New Algebra Set!!!

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

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Updated on: 20 Mar 2013, 08:40

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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Originally posted by ConnectTheDots on 18 Mar 2013, 11:25.
Last edited by ConnectTheDots on 20 Mar 2013, 08:40, edited 1 time in total.

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

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18 Mar 2013, 11:55

@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]

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18 Mar 2013, 13:00

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]

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18 Mar 2013, 13:05

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]

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18 Mar 2013, 13:11

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]

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18 Mar 2013, 13:17

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]

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18 Mar 2013, 13:33

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]

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18 Mar 2013, 13:42

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]

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18 Mar 2013, 13:47

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

Lets just start with the bottom and solve
-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]

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18 Mar 2013, 13:50

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]

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18 Mar 2013, 13:53

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]

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18 Mar 2013, 16:15

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]

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19 Mar 2013, 01:15

souvik101990 wrote:

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

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

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19 Mar 2013, 01:48

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]

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19 Mar 2013, 01:57

Archit143 wrote:

souvik101990 wrote:

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]

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Updated on: 19 Mar 2013, 05:40

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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Originally posted by mau5 on 19 Mar 2013, 02:17.
Last edited by mau5 on 19 Mar 2013, 05:40, edited 1 time in total.

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

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19 Mar 2013, 02:25

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]

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19 Mar 2013, 02:30

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]

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Updated on: 19 Mar 2013, 17:10

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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Originally posted by WoundedTiger on 19 Mar 2013, 07:05.
Last edited by WoundedTiger on 19 Mar 2013, 17:10, edited 1 time in total.

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

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19 Mar 2013, 10:57

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?

Re: New Algebra Set!!! &nbs [#permalink] 19 Mar 2013, 10:57

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