) = 2n^2  1\) \(g(n+12) = (n+12)^2\) \(= n^2 + 24n +144\) So if they are equal then \(2n^2  1 = n^2 +24n+144\) OR \(n^224n145=0\) Product of the roots of this is 145 A it is
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Re: New Algebra Set!!!
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18 Mar 2013, 16:15
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 lets say \(mn=a\) then \(a^2+a=12\) or \((a3)(a+4)=0\) or \(a=3, 4\) Or \(mn=3, 4\) or \(m=3/n\) or \(4/n\) E
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Re: New Algebra Set!!!
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19 Mar 2013, 01:15
souvik101990 wrote: 4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?
A. 39 B. 9 C. 0 D. 9 E. 39
differentiating with respect to x and equating with 0
\(6x + 12 = 0, OR x=2\)
differentiating with respect to y and equating with 0
\(4y  12 =0, OR y=3\)
So max value of the expression \(3x^2 + 12x 2y^2  12y  39\) is at (2, 3) i.e 12+2418+3639 = 9 B Hi Souvik I liked ur approach cn u pls expand on it...i mean is there any hard and fast rule that u r following.....completely missed this question.... Archit



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Re: New Algebra Set!!!
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19 Mar 2013, 01:48
4. 3x^2 + 12x 2y^2  12y  39 = 3x^2+12x122y^212y1839+18+12 = 3(x2)^22(y+3)^2 9. Now a negative sign is attached to both the squares. Thus the maximum value is at x=2 and y=3 and equals 9. B.
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Re: New Algebra Set!!!
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19 Mar 2013, 01:57
Archit143 wrote: souvik101990 wrote: 4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?
A. 39 B. 9 C. 0 D. 9 E. 39
differentiating with respect to x and equating with 0
\(6x + 12 = 0, OR x=2\)
differentiating with respect to y and equating with 0
\(4y  12 =0, OR y=3\)
So max value of the expression \(3x^2 + 12x 2y^2  12y  39\) is at (2, 3) i.e 12+2418+3639 = 9 B Hi Souvik I liked ur approach cn u pls expand on it...i mean is there any hard and fast rule that u r following.....completely missed this question.... Archit I used calculus maxima minima. However you are free to use the "completing the square" method where you are left with 2 perfect squares and a 9 as the guy above me did.
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Re: New Algebra Set!!!
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Updated on: 19 Mar 2013, 05:40
8.\(m^3 + 380 = 381m\) or m^3m = 380(m1) or m(m1)(m+1) = 380(m1) AS m !=1, thus, m(m+1) = 380 = 19*20 Thus, as m<0, only m=20 satisfies. B.
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Originally posted by mau5 on 19 Mar 2013, 02:17.
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Re: New Algebra Set!!!
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19 Mar 2013, 02:25
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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Re: New Algebra Set!!!
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19 Mar 2013, 02:30
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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Re: New Algebra Set!!!
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Updated on: 19 Mar 2013, 17:10
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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Re: New Algebra Set!!!
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19 Mar 2013, 10:57
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?




Re: New Algebra Set!!!
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