If ab>0 and (a^2)*(b^2) + 2ab - 3 = 0, which of the following could be a value of a?
I. -1
II. 1
III. 3
(A) I only
(B) II only
(C) I and II only
(D) II and III
(E) I, II and III
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If ab>0 and (a^2)*(b^2) + 2ab – 3 = 0
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TheGmatTutor
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Given: (a²)(b²) + 2ab - 3 = 0TheGmatTutor wrote:If ab>0 and (a^2)*(b^2) + 2ab - 3 = 0, which of the following could be a value of a?
I. -1
II. 1
III. 3
(A) I only
(B) II only
(C) I and II only
(D) II and III
(E) I, II and III
Rewrite as: (ab)² + 2ab - 3 = 0
Let's use a technique called u-substitution.
Let u = ab
So, take (ab)² + 2ab - 3 = 0 and replace ab with u to get: u² + 2u - 3 = 0
Factor: (u + 3)(u - 1) = 0
Solve: u = -3 or u = 1
Since u = ab, we can conclude that ab = -3 or ab = 1
Since we're told that ab > 0, we can eliminate the case that ab = -3, which means ab = 1
Now check the options:
I. a = -1. If ab = 1, and a = -1, then b = -1 (perfect - this adheres to the given info)
If we plug those values of a and b into the original equation, everything works
II. a = 1. If ab = 1, and a = 1, then b = 1 (perfect - this adheres to the given info)
If we plug those values of a and b into the original equation, everything works
III. a = 3. If ab = 1, and a = 3, then b = 1/3 (perfect - this adheres to the given info)
If we plug those values of a and b into the original equation, everything works
So, a can equal any of those values.
Answer: E
Cheers,
Brent
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vineet.nitd
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Solve the quadratic in ab to get the values of ab as -3 and 1. However, as per problem ab>0, hence ab=-3 can be discarded.
So, ab = 1.
Now, since there is no restriction on a or b belonging only to set of integers, we can assign any value to 'a' (except 0), such that we always have its multiplicative inverse in b, such that ab=1 is always satisfied.
Hence, option E is the right choice.
So, ab = 1.
Now, since there is no restriction on a or b belonging only to set of integers, we can assign any value to 'a' (except 0), such that we always have its multiplicative inverse in b, such that ab=1 is always satisfied.
Hence, option E is the right choice.
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GMATGuruNY
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An alternate approach is to test I, II and III in the original quadratic.TheGmatTutor wrote:If ab>0 and (a^2)*(b^2) + 2ab - 3 = 0, which of the following could be a value of a?
I. -1
II. 1
III. 3
(A) I only
(B) II only
(C) I and II only
(D) II and III
(E) I, II and III
I: a=-1
Plugging a=-1 into a²b² + 2ab - 3 = 0, we get:
(-1)²b² + 2(-1)b - 3 = 0
b² - 2b - 3 = 0
(b-3)(b+1) = 0.
Here, it's possible that b+1=0, implying that b=-1 and that ab = (-1)(-1) = 1.
II: a=1
Plugging a=1 into a²b² + 2ab - 3 = 0, we get:
1²b² + 2(1)b - 3 = 0
b² + 2b - 3 = 0
(b+3)(b-1) = 0.
Here, it's possible that b-1=0, implying that b=1 and that ab = 1*1 = 1.
III: a=3
Plugging a=3 into a²b² + 2ab - 3 = 0, we get:
3²b² + 2(3)b - 3 = 0
3b² + 2b - 1 = 0
(3b-1)(b+1) = 0.
Here, it's possible that 3b-1=0, implying that b=1/3 that ab = (3)(1/3) = 1.
Since I, II and III are all possible, the correct answer is E.
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Hi All,
The above prompt is remarkably similar to the following GMAC question:
If xy ≠0 and x^2*y^2 - xy = 6, which of the following could be y in terms of x?
I. 1/(2x)
II. - 2/x
III. 3/x
A. I only
B. II only
C. I and II
D. I and III
E. II and III
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
The above prompt is remarkably similar to the following GMAC question:
If xy ≠0 and x^2*y^2 - xy = 6, which of the following could be y in terms of x?
I. 1/(2x)
II. - 2/x
III. 3/x
A. I only
B. II only
C. I and II
D. I and III
E. II and III
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
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Here, we can plug in a value for x and solve for y.If xy ≠0 and x²y² - xy = 6, which of the following could be y in terms of x?
A. I only
B. II only
C. I and II
D. I and III
E. II and III
Let x=1.
Plugging x=1 into x²y² - xy = 6, we get:
1²y² - 1y = 6
y² - y - 6 = 0
(y+2)(y-3) = 0
y=-2 or y=3.
Since the question stem asks for possible values of y, our targets for y are -2 and 3.
Now plug x=1 into I, II and III to see whether any are equal to -2 or 3.
I. 1/(2x) = 1/(2*1) = 1/2.
II. - 2/x = -2/1 = -2.
III. 3/x = 3/1 = 3.
Since II and III yield our target values for y, the correct answer is E.
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Hi All,
This question is quirky in that it tests you on math rules and patterns that you probably know, but in ways that you're not used to thinking about...
We're told that neither X nor Y are equal to 0. We're also told that (X^2)(Y^2) - XY = 6. We're asked which of the following COULD be the value of Y in terms of X...
The first interesting thing about this question is the use of the word COULD....that word implies that there's MORE THAN ONE possible solution.
The second interesting thing is that the 'term' (XY) can be factored out of the 'left side' of the equation. Normally, you look to factor our a single variable or number, but here, it's the product of two variables that you can factor out. Doing so gives us...
XY(XY - 1) = 6
While this looks complicated, there's an easy pattern here:
(number)(number - 1) = 6
Can you think of 2 numbers, that differ by 1, that you can multiply to get 6?
You should be thinking 2 and 3... because (3)(3-1) = 6
So XY = 3 is a possible solution. In this case, Y = 3/X. The wording of the prompt makes me think that there should be another solution though, so is there ANOTHER pair of numbers, that differ by 1, that you can multiply together to get 6? Hint: the numbers do NOT have to be positive....
How about -2 and -3....
(-2)(-2-1) = 6
So XY = -2 is another possible solution. In this case, Y = -2/X
There's only one answer that includes both of those solutions...
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
This question is quirky in that it tests you on math rules and patterns that you probably know, but in ways that you're not used to thinking about...
We're told that neither X nor Y are equal to 0. We're also told that (X^2)(Y^2) - XY = 6. We're asked which of the following COULD be the value of Y in terms of X...
The first interesting thing about this question is the use of the word COULD....that word implies that there's MORE THAN ONE possible solution.
The second interesting thing is that the 'term' (XY) can be factored out of the 'left side' of the equation. Normally, you look to factor our a single variable or number, but here, it's the product of two variables that you can factor out. Doing so gives us...
XY(XY - 1) = 6
While this looks complicated, there's an easy pattern here:
(number)(number - 1) = 6
Can you think of 2 numbers, that differ by 1, that you can multiply to get 6?
You should be thinking 2 and 3... because (3)(3-1) = 6
So XY = 3 is a possible solution. In this case, Y = 3/X. The wording of the prompt makes me think that there should be another solution though, so is there ANOTHER pair of numbers, that differ by 1, that you can multiply together to get 6? Hint: the numbers do NOT have to be positive....
How about -2 and -3....
(-2)(-2-1) = 6
So XY = -2 is another possible solution. In this case, Y = -2/X
There's only one answer that includes both of those solutions...
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
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mallika hunsur
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Hi Rich, Mitch,[email protected] wrote:Hi All,
This question is quirky in that it tests you on math rules and patterns that you probably know, but in ways that you're not used to thinking about...
We're told that neither X nor Y are equal to 0. We're also told that (X^2)(Y^2) - XY = 6. We're asked which of the following COULD be the value of Y in terms of X...
The first interesting thing about this question is the use of the word COULD....that word implies that there's MORE THAN ONE possible solution.
The second interesting thing is that the 'term' (XY) can be factored out of the 'left side' of the equation. Normally, you look to factor our a single variable or number, but here, it's the product of two variables that you can factor out. Doing so gives us...
XY(XY - 1) = 6
While this looks complicated, there's an easy pattern here:
(number)(number - 1) = 6
Can you think of 2 numbers, that differ by 1, that you can multiply to get 6?
You should be thinking 2 and 3... because (3)(3-1) = 6
So XY = 3 is a possible solution. In this case, Y = 3/X. The wording of the prompt makes me think that there should be another solution though, so is there ANOTHER pair of numbers, that differ by 1, that you can multiply together to get 6? Hint: the numbers do NOT have to be positive....
How about -2 and -3....
(-2)(-2-1) = 6
So XY = -2 is another possible solution. In this case, Y = -2/X
There's only one answer that includes both of those solutions...
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
I am thinking-
xy(xy-1)=6 can be
1) xy=6
2) xy-1=6 or xy=7
so y=6/x or 7/x
Why is this line of reasoning not working..?
Thanks,
Mallika
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If we substitute xy=6 into xy(xy-1)=6, we get:mallika hunsur wrote:I am thinking-
xy(xy-1)=6 can be
1) xy=6
2) xy-1=6 or xy=7
Why is this line of reasoning not working..?
6(6-1) = 6
30 = 6.
Doesn't work.
If we substitute xy=7 into xy(xy-1)=6, we get:
7(7-1) = 6
42 = 6.
Doesn't work.
One way to avoid incorrect solutions is to solve as I did in my post above -- by plugging in an easy value for x and solving for y.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Hi Mallika,
Mitch has properly explained why the (X)(Y) cannot be 6 or 7. I'd like to know more about the logic you used to get to that point though. What did you 'see' in that original equation that made you think that (X)(Y) might equal 6?
GMAT assassins aren't born, they're made,
Rich
Mitch has properly explained why the (X)(Y) cannot be 6 or 7. I'd like to know more about the logic you used to get to that point though. What did you 'see' in that original equation that made you think that (X)(Y) might equal 6?
GMAT assassins aren't born, they're made,
Rich
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mallika hunsur
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Hi Rich,[email protected] wrote:Hi Mallika,
Mitch has properly explained why the (X)(Y) cannot be 6 or 7. I'd like to know more about the logic you used to get to that point though. What did you 'see' in that original equation that made you think that (X)(Y) might equal 6?
GMAT assassins aren't born, they're made,
Rich
I have no clue now why i did that. I should've just assigned xy=a and solved a quadratic equation.
Need to get back into the groove
Thanks,
Mallika
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