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by GMATinsight » Sat Jul 19, 2014 3:58 am
[email protected] wrote:Dint get how the answer is C
Question : What is the Value of x?

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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by GMATGuruNY » Sat Jul 19, 2014 7:02 am
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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by [email protected] » Sat Jul 19, 2014 9:48 am
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

GMATinsight wrote:
[email protected] wrote:Dint get how the answer is C
Question : What is the Value of x?

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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by GMATinsight » Sat Jul 19, 2014 6:34 pm
[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
substituting the value of y in other equation obtained from statement 2
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)
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by [email protected] » Sat Jul 19, 2014 8:27 pm
Hi Mitch,

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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by GMATGuruNY » Sun Jul 20, 2014 6:11 am
[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?
z can be any value, but the LEAST possible value for z² is 0.
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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