Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The area of the circle is \(\pi 2^2 = 4\pi\) . The area of the sector = \(\text{(the area of the circle)}*\frac{90}{360} = \pi\) . The correct answer is C.

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

The area of the circle is \(\pi 2^2 = 4\pi\) . The area of the sector = \(\text{(the area of the circle)}*\frac{90}{360} = \pi\) . The correct answer is C.

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

\(x^2 + y^2 = 4\) is an equation of a circle centered at the origin and the radius \(\sqrt{4}=2\).

Graph of the function \(y = |x|\) is given below:

Attachment:

graph_modulus.png [ 7.61 KiB | Viewed 8538 times ]

Top part of the sector would be a sector with 90 degrees angle, and its area would be 1/4 of that of the circle (circle 360 degrees).

Area of the circle =\({\pi}{r^2}=4\pi\), 1/4 of this value = \(\pi\). _________________

I know these are supposed to be the toughest challenges, but is this something which is included in the scope of the GMAT? I thought coordinate plane on GMAT was restricted to straight line equations only....

From x^2 + y^2 = r^2 => radius = 2 Area of the circle = [img]\pi[/img]r^2 The line |x| has a reflection about the origin at [img]90^{\circ}[/img] Area of sector is 1/4 x area of circle = [img]\pi[/img]4/4 = [img]\pi[/img] Therefore ans is C _________________

KUDOS me if you feel my contribution has helped you.

I did not know the equation for a circle, so had to spend extra seconds figuring out that x^2 + Y^2 = 4 was indeed the equation of a circle with radius 2. I ended up with C, but probably took 30-60s more than I should have.

Question - do these type of problems represent the problems at the 700-800 level on the GMAT? I thought the equation for a circle was pretty much out of scope for the GMAT. I hear that the GMAT quant section is getting tougher but is this (problems like this) the level where its headed?

I saw a lot of questions with circle formulas (both in paper and digital tests) and suppose it's 650 question maximum.

Well, then I'm glad I came across this problem Could you tell me in which paper and digital tests you saw questions with circle formulas? I've pretty much based my Math theory on the Manhattan guides and don't remember seeing the circle formula there (or for that matter OG11).

I saw a lot of questions with circle formulas (both in paper and digital tests) and suppose it's 650 question maximum.

Well, then I'm glad I came across this problem Could you tell me in which paper and digital tests you saw questions with circle formulas? I've pretty much based my Math theory on the Manhattan guides and don't remember seeing the circle formula there (or for that matter OG11).

If we're talking about must-have tests, it definitely was in GMAT Prep.

y = |x| ---- the slope of this line is (+1) if X is positive and (-1) if X is negative...which means the lines make a 45 degree angle with the x axis...and both the lines sweep a total of 90 degrees...

The circle is 360 degrees -- so the area swept is 1/4 of the total area.

The area of the circle is \(\pi 2^2 = 4\pi\) . The area of the sector = \(\text{(the area of the circle)}*\frac{90}{360} = \pi\) . The correct answer is C.

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

\(x^2 + y^2 = 4\) is an equation of a circle centered at the origin and the radius \(\sqrt{4}=2\).

Graph of the function \(y = |x|\) is given below:

Attachment:

graph_modulus.png

Top part of the sector would be a sector with 90 degrees angle, and its area would be 1/4 of that of the circle (circle 360 degrees).

Area of the circle =\({\pi}{r^2}=4\pi\), 1/4 of this value = \(\pi\).

How did you find the radius "\(x^2 + y^2 = 4\) is an equation of a circle centered at the origin and the radius \(\sqrt{4}=2\)." can you please explain?

The area of the circle is \(\pi 2^2 = 4\pi\) . The area of the sector = \(\text{(the area of the circle)}*\frac{90}{360} = \pi\) . The correct answer is C.

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

\(x^2 + y^2 = 4\) is an equation of a circle centered at the origin and the radius \(\sqrt{4}=2\).

Graph of the function \(y = |x|\) is given below:

Attachment:

graph_modulus.png

Top part of the sector would be a sector with 90 degrees angle, and its area would be 1/4 of that of the circle (circle 360 degrees).

Area of the circle =\({\pi}{r^2}=4\pi\), 1/4 of this value = \(\pi\).

How did you find the radius "\(x^2 + y^2 = 4\) is an equation of a circle centered at the origin and the radius \(\sqrt{4}=2\)." can you please explain?

The area of the circle is \(\pi 2^2 = 4\pi\) . The area of the sector = \(\text{(the area of the circle)}*\frac{90}{360} = \pi\) . The correct answer is C.

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

BELOW IS REVISED VERSION OF THIS QUESTION:

What is the area of the smaller sector enclosed by the graphs \(x^2 + y^2 = 4\) and \(y = |x|\)?

A. \(\frac{\pi}{4}\) B. \(\frac{\pi}{2}\) C. \(\pi\) D. \(2\pi\) E. \(3\pi\)

\(x^2 + y^2 = 4\) is an equation of a circle centered at the origin and with the radius of \(\sqrt{4}=2\).

The graph of \(y = |x|\) is given below (blue):

Attachment:

Curves.png [ 13.45 KiB | Viewed 4283 times ]

The smaller sector enclosed by these graphs would be the top part of the circle, so the yellow sector. Since the central angle of this sector is 90 degrees then its area would be 1/4 of that of the circle (since circle is 360 degrees).

The area of the circle is \({\pi}{r^2}=4\pi\), 1/4 of this value is \(\pi\).

The smaller sector enclosed by these graphs would be the top part of the circle, so the yellow sector. Since the central angle of this sector is 90 degrees then its area would be 1/4 of that of the circle (since circle is 360 degrees).

The area of the circle is \({\pi}{r^2}=4\pi\), 1/4 of this value is \(\pi\).

Answer: C.[/quote]

Please, could you explain what does the "smaller sector" mean? It's smaller than what? The greater part of the circle, not enclosed by y=|x|, is considered a sector too? Thanks in advance.

X^2+y^2= 4 is a equation of circle centered at origin. this circle has a radius of 2. y= lxl is a equation of straight line in the first and the second quadrant and makes an angle of 45 degree with y axis in both the quadrant. So a total of 90 degree. It means that line occupies 1/4 th of a circle so net area= (pi (2)^2)/4 =pi. _________________

You have to have the darkness for the dawn to come.