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# New Algebra Set!!!

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New Algebra Set!!! [#permalink]  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

Kudos points for each correct solution!!!
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Re: New Algebra Set!!! [#permalink]  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]  18 Mar 2013, 08:50
1
KUDOS
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]  18 Mar 2013, 08:54
1
KUDOS
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]  18 Mar 2013, 10:25
1
KUDOS
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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Last edited by ConnectTheDots on 20 Mar 2013, 07:40, edited 1 time in total.
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Re: New Algebra Set!!! [#permalink]  18 Mar 2013, 10:55
2
KUDOS
@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]  18 Mar 2013, 12:00
1
KUDOS
Expert's post
1. $$x^4=x^3 + 6x^2$$

So $$x^2(x^2-x-6)=0$$
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!!! [#permalink]  18 Mar 2013, 12:05
1
KUDOS
Expert's post
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]  18 Mar 2013, 12:11
1
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Expert's post
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 = m-n

4a^2b^2= m^2 - n^2
or 2ab= √(m^2-n^2)
C
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Re: New Algebra Set!!! [#permalink]  18 Mar 2013, 12:17
2
KUDOS
Expert's post
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]  18 Mar 2013, 12:33
2
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Expert's post
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]  18 Mar 2013, 12:42
Expert's post
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]  18 Mar 2013, 12:47
Expert's post
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

-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]  18 Mar 2013, 12:50
Expert's post
9. (√5 - √7)^2 = 5+7-2√35
√35 is almost equal to √36 i.e 6

So the expression is approximated as 5+7-12 = 0
A
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Re: New Algebra Set!!! [#permalink]  18 Mar 2013, 12:53
Expert's post
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]  18 Mar 2013, 12:54
Expert's post
Hey Bunuel!
I touched quant after 3 months and I am really out of shape and its 2:30 in the morning!
So If i answer incorrectly its partly my brain's fault!
Oh well, when it isn't!
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I missed this one [#permalink]  18 Mar 2013, 15:15
Expert's post
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]  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]  19 Mar 2013, 00:48
1
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Expert's post
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]  19 Mar 2013, 00:57
Expert's post
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]  19 Mar 2013, 01:17
1
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Expert's post
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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