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New Algebra Set!!!

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New Algebra Set!!! [#permalink] New post 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: new-algebra-set-149349-60.html#p1200948


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: new-algebra-set-149349-60.html#p1200950


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{m-n}}{2}
B. \frac{\sqrt{mn}}{2}
C. \frac{\sqrt{m^2-n^2}}{2}
D. \frac{\sqrt{n^2-m^2}}{2}
E. \frac{\sqrt{m^2+n^2}}{2}

Solution: new-algebra-set-149349-60.html#p1200956


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: new-algebra-set-149349-60.html#p1200962


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: new-algebra-set-149349-60.html#p1200970


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: new-algebra-set-149349-60.html#p1200973


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: new-algebra-set-149349-60.html#p1200975


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: new-algebra-set-149349-60.html#p1200980


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: new-algebra-set-149349-60.html#p1200982


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: new-algebra-set-149349-80.html#p1200987


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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:45
4) -3x^2 + 12x -2y^2 - 12y - 39 MAX
-3x^2 + 12x has max in (2,12)
-2y^2 - 12y - 39 has max in (-6,-3)
-3x^2 + 12x -2y^2 - 12y - 39 MAX = 12 - 3 =9
D
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:50
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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=5+7-2\sqrt{35}
\sqrt{35}=6 almost...
x=5+7-2*6
x=0

A
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:54
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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

2n^2-1=(n+12)^2

2n^2-1=n^12+144+24n

n^2-24n-145=0

n1=29, n2=-5

n1*n2=-145

A
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 10:25
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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

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
For16 - m = 16 => m = 0
For16 - 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!!! [#permalink] New post 18 Mar 2013, 10:55
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@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(m-15)>0 and 4-4(m-15) must be a perfect square
=> 4 (16-m) must be a perfect square and >0
=> 16-m 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!!! [#permalink] New post 18 Mar 2013, 12:00
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 12:05
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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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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 12:11
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 12:17
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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+24-18+36-39 = -9
B
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 12:33
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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

x^2 + 2x - 15 =-m or x^2 + 2x + (m-15) = 0
or (x+5)(x-3) = -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!!! [#permalink] New post 18 Mar 2013, 12:42
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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 + 0x^3 - 29x^2 + 0x + 100 =0
this equation reduces to
(x^2-25)(x^2-4)=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!!! [#permalink] New post 18 Mar 2013, 12:47
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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

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(19^2-20)

=-19(361-20)
Not working

Lets try -20

-20^3 + 380

= -20(400-19)

=-20(381)

=381m

BINGO
B
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 12:50
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 12:53
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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^2-24n-145=0

Product of the roots of this is -145

A it is
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 12:54
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I missed this one [#permalink] New post 18 Mar 2013, 15:15
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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 (a-3)(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!!! [#permalink] New post 19 Mar 2013, 00:15
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+24-18+36-39 = -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....

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Re: New Algebra Set!!! [#permalink] New post 19 Mar 2013, 00:48
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4. -3x^2 + 12x -2y^2 - 12y - 39 = -3x^2+12x-12-2y^2-12y-18-39+18+12 = -3(x-2)^2-2(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!!! [#permalink] New post 19 Mar 2013, 00:57
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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+24-18+36-39 = -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!!! [#permalink] New post 19 Mar 2013, 01:17
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8.m^3 + 380 = 381m
or m^3-m = 380(m-1)
or m(m-1)(m+1) = 380(m-1)
AS m !=1, thus,

m(m+1) = 380 = 19*20

Thus, as m<0, only m=-20 satisfies.

B.
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Last edited by mau5 on 19 Mar 2013, 04:40, edited 1 time in total.
Re: New Algebra Set!!!   [#permalink] 19 Mar 2013, 01:17
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