Algebra Equation
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Question : What is the Value of x?[email protected] wrote:Dint get how the answer is C
Statement 1) x+y = 2
No clues about y therefore the value of x can't be calculated
Insufficient
Statement 2)xy = z^2+1
No clues about y and z therefore the value of x can't be calculated
Insufficient
Combining the two statements
[xy = z^2+1] and [x+y=2]
x+y=2
i.e. y = 2-x
substituting the value of y in other equation obtained from statement 2
x(2-x) = z^2+1
x(x-2) = -(z^2+1)
i.e. product of two numbers separated by 2 = Negative number with absolute value Greater than or equal to 1 [because (z^2+1)>1]
Also
One of the two numbers (x) must be positive and other one of them (x-2) must be -ve
which is true only if x = 1
[Because if you take any value of x which is between 0 and 1 then the product of x and x-2 will be negative but absolute value of their product will not be greater than 1]
therefore SUFFICIENT
Answer: Option C
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What is the value of x?
Statement 1: x+y = 2
Case 1: x=1, y=1
Case 2: x=0, y=2
Since x can be different values, INSUFFICIENT.
Statement 2: xy = z²+1
The smallest possible value for z² is 0.
If z²=0, then xy = 1.
Case 1 also satisfies statement 2.
In this case, x=1.
Case 3: x=2, y=1/2
Since x can be different values, INSUFFICIENT.
Statements combined:
Case 1 satisfies both statements.
In this case, x=1.
Statement 2 implies the following:
Since z² ≥ 0, xy ≥ 1.
Thus, when the statements are combined, the following constraints must both be satisfied:
x+y = 2.
xy ≥ 1.
Other than Case 1, no other combination for x and y will satisfy both constraints.
If x=3/2 and y=1/2, then xy = 3/4, which is not greater than or equal to 1.
If x=1/4 and y=7/4, then xy = 7/16, which is not greater than or equal to 1.
Since only Case 1 is possible, x=1.
The correct answer is C.
Statement 1: x+y = 2
Case 1: x=1, y=1
Case 2: x=0, y=2
Since x can be different values, INSUFFICIENT.
Statement 2: xy = z²+1
The smallest possible value for z² is 0.
If z²=0, then xy = 1.
Case 1 also satisfies statement 2.
In this case, x=1.
Case 3: x=2, y=1/2
Since x can be different values, INSUFFICIENT.
Statements combined:
Case 1 satisfies both statements.
In this case, x=1.
Statement 2 implies the following:
Since z² ≥ 0, xy ≥ 1.
Thus, when the statements are combined, the following constraints must both be satisfied:
x+y = 2.
xy ≥ 1.
Other than Case 1, no other combination for x and y will satisfy both constraints.
If x=3/2 and y=1/2, then xy = 3/4, which is not greater than or equal to 1.
If x=1/4 and y=7/4, then xy = 7/16, which is not greater than or equal to 1.
Since only Case 1 is possible, x=1.
The correct answer is C.
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- Master | Next Rank: 500 Posts
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Hello Bhoopendra,
How did we derive the below:
substituting the value of y in other equation obtained from statement 2
x(2-x) = z^2+1
x(x-2) = -(z^2+1)
Thanks
How did we derive the below:
substituting the value of y in other equation obtained from statement 2
x(2-x) = z^2+1
x(x-2) = -(z^2+1)
Thanks
GMATinsight wrote:Question : What is the Value of x?[email protected] wrote:Dint get how the answer is C
Statement 1) x+y = 2
No clues about y therefore the value of x can't be calculated
Insufficient
Statement 2)xy = z^2+1
No clues about y and z therefore the value of x can't be calculated
Insufficient
Combining the two statements
[xy = z^2+1] and [x+y=2]
x+y=2
i.e. y = 2-x
substituting the value of y in other equation obtained from statement 2
x(2-x) = z^2+1
x(x-2) = -(z^2+1)
i.e. product of two numbers separated by 2 = Negative number with absolute value Greater than or equal to 1 [because (z^2+1)>1]
Also
One of the two numbers (x) must be positive and other one of them (x-2) must be -ve
which is true only if x = 1
[Because if you take any value of x which is between 0 and 1 then the product of x and x-2 will be negative but absolute value of their product will not be greater than 1]
therefore SUFFICIENT
Answer: Option C
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substituting the value of y in other equation obtained from statement 2[email protected] wrote:Hello Bhoopendra,
How did we derive the below:
substituting the value of y in other equation obtained from statement 2
x(2-x) = z^2+1
x(x-2) = -(z^2+1)
Thanks
x(2-x) = z^2+1
If you take Negative sign common from (2-x)
you get
(x)[-(x-2) = z^2+1
- x(x-2) = z^2+1
Change sign on both sides
i.e. x(x-2) = -(z^2+1)
"GMATinsight"Bhoopendra Singh & Sushma Jha
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Whatsapp/Mobile: +91-9999687183 l [email protected]
Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
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Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
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Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
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Hi Mitch,
I dint get why we will assume z to be 0 or why should we take the minimum value for z?
I dint get why we will assume z to be 0 or why should we take the minimum value for z?
GMATGuruNY wrote:What is the value of x?
Statement 1: x+y = 2
Case 1: x=1, y=1
Case 2: x=0, y=2
Since x can be different values, INSUFFICIENT.
Statement 2: xy = z²+1
The smallest possible value for z² is 0.
If z²=0, then xy = 1.
Case 1 also satisfies statement 2.
In this case, x=1.
Case 3: x=2, y=1/2
Since x can be different values, INSUFFICIENT.
Statements combined:
Case 1 satisfies both statements.
In this case, x=1.
Statement 2 implies the following:
Since z² ≥ 0, xy ≥ 1.
Thus, when the statements are combined, the following constraints must both be satisfied:
x+y = 2.
xy ≥ 1.
Other than Case 1, no other combination for x and y will satisfy both constraints.
If x=3/2 and y=1/2, then xy = 3/4, which is not greater than or equal to 1.
If x=1/4 and y=7/4, then xy = 7/16, which is not greater than or equal to 1.
Since only Case 1 is possible, x=1.
The correct answer is C.
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z can be any value, but the LEAST possible value for z² is 0.[email protected] wrote:Hi Mitch,
I dint get why we will assume z to be 0 or why should we take the minimum value for z?
When a statement implies a least possible value, we should consider how this value might constrain the problem.
Thus, when we evaluate statement 2, we should determine how the least possible value for z² constrains the value of x.
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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].
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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].
Student Review #1
Student Review #2
Student Review #3