Oct 22 08:00 AM PDT  09:00 AM PDT Join to learn strategies for tackling the longest, wordiest examples of Counting, Sets, & Series GMAT questions Oct 22 09:00 AM PDT  10:00 AM PDT Watch & learn the Do's and Don’ts for your upcoming interview Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss! Oct 26 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes. Oct 27 07:00 AM EDT  09:00 AM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Current Student
Joined: 19 Mar 2012
Posts: 4263
Location: India
GPA: 3.8
WE: Marketing (NonProfit and Government)

Re: New Algebra Set!!!
[#permalink]
Show Tags
19 Mar 2013, 11:18
NeverQuit wrote: GyanOne wrote: 4. 3x^2 + 12x 2y^2  12y  39 = 3(x^2 + 4x  13) + 2(y^26y)
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? \(3x^2 + 12x 2y^2  12y  39\) \(=3(x^24x)2(y^2+6y) 39\) \(=3(x^24x+44)2(y^2+6y+99)39\) \(=3(x2)^2 +12 2(y+3)^2 +18 39\) NOW The above expression will be minimum IFF \((x2)^2=0\) and \((y+3)^2=0\) which leaves the maximum value to be \(12+1839=9\)
_________________



Director
Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Algebra Set!!!
[#permalink]
Show Tags
19 Mar 2013, 17:15
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
_________________
“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.”



Director
Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Algebra Set!!!
[#permalink]
Show Tags
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
_________________
“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.”



Director
Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Algebra Set!!!
[#permalink]
Show Tags
19 Mar 2013, 17:20
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.”



Director
Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Algebra Set!!!
[#permalink]
Show Tags
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
_________________
“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.”



Director
Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Algebra Set!!!
[#permalink]
Show Tags
19 Mar 2013, 17:32
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^24ac}/ 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.”



Director
Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Algebra Set!!!
[#permalink]
Show Tags
19 Mar 2013, 17:39
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{mn/2} Ans Option C
_________________
“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.”



Director
Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Algebra Set!!!
[#permalink]
Show Tags
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 8xb=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.”



Intern
Joined: 18 Feb 2013
Posts: 29

Re: New Algebra Set!!!
[#permalink]
Show Tags
19 Mar 2013, 18:03
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^429x^2+100=0 (x^225)(x^24)=0 (x5)(x+5)(x+2)(x2)=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



Intern
Joined: 04 Sep 2012
Posts: 14
Location: India
WE: Marketing (Consumer Products)

Re: New Algebra Set!!!
[#permalink]
Show Tags
20 Mar 2013, 02:00
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)



Intern
Joined: 04 Sep 2012
Posts: 14
Location: India
WE: Marketing (Consumer Products)

Re: New Algebra Set!!!
[#permalink]
Show Tags
20 Mar 2013, 02:32
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)



Intern
Joined: 04 Sep 2012
Posts: 14
Location: India
WE: Marketing (Consumer Products)

Re: New Algebra Set!!!
[#permalink]
Show Tags
20 Mar 2013, 02:40
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)



Intern
Joined: 14 Jan 2011
Posts: 3

Re: New Algebra Set!!!
[#permalink]
Show Tags
20 Mar 2013, 08:58
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 * (x3) * (x+2) = 0 So Sum of all the possible solution of x = 1 Quote: Thanks



Intern
Joined: 29 Jan 2013
Posts: 21
Location: United States
Concentration: Strategy
GPA: 3.4
WE: Analyst (Consulting)

Re: New Algebra Set!!!
[#permalink]
Show Tags
21 Mar 2013, 10:28
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



Intern
Joined: 29 Jan 2013
Posts: 21
Location: United States
Concentration: Strategy
GPA: 3.4
WE: Analyst (Consulting)

Re: New Algebra Set!!!
[#permalink]
Show Tags
21 Mar 2013, 10:30
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 (n29)(n+5)=0 n= 29 or 5 29*(5) = 145  Answer A



Intern
Joined: 29 Jan 2013
Posts: 21
Location: United States
Concentration: Strategy
GPA: 3.4
WE: Analyst (Consulting)

Re: New Algebra Set!!!
[#permalink]
Show Tags
21 Mar 2013, 10:37
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 ~= 1212 = 0 Answer A



Intern
Joined: 22 Aug 2012
Posts: 17
GMAT Date: 09022013
GPA: 3
WE: Information Technology (Computer Software)

Re: New Algebra Set!!!
[#permalink]
Show Tags
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
\(x^4 = 29x^2  100\) \(=> x^429x^2+100 = 0\) \(=> x^4  25x^2  4x^2 + 100=0\) \(=> x^2(x^225)  4(x^225) = 0\) \(=>(x^2  4)(x^225) = 0\) => 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 and answer is 25



Manager
Joined: 20 Dec 2011
Posts: 73

Re: New Algebra Set!!!
[#permalink]
Show Tags
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(m15)>0 and 44(m15) must be a perfect square => 4 (16m) must be a perfect square and >0 => 16m 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.



Intern
Joined: 24 Dec 2012
Posts: 45

Re: New Algebra Set!!!
[#permalink]
Show Tags
21 Mar 2013, 22:11
7  B
x^4=29x^2100 x^429x^2+100=0 which is equal to
(x2)(x+2)(x5)(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.



Math Expert
Joined: 02 Sep 2009
Posts: 58398

Re: New Algebra Set!!!
[#permalink]
Show Tags
22 Mar 2013, 05:10
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: \(x^28xb=0\). Now, we know that it has equal roots thus its discriminant must equal to zero: \(d=(8)^2+4b=0\). Solving for \(b\) gives \(b=16\). Answer: B.
_________________




Re: New Algebra Set!!!
[#permalink]
22 Mar 2013, 05:10



Go to page
Previous
1 2 3 4 5 6 7 8
Next
[ 145 posts ]



