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BrentGMATPrepNow
GeorgeKo111
Is y < z ?

(1) y + z = 1
(2) y² < z²

Target question: Is y < z ?

Statement 1: y + z = 1
Let's TEST some values.
There are several values of y and z that satisfy statement 1. Here are two:
Case a: y = 0 and z = 1. In this case, the answer to the target question is YES, y is less than z
Case b: y = 1 and z = 0. In this case, the answer to the target question is NO, y is not less than z
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y² < z²
There are several values of y and z that satisfy statement 2. Here are two:
Case a: y = 0 and z = 1. In this case, the answer to the target question is YES, y is less than z
Case b: y = 0 and z = -1. In this case, the answer to the target question is NO, y is not less than z
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 2 tells us that: y² < z²
Subtract z² from both sides of the inequality to get: y² - z² < 0
Factor the left side to get: (y + z)(y - z) < 0
Since statement 1 tells us that y + z = 1, we can replace (y + z) with 1 to get: (1)(y - z) < 0
Simplify: y - z < 0
Add z to both sides to get: y < z
The answer to the target question is YES, y is less than z
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

RELATED VIDEO FROM MY COURSE]

Hi Brent,

Can't we answer the question from statement 2 itself?

y^2-z^2<0
(y+z)(y-z)<0
-z<y<z
So y in any case is less than z from second statement itself, isn't that so?
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Quote:
Is y < z ?
Step 1: Understanding statement 1 alone
(1) y + z = 1
When y = 0.4, z = 0.6; y < z
When y = 0.6, x = 0.4; y > z
Insufficient

Step 2: Understanding statement 2 alone
(2) y^2 < z^2
Taking square root on both the sides
\(\sqrt{y^2}\) < \(\sqrt{z^2}\)
|y| < |z|
When y = -1, z = 2; y < z
When y = 1, z = -2; y > z
Insufficient

Step 3: Combining statement 1 and 2
y^2 < z^2
y^2 - z^2 < 0
(y - z) ( y+z) < 0
As (y + z) = 1; (y - z) < 0
Hence, y < z
Sufficient

C is correct
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Alternate approach:

GeorgeKo111
Is y < z ?

(1) y + z = 1

(2) y^2 < z^2

Statement 1:
Clearly INSUFFICIENT.

Statement 2, rephrased: |y| < |z|
Case 1: y=0 and z=1
In this case, the answer to the question stem is YES.
Case 2: y=0 and z=-1
In this case, the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

Statements combined:
Substituting z=1-y into |y| < |z|, we get:
|y| < |1-y|
Implication:
The distance between y and 0 is less than the distance between y and 1.
In other words, Y IS CLOSER TO 0 THAN TO 1, implying that y < 1/2.

Substituting y=1-z into |y| < |z|, we get:
|1-z| < |z|
Implication:
The distance between z and 1 is less than the distance between z and 0.
In other words, Z IS CLOSER TO 1 THAN TO 0, implying that z > 1/2.

Since y < 1/2 and z > 1/2, we get:
y < 1/2 < z
y < z
Thus, the answer to the question stem is YES.
SUFFICIENT.

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I have solved this problem like this
Statement 1 says y+z=1, y=1-z or z=1-y Not Sufficient at all.

Statement 2 says y^2 < z^2

Putting value of y(1-z) in y^2<z^2 = 1-2z+z^2, 1/2<z

Similarly putting value of Z(z=1-y) in y^2<z^2, the result will be y<1/2.

So now after joining them it will give y<1/2<z. Therefore the statements combined are sufficient.

Posted from my mobile device
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We want to know if z-y is positive. Statement 1 tells us that z+y is positive, and Statement 2 tells us that z^2 - y^2 is positive. But z^2 - y^2 = (z + y)(z-y), and if z+y is positive, then z-y must also be positive for it to be true that the product (z-y)(z+y) is positive. So the two Statements together are sufficient.

Statement 1 is clearly insufficient alone, because it's symmetric in y and z (if you can find numbers that work for y and z where y < z, you can just flip the numbers and you'll automatically have a situation where z < y). Statement 2 is clearly insufficient alone also, since if y = 0, z can be -100 or +100. So the answer is C.
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Is \(y < z\) ?

(1) \(y + z = 1\\
\)
Clearly insufficient.

(2) \(y^2 < z^2\)

\(y^2 - x^2 < 0\)
\((y+z)(y-z) < 0\)

\((y+z) and (y-z)\) have different signs. We can't conclude y <z, however. INSUFFICIENT.

(1&2)
If \(y+z = 1, then y-z < 0\)
\(y < z\).

SUFFICIENT.

Answer is C.
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