harder problem solving questions

[bek_gmat]

I need help with following questions. Explanations are appreciated.


1. If x, y, and z are nonzero integers and x > yz, which of the following must be true?

I. x/y > z
II. x/z > 1
III. x/yz > 1

A None of the above
B I only
C III only
D I and II only
E I, II, and III

[Frankenstein]

Hi,
x>yz. Dividing both sides with y gives (I) only if y is positive.
Similarly, we can check for II and III.

You can pick a counter example as well.Consider x=1,y=-2,z=3. None of the 3 equations satisfy.

Hence, A

[bek_gmat]

Hi, thanks Frankenstein,

Here I'm posting the second problematic question:

2. A circle is inscribed in triangle ABC such that point D lies on the circle and on line segment AC, point E lies on the circle and on line segment AB, and point F lies on the circle and on line
segment BC. If line segment AB = 6, what is the area of the figure created by line segments AD, AE, and minor arc DE?
A 3√3- 9π/4
B 3√3- π
C 6√3- π
D 9√3- 3π
E It cannot be determined from the information given.

[GMATGuruNY]

A circle is inscribed in EQUILATERAL triangle ABC such that point D lies on the circle and on line segment AC, point E lies on the circle and on line segment AB, and point F lies on the circle and on line segment BC. If line segment AB = 6, what is the area of the figure created by line segments AD, AE, and minor arc DE?

A 3√3- 9π/4
B 3√3- π
C 6√3- π
D 9√3- 3π
E It cannot be determined from the information given.

The question is missing a key word: equilateral. I've amended the question to reflect its original wording.

Here is a drawing:


Region ADE = (triangle - circle)/3.

Triangle:
The formula for the area of an equilateral triangle is A=(s^2)/4 * √3.
A = (6^2)/4 * √3 = 9√3.

Circle:
The sides of a 30-60-90 triangle are proportioned x : x√3 : 2x.
In the 30-60-90 triangle shown above, x√3=3, so x=√3.
x=√3 is also the radius of the circle.
Area = πr^2 = π(√3^2) = 3π.

Region ADE:
Area = (triangle - circle)/3 = (9√3 - 3π)/3 = 3√3 - π.

The correct answer is B.