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.

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:

Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
01 Sep 2008, 03:33

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (medium)

Question Stats:

30% (02:19) correct
70% (01:34) wrong based on 27 sessions

Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

Re: CIRCLE ... NICE ONE [#permalink]
02 Sep 2008, 08:30

arjtryarjtry wrote:

Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute 2. \angle ADB > \angle CAD

The answer is not B. If we use only that statement, we can draw the following (no need to use co-ordinate geometry, but it's easier to explain if I do, since I can't draw a picture):

-draw the circle in the co-ordinate plane, with centre at (0,0). The radius is 1. -make AC a diameter: put A at (-1, 0), and C at (1,0). -draw D and B in the third quadrant (i.e. where x and y are both negative), and so that when read counterclockwise, the points are in the sequence ADBC. That is, minor arc AD should be shorter than minor arc AB.

Drawn this way, ADB should be (very) obtuse, and much larger than CAD, which must be less than 90 degrees (angle CDA must be 90, since AC is a diameter so CAD and DCA are both less than 90). The length of minor arc AB is clearly less than one quarter of the circumference, while minor arc CD is clearly greater than one quarter of the circumference, so it is possible for x to be less than y. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
27 Sep 2012, 10:18

I also fell for trap B, then realized.

I think answer must be C only, because order of the points on the circle were not mentioned, only if you consider that ADB<90 then only we know B is between A and D.

Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
01 Oct 2012, 02:03

voodoochild wrote:

If the angle subtended by the arc is greater, the length of the arc has to be greater. I am not sure why C) is OA....help?

(2) alone is not sufficient. Remember the condition in (1) and check the case when that condition doesn't hold. In my attached drawing, in the isosceles trapezoid, x=y.

Attachments

ArcsComparison.jpg [ 14.92 KiB | Viewed 1832 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
06 Oct 2012, 00:24

study wrote:

Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute 2. \angle ADB > \angle CAD

Well We can make a comparison between the 2 arcs only if its given explicitly.

Statement 1 says Angle ADB is acute and we cannot say here if x is greater than y. Statement 2 says one angle is greater than the other. So this must be sufficient

Is there an explanation for Y it is C. _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
08 Oct 2012, 00:25

154238 wrote:

OA is B for me . Kindly Confirm Bunuel.

cheers

No, B is not the answer. See my previous post above points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1126726 In the drawing, an isosceles trapezoid is depicted, and obviously x=y. For example, by moving point B, you can increase or decrease the arc AB, while arc DC stays the same. So, you can have both x>y and x<y, after you already had x=y. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
08 Oct 2012, 02:14

1

This post received KUDOS

rajathpanta wrote:

study wrote:

Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute 2. \angle ADB > \angle CAD

Well We can make a comparison between the 2 arcs only if its given explicitly.

Statement 1 says Angle ADB is acute and we cannot say here if x is greater than y. Statement 2 says one angle is greater than the other. So this must be sufficient

Any two points on the circumference of the circle define two arcs. Only if the two points are the endpoints of a diameter, the two arcs have equal measure. Otherwise, one of the arcs is less than the other. In our case, the circumference of the circle is 2\pi, therefore x and y being less than \pi are both the shorter arcs defined by A,B and C,D respectively.

If we don't know that \angle{ADB} is acute, we don't know whether point D is on the shorter or the longer arc defined by A and B. From (1), we have that \angle{ADB} is acute. In the attached drawing, D can be only above the chord AB. Therefore to \angle{ADB} corresponds the shorter arc x. From (2) we have that \angle{ADB}>\angle{CAD}, meaning that \angle{CAD} is also acute. Therefore the corresponding arc to \angle{CAD} is the shorter arc y. In the attached drawing C can be only on the right of the diameter DC_1. Using (2), now we can conclude that x>y. So, (1) and (2) together is sufficient.

Answer C.

Attachments

ArcsComparison-more.jpg [ 16.77 KiB | Viewed 1624 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
09 Oct 2012, 00:28

EvaJager wrote:

rajathpanta wrote:

study wrote:

Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute 2. \angle ADB > \angle CAD

Well We can make a comparison between the 2 arcs only if its given explicitly.

Statement 1 says Angle ADB is acute and we cannot say here if x is greater than y. Statement 2 says one angle is greater than the other. So this must be sufficient

Any two points on the circumference of the circle define two arcs. Only if the two points are the endpoints of a diameter, the two arcs have equal measure. Otherwise, one of the arcs is less than the other. In our case, the circumference of the circle is 2\pi, therefore x and y being less than \pi are both the shorter arcs defined by A,B and C,D respectively.

If we don't know that \angle{ADB} is acute, we don't know whether point D is on the shorter or the longer arc defined by A and B. From (1), we have that \angle{ADB} is acute. In the attached drawing, D can be only above the chord AB. Therefore to \angle{ADB} corresponds the shorter arc x. From (2) we have that \angle{ADB}>\angle{CAD}, meaning that \angle{CAD} is also acute. Therefore the corresponding arc to \angle{CAD} is the shorter arc y. In the attached drawing C can be only on the right of the diameter DC_1. Using (2), now we can conclude that x>y. So, (1) and (2) together is sufficient.

Answer C.

Hi Eva,

we have that \angle{ADB}>\angle{CAD}, how can you conclude that that [m]\angle{CAD} is also acute. Let say you make a square inside the circle a---------d | | | | c---------b

now, if \angle{ADB}>\angle{CAD}, this doesnt mean that angle{CAD} is acute. Right ?

Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
09 Oct 2012, 01:03

EvaJager wrote:

rajathpanta wrote:

study wrote:

Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

Any two points on the circumference of the circle define two arcs. Only if the two points are the endpoints of a diameter, the two arcs have equal measure. Otherwise, one of the arcs is less than the other. In our case, the circumference of the circle is 2\pi, therefore x and y being less than \pi are both the shorter arcs defined by A,B and C,D respectively.

If we don't know that \angle{ADB} is acute, we don't know whether point D is on the shorter or the longer arc defined by A and B. From (1), we have that \angle{ADB} is acute. In the attached drawing, D can be only above the chord AB. Therefore to \angle{ADB} corresponds the shorter arc x. From (2) we have that \angle{ADB}>\angle{CAD}, meaning that \angle{CAD} is also acute. Therefore the corresponding arc to \angle{CAD} is the shorter arc y. In the attached drawing C can be only on the right of the diameter DC_1. Using (2), now we can conclude that x>y. So, (1) and (2) together is sufficient.

Answer C.

Hi Eva,

we have that \angle{ADB}>\angle{CAD}, how can you conclude that that [m]\angle{CAD} is also acute. Let say you make a square inside the circle a---------d | | | | c---------b

now, if \angle{ADB}>\angle{CAD}, this doesnt mean that angle{CAD} is acute. Right ?

Here, in fact all angles are equal and are right angles. This cannot be the situation in the present question. In addition, if both angles ADB and CAD are right angles, then the inequality \angle{ADB}>\angle{CAD} doesn't hold! See below.

When taking (1) and (2) together, from (1) we have that \angle{ADB} is acute, meaning its measure is less than 90 degrees. Then using (2) we get that \angle{CAD}<\angle{ADB}<90^o, so necessarily \angle{CAD} is also acute. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
06 Oct 2013, 03:22

EvaJager wrote:

voodoochild wrote:

If the angle subtended by the arc is greater, the length of the arc has to be greater. I am not sure why C) is OA....help?

(2) alone is not sufficient. Remember the condition in (1) and check the case when that condition doesn't hold. In my attached drawing, in the isosceles trapezoid, x=y.

In your diagram Angel ADB is not acute. How can we use that to prove that we can have x = y with both (1) & (2) statements.

Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]
20 Apr 2014, 18:59

1

This post received KUDOS

Expert's post

arjtryarjtry wrote:

Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute 2. \angle ADB > \angle CAD

Responding to a pm:

A, B, C and D could lie anywhere on the circle. There is nothing given to say that they must lie in that order. What is the relevance of x \lt \pi and y \lt \pi? The total circumference of the circle of radius 1 is 2\pi. So this is to tell you that both arcs are less than semi circles so we are talking about the minor arcs of AB and CD.

Question: Is x > y?

1. \angle ADB is acute

This tells us that D lies on the major arc of AB, not on minor arc i.e. it doesn't lie on the red part which is x. Had D been on x, the angle ADB would have been obtuse.

Attachment:

Ques3.jpg [ 9.91 KiB | Viewed 418 times ]

But y may be smaller than x or greater depending on where C is. Hence this statement alone is not sufficient.

2. \angle ADB > \angle CAD Angles ADB and CAD could be inscribed angles of arcs x and y on their major arcs in which case this implies that x > y. Or angle ADB could be obtuse and arc x could be less than arc y.

Attachment:

Ques4.jpg [ 24.37 KiB | Viewed 418 times ]

Note that in both the cases above, angle ADB > angle CAD. Angle ADB is obtuse since the inscribed angle is in the minor arc. Angle CAD is acute since the angle is in the major arc. But in the first case x > y and in the second case x < y. Hence this statement alone is not sufficient.

Using both statements together, angle ADB is acute and greater than angle CAD so angle CAD is also acute. This means we are talking about inscribed angles in the major arc in both cases i.e. the first case in the figure given above. Then, if inscribed angle of x is greater than inscribed angle of y, it means central angle of x is greater than central angle of y (Central angle = 2*Inscribed angle in major arc) and hence arc x is greater than arc y.