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# New Algebra Set!!!

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Current Student
Joined: 19 Mar 2012
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19 Mar 2013, 11:18
1
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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19 Mar 2013, 17:15
1
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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19 Mar 2013, 17:18
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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19 Mar 2013, 17:20
1
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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19 Mar 2013, 17:22
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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19 Mar 2013, 17:32
1
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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19 Mar 2013, 17:39
1
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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19 Mar 2013, 17:40
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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19 Mar 2013, 18:03
1
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)

B. II only
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20 Mar 2013, 02:00
1
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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20 Mar 2013, 02:32
1
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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20 Mar 2013, 02:40
1
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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20 Mar 2013, 08:58
1
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:
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21 Mar 2013, 10:28
1
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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21 Mar 2013, 10:30
1
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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21 Mar 2013, 10:37
1
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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21 Mar 2013, 12:58
[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
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21 Mar 2013, 14:10
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, 22:11
1
1
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, 05:10
5
25
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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Re: New Algebra Set!!! &nbs [#permalink] 22 Mar 2013, 05:10

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