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# New Algebra Set!!!

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New Algebra Set!!! [#permalink]

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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]

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21 Mar 2013, 08:17
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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21 Mar 2013, 08:39
surya167 wrote:
1 - C
2 - b
3 - c
4 - ?
5 - A
6 - E
7 - B
8 - C ?
9 - 1
10 -A

Please do publish the OA

OA's and solutions will be published on Friday.
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Re: New Algebra Set!!! [#permalink]

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21 Mar 2013, 09:28
1
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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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Re: New Algebra Set!!! [#permalink]

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21 Mar 2013, 09:30
1
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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
(n-29)(n+5)=0
n= 29 or -5
29*(-5) = -145 - Answer A
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Re: New Algebra Set!!! [#permalink]

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21 Mar 2013, 09:37
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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 ~= 12-12 = 0
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Re: New Algebra Set!!! [#permalink]

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21 Mar 2013, 10:23
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, 11:58
[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

$$x^4 = 29x^2 - 100$$
$$=> x^4-29x^2+100 = 0$$
$$=> x^4 - 25x^2 - 4x^2 + 100=0$$
$$=> x^2(x^2-25) - 4(x^2-25) = 0$$
$$=>(x^2 - 4)(x^2-25) = 0$$
=> 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, 12:01
surya167 wrote:
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, 12:10
1.C
2.B
3.C
4.B
5.B
6.E
7B
8B
9A
10 A

if 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, 13:10
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(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

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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21 Mar 2013, 21:11
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7 - B

x^4=29x^2-100
x^4-29x^2+100=0
which is equal to

(x-2)(x+2)(x-5)(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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22 Mar 2013, 04:01
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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: $$x^4=x^3+6x^2$$;

Re-arrange and factor out x^2: $$x^2(x^2-x-6)=0$$;

Factorize: $$x^2(x-3)(x+2)=0$$;

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.

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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

Since one of the roots of the equation $$x^2 + ax + 15 = 0$$ is 3, then substituting we'll get: $$3^2+3a+15=0$$. Solving for $$a$$ gives $$a=-8$$.

Substitute $$a=-8$$ in the first equation: $$x^2-8x-b=0$$.

Now, we know that it has equal roots thus its discriminant must equal to zero: $$d=(-8)^2+4b=0$$. Solving for $$b$$ gives $$b=-16$$.

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22 Mar 2013, 04:16
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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. $$\frac{\sqrt{m^2+n^2}}{2}$$

Sum the two equations: $$2a^2=m+n$$;
Subtract the two equations: $$2b^2=m-n$$;

Multiply: $$4a^2b^2=m^2-n^2$$;

Solve for $$ab$$: $$ab=\frac{\sqrt{m^2-n^2}}{2}$$

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22 Mar 2013, 04:47
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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

$$-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=-3(x^2-4x+4)-2(y^2+6y+9)-9=$$
$$=-3(x-2)^2-2(y+3)^2-9$$.

So, we need to maximize the value of $$-3(x-2)^2-2(y+3)^2-9$$.

Since, the maximum value of $$-3(x-2)^2$$ and $$-2(y+3)^2$$ is zero, then the maximum value of the whole expression is $$0+0-9=-9$$.

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22 Mar 2013, 05:08
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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

Re-arrange the given equation: $$-x^2-2x+15=m$$.

Given that $$x$$ is an integer from -10 and 10, inclusive (21 values) we need to find the probability that $$-x^2-2x+15$$ is greater than zero, so the probability that $$-x^2-2x+15>0$$.

Factorize: $$(x+5)(3-x)>0$$. 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.

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22 Mar 2013, 05:13
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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

Re-arrange: $$(mn)^2 + mn - 12=0$$.

Factorize for mn: $$(mn+4)(mn-3)=0$$. Thus $$mn=-4$$ or $$mn=3$$.

So, we have that $$m=-\frac{4}{n}$$ or $$m=\frac{3}{n}$$.

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22 Mar 2013, 05:20
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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

Re-arrange and factor for x^2: $$(x^2-25)(x^2-4)=0$$.

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

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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

Given $$m^3 + 380 = 380m+m$$.

Re-arrange: $$m^3-m= 380m-380$$.

$$m(m+1)(m-1)=380(m-1)$$. Since m is a negative integer, then $$m-1\neq{0}$$ and we can safely reduce by $$m-1$$ to get $$m(m+1)=380$$.

So, we have that 380 is the product of two consecutive negative integers: $$380=-20*(-19)$$, hence $$m=-20$$.

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22 Mar 2013, 05:35
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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

$$x=(\sqrt{5}-\sqrt{7})^2=5-2\sqrt{35}+7=12-2\sqrt{35}$$.

Since $$\sqrt{35}\approx{6}$$, then $$12-2\sqrt{35}\approx{12-2*6}=0$$.

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# New Algebra Set!!!

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