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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]  19 Mar 2013, 01:25
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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!!! [#permalink]  19 Mar 2013, 01:30
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7. $$x^4 = 29x^2 - 100$$

or $$x^4 - 29x^2+100 = 0$$

Let $$x^2 = t$$, thus

$$t^2-29+100 = 0$$

or (t-25)(t-4) = 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!!! [#permalink]  19 Mar 2013, 06:05
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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^2-1= (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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Last edited by WoundedTiger on 19 Mar 2013, 16:10, edited 1 time in total.
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 07:54
my first post
1-C
2-B
3-C
4-B
5-E
6-D
7-C
8-D
9-A
10-A
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 09:57
GyanOne wrote:
4. -3x^2 + 12x -2y^2 - 12y - 39 = 3(-x^2 + 4x - 13) + 2(-y^2-6y)

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?
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 10:18
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NeverQuit wrote:
GyanOne wrote:
4. -3x^2 + 12x -2y^2 - 12y - 39 = 3(-x^2 + 4x - 13) + 2(-y^2-6y)

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?

$$-3x^2 + 12x -2y^2 - 12y - 39$$

$$=-3(x^2-4x)-2(y^2+6y) -39$$

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

$$=-3(x-2)^2 +12 -2(y+3)^2 +18 -39$$

NOW The above expression will be minimum IFF $$(x-2)^2=0$$ and $$(y+3)^2=0$$

which leaves the maximum value to be $$12+18-39=-9$$
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 16:15
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9. If x=(\sqrt{5}-\sqrt{7})^2, then the best approximation of x is
Ans: \sqrt{5} is aproximately 2.236
and \sqrt{7}) will be around ~ 2.6 therefore

we have x= (2.236 - 2.6)^2 -----> (-0.364)^2-------> Approximation of X is 0.
Ans Option A
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 16:18
8.If m is a negative integer and m^3 + 380 = 381m , then what is the value of m?

Ans: Plug in in the answer choices and only option m=-21 satisfies.
So option B
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 16:20
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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

Sol: In the given eqn let X^2= y.The eqn becomes

y^2 - 29y +100=0, Solving for "y" we get
y= 25 or y= 4
This implies
X^2 = 25 or X^2= 4
Possible values of x will be -5,5 -2 or 2.
Only 25 is not possible as product of three factors and hence Ans should be B
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 16:22
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

Sol: The given eqn can be written as mn (mn + 1) = 12
Now mn, mn+ will have to be consecutive Integers and possible combinations are

mn=3 and mn+ 1= 4 ------> possible value of m will be 3/n
or mn+1 = -4 and mn= -3 -------. possible value of m will be -3/n

Ans should be III only i.e Option C
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 16:32
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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

This was little trickier but here is my approach

We need to know Probability that m is greater than 0.
Integer is from -10 to 10 inclusive.
Consider value of m greater than 0 only and put in the value in the given eqn

For ex let us say m= 2, the eqn becomes
x^2+2x -15= -2----. x^2+ 2x -13=0 -----> Eqn will have roots of the form -b+/- \sqrt{b^2-4ac}/ 2

We see that for m= 7, we get the eqn as x^2+ 2x -8 =0, Solving for x we get,
x = 4 or -2.

This seems to be the only case where the eqn gives us 2 real roots and therefore looking at option choices selected B (assuming other 2 cases will be for value of m between -10 to 0)

Ans Choice B
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 16:39
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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

Sol: Adding the 2 equations we get a= \sqrt{(m+n)/2}
Subtracting Eqn 2 from 1, we get b= \sqrt{m-n/2}

Ans Option C
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 16:40
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

Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get

3^2 +3a+15=0----> Solving for a we get a=-8.

Putting the value if a in Eqn 1 we get
x^2 -8x-b=0
We know for Eqn ax^2 +bx+c=-0
Sum of roots is given by -b/a
Product of roots is c/a

Therefore from Eqn1 (after subsitution of value of a) we get
Sum of roots = -8 and roots are equal ie. -4 and -4
Product of roots will be 16. Therefore Option B
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 17:03
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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 = 29x^2 - 100
x^4-29x^2+100=0
(x^2-25)(x^2-4)=0
(x-5)(x+5)(x+2)(x-2)=0
roots=5, -5, -2, 2

I. -50 =(2, 5, -5)
II. 25 not possible
III. 50 =(-2, 5, -5)

Answer is
B. II only
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 17:14
GyanOne wrote:
Edited

1. x^4 = x^3 + 6x^2
=> x^2 (x^2 - x - 6) = 0
The roots of x^2 - x - 6 are -2 and 3, but -2 cannot be the value of x. So 3 and 0 are the only possible roots.
=> Sum of all possible solutions = 3.

Option D

why -2 cannot equal to x?
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Re: New Algebra Set!!! [#permalink]  19 Mar 2013, 17:38
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Bunuel wrote:
Please suggest on what category would you like the next set to be. Thank you!

Hello Bunuel,

I do like to see Questions on Geometry or Time Speed and Distance

Thanks
Mridul
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Re: New Algebra Set!!! [#permalink]  20 Mar 2013, 01:00
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1. If , then the sum of all possible solutions for x is:

A. -2
B. 0
C. 1
D. 3
E. 5

Soln: Possible solutions are x = -2 & 3, Ans(C)

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

Soln: Solving for a = -8, Ans (B)
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Re: New Algebra Set!!! [#permalink]  20 Mar 2013, 01:32
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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.
B.
C.
D.
E.

Soln: Take equation 2, square on both sides, substitute for a^2 + b^2 with m, Ans (C)

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

Soln: Used a 'gestimation' method. Since the unit digit of the resulting values on both sides has to be same, the only number from the options that will fit this condition is -20. Ans (B)
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Re: New Algebra Set!!! [#permalink]  20 Mar 2013, 01:40
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9. If , then the best approximation of x is:

A. 0
B. 1
C. 2
D. 3
E. 4

Soln: x = 12 - 2√35, if √35 =~ 6, then x =~ 0, Ans (A)

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

Soln: Solving for f(x) & g(x), 2n^2 - 1 = (n+12)^2, Solving for n gives n = 29 or -5, Ans (A)
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Re: New Algebra Set!!! [#permalink]  20 Mar 2013, 07:58
1. If x = (x^3 + 6x^2)^1/4, then the sum of all possible solutions for x is:

A. -2
B. 0
C. 1
D. 3
E. 5

My take ==> C

Approach:

x = (x^3 + 6x^2)^1/4
x^4 = x^3 + 6x^2
x^4 - x^3 - 6x^2 = 0
x^2(x^2 - x - 6) = 0
x^2 * (x-3) * (x+2) = 0

So Sum of all the possible solution of x = 1

Quote:
Thanks
Re: New Algebra Set!!!   [#permalink] 20 Mar 2013, 07:58

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