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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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22 Mar 2013, 05:44
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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

$$f(x) = 2x - 1$$, hence $$f(n^2)=2n^2-1$$.
$$g(x) = x^2$$, hence $$g(n+12)=(n+12)^2=n^2+24n+144$$.

Since given that $$f(n^2)=g(n+12)$$, then $$2n^2-1=n^2+24n+144$$. Re-arranging gives $$n^2-24n-145=0$$.

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 $$n_1*n_2=-145$$.

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Re: New Algebra Set!!! [#permalink]

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22 Mar 2013, 05:49
Links to the solutions are in the original post here: new-algebra-set-149349.html#p1198114
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Re: New Algebra Set!!! [#permalink]

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22 Mar 2013, 06:05
Kudos points given for each correct solution.

Note that I cannot award more than 5 Kudos to the same person per day, so those of you who have more than 5 correct solutions please PM me tomorrow the links for which I owe you kudos points.

Thank you.

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Re: New Algebra Set!!! [#permalink]

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22 Mar 2013, 09:20
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For those who are interested to ALL Bunuel's sets (each of them are in its signature), you can download them in a single RAR file with solutions

Now the same has been updated with this last Algebra set.

bunuel-signature-collection-all-in-one-with-solutions-146628.html#p1176711

bunuel-signature-collection-all-in-one-with-solutions-146628.html#p1201116

regards
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Re: New Algebra Set!!! [#permalink]

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22 Mar 2013, 09:25
carcass wrote:
For those who are interested to ALL Bunuel's sets (each of them are in its signature), you can download them in a single RAR file with solutions

Now the same has been updated with this last Algebra set.

bunuel-signature-collection-all-in-one-with-solutions-146628.html#p1176711

bunuel-signature-collection-all-in-one-with-solutions-146628.html#p1201116

regards

Why are you so awesome carcass?
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Re: New Algebra Set!!! [#permalink]

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23 Mar 2013, 02:54
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:
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 $$m=-20$$.

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 (m-1) terms, instead of cancelling it without given a thought.

Thanks
H
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Re: New Algebra Set!!! [#permalink]

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24 Mar 2013, 02:44
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imhimanshu wrote:
Quote:
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 $$m=-20$$.

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 (m-1) 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 --> xy-x=0 --> x(y-1)=0 --> x=0 or y=1.

Now, in the question we have $$m(m+1)(m-1)=380(m-1)$$ and we know that m is a negative number, thus m-1 does not equal to 0, which means that we can safely reduce by m-1.

Hope it's clear.
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Re: New Algebra Set!!! [#permalink]

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24 Mar 2013, 14:07
Bunuel wrote:
Please suggest on what category would you like the next set to be. Thank you!

Number properties please !!!
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Re: R: New Algebra Set!!! [#permalink]

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24 Mar 2013, 15:01
I know the benefits of a single set in a specific content area.... but a set with two questions from the main contents?

We already have specific sets

Nexus 4 con Tapatalk 2
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Re: New Algebra Set!!! [#permalink]

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25 Mar 2013, 03:51
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pleaman wrote:
Bunuel wrote:
Please suggest on what category would you like the next set to be. Thank you!

Number properties please !!!

Done! New set on number properties is here: new-set-number-properties-149775.html
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Re: New Algebra Set!!! [#permalink]

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25 Mar 2013, 10:22
My answers for the above questionnaire:
1- C
2- B
3- C
4- B
5- Wasn't able to solve.
6- E
7- B
8- B
9- A
10- E
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Re: New Algebra Set!!! [#permalink]

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25 Mar 2013, 11:31
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: $$x^4=x^3+6x^2$$;

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.

Hi Bunuel,

Query here.

First of all we don't necessarily have to solve for all the three roots. Since, it is asking for a composite value i.e. the sum, we can directly obtain it.

We can rearrange to get:-

$$x^4-x^3-6x^2=0$$
Over here, by the system of poly equations Sum of roots = - ($$coeff ofx^3/coeff of x^4$$)
Similarly, Sum of roots taken two at a time = ($$coeff ofx^2/coeff of x^4$$)
Product of roots = "Zero" as there is no constant term.

Also, -2 satisfies the equation $$x^4-x^3-6x^2=0$$

So, shouldn't the answer be -1, with roots as 0,-2,3?
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Re: New Algebra Set!!! [#permalink]

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26 Mar 2013, 01:07
Sinner1706 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: $$x^4=x^3+6x^2$$;

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.

Hi Bunuel,

Query here.

First of all we don't necessarily have to solve for all the three roots. Since, it is asking for a composite value i.e. the sum, we can directly obtain it.

We can rearrange to get:-

$$x^4-x^3-6x^2=0$$
Over here, by the system of poly equations Sum of roots = - ($$coeff ofx^3/coeff of x^4$$)
Similarly, Sum of roots taken two at a time = ($$coeff ofx^2/coeff of x^4$$)
Product of roots = "Zero" as there is no constant term.

Also, -2 satisfies the equation $$x^4-x^3-6x^2=0$$

So, shouldn't the answer be -1, with roots as 0,-2,3?

It's explained in the solution above: $$x$$ 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 $$x=3$$.
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Re: New Algebra Set!!! [#permalink]

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26 Mar 2013, 04:03
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

Kudos points for each correct solution!!![/quote]

X^4 = X^3 + 6x^2.Solving this we get teh values of x as 0, 3 , -2.
But X cannot be negative as X is the positive fourth root of an expression(\sqrt{Expression}>= 0)
So X = 0 and 3
Sum of the values = 0+3 = 3
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Re: New Algebra Set!!! [#permalink]

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26 Mar 2013, 04:12
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
Kudos points for each correct solution!!![/quote]

Since 3 is the root of the equation x^2 + ax + 15 = 0, therefore 3 must satisfy the equation x^2 + ax + 15 = 0 .
On putting x=3 in the equation x^2 + ax + 15 = 0 , we get a= -8
On putting the value of a in the original equation x^2 + ax - b = 0, we get x^2 - 8x - b = 0. As per question x^2 - 8x - b = 0 has equal roots so Discriminant(D) = b^2 -4ac = 0 = (-8)^2 - 4.1.(-b)= 0. From this we get the value of b as -16
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Re: New Algebra Set!!! [#permalink]

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26 Mar 2013, 04:17
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. $$\frac{\sqrt{m^2+n^2}}{2}$$
Kudos points for each correct solution!!![/quote]

Adding the given two equations we get 2a^2 = (m+n)/2
Subtracting the given two equations we get 2b^2 = (m-n)/2

Multiplying above two equations we get 4a^2b^2 = (m^2 - n^2) => ab = $$\frac{\sqrt{m^2-n^2}}{2}$$
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Re: New Algebra Set!!! [#permalink]

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26 Mar 2013, 04:29
4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?

A. -39
B. -9
C. 0
D. 9
E. 39
Kudos points for each correct solution!!![/quote]

Rearranging the given equation we get
-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 when x= 2 and y = -3, then the maximum value of the whole expression is 0 + 0 -9 = -9
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Re: New Algebra Set!!! [#permalink]

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26 Mar 2013, 04:39
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
Kudos points for each correct solution!!![/quote]

On Re-arranging the given equation we get m = - x^2 - 2x +15

For m to be positive - x^2 - 2x +15 should be greater then 0. Solving we get (x + 5)(x - 3)<0. Since x is an integer we get the values of x as -4 , -3, -2, -1, 0, 1, 2 (total 7 values)

Given that X is an integer from -10 and 10, inclusive (21 values)

So, the probability is 7/21=1/3.

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Re: New Algebra Set!!! [#permalink]

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26 Mar 2013, 04:44
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
Kudos points for each correct solution!!![/quote]

Rearranging we get m^2n^2 + mn - 12 = 0 => (mn + 4)(mn - 3)= 0 =>mn = -4 or mn = 3 => m= -4/n or 3/n
E. I and III only
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Re: New Algebra Set!!! [#permalink]

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26 Mar 2013, 04:55
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
Kudos points for each correct solution!!![/quote]

Rearranging the equation we get x^4 - 29x^2 + 100 = 0 => (x^2 - 25)(x^ - 4) =0 => x= +5, -5, +2, -2
From these values of x we can get -50 (-5 * +5 * +2), +50(-5 * +5 * -2) but we can't get 25

B. II only
Re: New Algebra Set!!!   [#permalink] 26 Mar 2013, 04:55

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