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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:

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 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Last edited by WoundedTiger on 19 Mar 2013, 16:10, edited 1 time in total.

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?

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 _________________

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 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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.

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)