arkham

07-21-2004, 02:23 AM

Hi guys,

Can someone please, help me out with these questions - they're driving me crazy.

thanks,

ArKham

Can someone please, help me out with these questions - they're driving me crazy.

thanks,

ArKham

View Full Version : Help - Practise Questions Powerprep

arkham

07-21-2004, 02:23 AM

Hi guys,

Can someone please, help me out with these questions - they're driving me crazy.

thanks,

ArKham

Can someone please, help me out with these questions - they're driving me crazy.

thanks,

ArKham

swetak28

07-21-2004, 02:51 AM

are the answers b & C?

sri830

07-21-2004, 03:01 AM

Answers are C and C??

arkham

07-21-2004, 03:01 AM

Yup that's right, could you tell me how you got those anwers, please.

thanks,

arK

thanks,

arK

arkham

07-21-2004, 03:05 AM

Sorry, I didn't see srini's post. Sweta is right, the answers are B & C.

arK

arK

swetak28

07-21-2004, 03:33 AM

sq.rt{(b-0)2+(c-a)2}=sq.rt{(d-0)2+(e-a)2}

as distance between the origin and those 2 set of points is same

as u simplify the equations they become

b2+c2=d2+e2-2a(e-c)

we know e-c is +ve as e>c

so the column B is a larger value.

as distance between the origin and those 2 set of points is same

as u simplify the equations they become

b2+c2=d2+e2-2a(e-c)

we know e-c is +ve as e>c

so the column B is a larger value.

arkham

07-21-2004, 03:51 AM

Yup makes sense, thanks a lot for that explanation. You meant the distance between (0,a) and the two points is the same, right? What about the second question?

thanks,

ArK

thanks,

ArK

greenglobs

07-21-2004, 07:19 PM

For the second question i will try to explain why the answer is c. it may be difficult without a picture. but lets try.

to start, i would draw a dotted line connecting DE, this will make the problem more visually appealing. now we have four triangles, we just need to find how much of the original area the two triangles occupy.

note, bd=1/2ba and be=1/2bc.... also note angle abc= angle dbe. therefore triangle dbe is similar to abc, using the side-angle-side.

now the base of the triangles is in the ratio 1:2, and thus the height of the triangles is in the ratio of 1:2. Since a=1/2bh the area of dbe=1/4 of abc.

since you can oreint this anyway you want, each of the four triangles is 1/4th the area of abc, and thus the area of quadrilateral dbef = adf + fec.

i think this is clear, and i am pretty sure it is accurate.

to start, i would draw a dotted line connecting DE, this will make the problem more visually appealing. now we have four triangles, we just need to find how much of the original area the two triangles occupy.

note, bd=1/2ba and be=1/2bc.... also note angle abc= angle dbe. therefore triangle dbe is similar to abc, using the side-angle-side.

now the base of the triangles is in the ratio 1:2, and thus the height of the triangles is in the ratio of 1:2. Since a=1/2bh the area of dbe=1/4 of abc.

since you can oreint this anyway you want, each of the four triangles is 1/4th the area of abc, and thus the area of quadrilateral dbef = adf + fec.

i think this is clear, and i am pretty sure it is accurate.

arkham

07-22-2004, 02:18 AM

Brilliant - it all makes sense now. Have to catch up on those pesky properties of similar triangles.

thanks,

ArK

thanks,

ArK

manwiththemission2005

10-22-2005, 03:01 AM

1. sq.rt{(b-0)2+(c-a)2}=sq.rt{(d-0)2+(e-a)2}

as distance between the origin and those 2 set of points is same

as u simplify the equations they become

b2+c2=d2+e2-2a(e-c)

we know e-c is +ve as e>c ,a is +ve

Isnt it A>B?

as distance between the origin and those 2 set of points is same

as u simplify the equations they become

b2+c2=d2+e2-2a(e-c)

we know e-c is +ve as e>c ,a is +ve

Isnt it A>B?

salamatchaudhry

06-28-2007, 03:02 PM

very simple, both points are on the circle so they indicate the radius of cirle so ans is C

GRE_Kickboxer

06-28-2007, 08:45 PM

Regarding the first problem:

The detailed solution to the first question you posted is right in PowerPrep. Isn't that where you got the problem? They are very thorough in their explanation. Have you studied that yet? I just want to make sure you've seen it.

Regarding the second problem:

I'm not sure the best way to solve the second problem is through mathematical techniques. I think that this geometry problem is designed to test spatial awareness, perceptual reasoning and common sense. There are many problems like this in the GRE BIG BOOK from ETS. I strongly suggest you practice developing an intuitive sense for these types of giveaway geometry problems, otherwise you could waste a lot of time on them during the test. This is one of the easiest things you can do to boost your score. Trust me.

The detailed solution to the first question you posted is right in PowerPrep. Isn't that where you got the problem? They are very thorough in their explanation. Have you studied that yet? I just want to make sure you've seen it.

Regarding the second problem:

I'm not sure the best way to solve the second problem is through mathematical techniques. I think that this geometry problem is designed to test spatial awareness, perceptual reasoning and common sense. There are many problems like this in the GRE BIG BOOK from ETS. I strongly suggest you practice developing an intuitive sense for these types of giveaway geometry problems, otherwise you could waste a lot of time on them during the test. This is one of the easiest things you can do to boost your score. Trust me.

amishera2007

07-06-2007, 04:41 PM

Should we assume any thing about the position of the two points? Arent we suppose anything unless mentioned specifically? In many cases, for instance, a triangle seem isoceles; but we are told not to assume anything. So why in this case we are assuming that the points b > d and e > c?

polkaparty

07-12-2007, 08:04 PM

Should we assume any thing about the position of the two points?

You're right, we can let the points move around then entire circle. I would have gotten this question wrong because I didn't make the implicit assumption that about a. That is, the picture is drawn such that a > 0, but they don't say anything about the value of a, so I assumed it could be any real number. That is, it could even be zero.

In this case, it's clear that the answer is D. Just consider the situation where (d,e) is at the top of the circle (d=0) and (b,c) is at the bottom (b=0). Now move the circle above the axis. So e > c which implies e^2 > c^2. Move the circle below the axis ad you get c^2 > e^2.

If you assume a>0 then the circle is above the axis. The same trick still works, just swap the positions of (d,e) and (b,c). First let (d,e) be at the top. Then e^2 > c^2. Now let (b,c) be at the top, then c^2 > e^2.

The main reason I didn't assume a>0 is because of a stupid ETS problem posted in this thread (http://www.urch.com/forums/gre-math/47470-confusing-official-gre-practice-problems-get-load.html). It's the one with the lines drawn through the center of the circle, but you can't assume they actually are at the center.

For #2, I drew the line connecting D and E and then just guessed that the areas are equal by trying to visualize changes in the lengths of BA, AC, and BC.

You're right, we can let the points move around then entire circle. I would have gotten this question wrong because I didn't make the implicit assumption that about a. That is, the picture is drawn such that a > 0, but they don't say anything about the value of a, so I assumed it could be any real number. That is, it could even be zero.

In this case, it's clear that the answer is D. Just consider the situation where (d,e) is at the top of the circle (d=0) and (b,c) is at the bottom (b=0). Now move the circle above the axis. So e > c which implies e^2 > c^2. Move the circle below the axis ad you get c^2 > e^2.

If you assume a>0 then the circle is above the axis. The same trick still works, just swap the positions of (d,e) and (b,c). First let (d,e) be at the top. Then e^2 > c^2. Now let (b,c) be at the top, then c^2 > e^2.

The main reason I didn't assume a>0 is because of a stupid ETS problem posted in this thread (http://www.urch.com/forums/gre-math/47470-confusing-official-gre-practice-problems-get-load.html). It's the one with the lines drawn through the center of the circle, but you can't assume they actually are at the center.

For #2, I drew the line connecting D and E and then just guessed that the areas are equal by trying to visualize changes in the lengths of BA, AC, and BC.

KBTA

07-13-2007, 02:45 PM

Dear arkham, amishera, polkaparty,

Here, positons are defined as the figure is under under coordinate system. In case of other geometrical figure we need not assume any conditions unless they are clearly defined. Please, inform us in case of any esception.</p>

.By the way, OA is B and Sweta's correct. Thanks to sweta for her nice explanations.

Rgds,

Jakir

Here, positons are defined as the figure is under under coordinate system. In case of other geometrical figure we need not assume any conditions unless they are clearly defined. Please, inform us in case of any esception.</p>

.By the way, OA is B and Sweta's correct. Thanks to sweta for her nice explanations.

Rgds,

Jakir

polkaparty

07-13-2007, 08:31 PM

Here, positons are defined as the figure is under under coordinate system. In case of other geometrical figure we need not assume any conditions unless they are clearly defined.

For #1, what if you wanted to draw a picture of the general case where a circle is centered at a point (0,a)? You'd have to place it on the Y axis somewhere, so your drawing will have an implicit value of a, just as the ETS drawing has implict values of (d,e) and (b,c). So simply drawing it such that a > 0 doesn't imply anything about its actual value. This is quite reasonable.

Similarly, if you wanted to draw the graph of a circle centered at say (s,t), you'd have to place it somewhere on the XY plane, but that placement says nothing about the actual values of s and t.

Of course ETS apparently means to say that a > 0, so that's what we should keep in mind when taking the GRE. I try not to hate this test, I try to stay positive....

For #1, what if you wanted to draw a picture of the general case where a circle is centered at a point (0,a)? You'd have to place it on the Y axis somewhere, so your drawing will have an implicit value of a, just as the ETS drawing has implict values of (d,e) and (b,c). So simply drawing it such that a > 0 doesn't imply anything about its actual value. This is quite reasonable.

Similarly, if you wanted to draw the graph of a circle centered at say (s,t), you'd have to place it somewhere on the XY plane, but that placement says nothing about the actual values of s and t.

Of course ETS apparently means to say that a > 0, so that's what we should keep in mind when taking the GRE. I try not to hate this test, I try to stay positive....