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AB=BC=CD=ED=EF=FG=GA , Find Angle DAE [#permalink]
08 Mar 2010, 14:04

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Difficulty:

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Question Stats:

86% (01:51) correct
14% (00:00) wrong based on 30 sessions

1. AB=BC=CD=ED=EF=FG=GA , Find Angle DAE

a.15 b.20 c.30 d.25 e.22

Donot have OA, mine answer is D

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File comment: Question 1.

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2. Triangle ABC is divided into four parts by straight lines from two of its vertices. Area of three triangular parts are 8 , 5 and 10. what is the area of remaining part?

they might not reflect true gmat , but solving them will surely help us learning new concept.

Trust me, forget them. Each Indian CAT question has a specific strategy and they cannot be solved by mere mortals with normal knowledge. The geometry questions you posed, am very sure, without using trigonometric formulas and concepts, you cannot quickly come up with an answer. And most CAT quant questions are solved backwards by substituting the answer options unlike GMAT quant questions that are solved forwards to get the correct answer within the stipulated time-frame. _________________

I dont want to divert anyone's attention from GMAT, but these question are to be solved to learn more concepts rather than practicing for GMAT.

All those who had prepared for Indian Cat , had score very well in GMAT by just getting quick review for the GMAT material. I m not motivating any one for this, but learning something is never harmful, all those who have less time for GMAT should not do such question but those who got spare time should definitely do tough problems to learn more application of the concepts.

This is just mine opinion, I welcome all criticism. _________________

angle GAB = angle ACB = x which makes angle CBF = 2x and hence angle CDB = 2x which means angle BCD = 180-4x now, ACB+BCD+DCE=180, therefore DCE =3x which makes DEC = 3x which makes EDC = 180-6x which again makes EFD= 180-4x now EAF + AEF = EFD which means, x + 3x = 180 - 4x x= 180/8= 22.5 degrees

Hello Kavesh, I checked your solution, you went from the same way of mine. But the place is where you are wrong is in the last line. It should be x + 2x = 180 - 4x, instead of (x+ 3x = 180 - 4x which is wrong).

which means, x + {3x--->Wrong place} = 180 - 4x x= 180/8= 22.5 degrees

you got it absolutely right! thanks for pointing out where I had gone wrong. i had misread last part of the question (=GA). having understood the question well enough it took under 30 seconds to arrive at an answer. it's so simple, all the points on AE and AD mirror each other.

Re: Tough geometry questions. 700+ [#permalink]
31 Aug 2011, 11:05

gurpreetsingh wrote:

3. Radius of quarter circle is 10 with center at C. If perimeter of rectangle CPQR is 26, then what is the perimeter of APRBQA.

a. 112+5pie b. 17+5pie c. 15+7pie d. 13+7pie e. none of the above.

Donot have OA, open for discussion. Mine ans came to be B, just by approximation though i was not able to solve Length of PR.

Attachment:

3.jpg

4. A square is inscribed in the triangle PQR with sides as PR = 10 , PQ = 17 , QR = 21. Find the perimeter of square ABCD.

a. 28 b. 23.2 c. 25.4 d. 28.8 e. none of the above

HEY IM REALLLLLLLLLLLLLLLLLY LATE I JUST STUMBLED UPON YOUR QUESTION NOW (2011) any way i dont think there is a point answering but why not. the answer is a perfect B PR=CQ(DIAGONALS) . CQ is the radius hence 10 so PQ is 10.total length required is circumference of the quarter circle + PR + AP + BR.The circumference of the quarter is 1/4*2pi*10(radius) = 5pi .Now AP + BR = (AC + AB) - (PC+CR) . PC+CR = Half perimeter of rectangle = 13. so that is 20 (AC+ AB) - 13(PC+CR) = 7. So now adding it all up we get 5PI + 10 + 7 = 17 + 5PI = OPTION B i hope this helps

Solution of question 2: I m short of time so I would just let you guys imagine the triangle.

Consider the line from B to AC intersecting Ac at E and similarly from C to AB at F. Their point of intersection as D and join FE.

Now. since area of BDC = 10 and DRC = 5 => ratio is 2.. so BD : DE = 2 as well. Reason: The line from the opposite vertex divided the triangle into ratio of its bases. Why? because altitude remains same. Do not progress further unless you grasp this concept because it will be used multiple times.

Similarly if we consider triangle FDB and EDE both have bases BD and DE in ratio 2 so area of FDB and FDE have ratio 2 since area FDB = 8 => area FDE = 4 ...why? read the concept above.

Similarly for triangle ABD and ADE areas are in ratio =2/1 Let area AFD = x and area ADE = y ( our requirement is actually x+y value) => since ratio = 2 => 8+x=2y --------------------------------1

similarly for triangle AFD and ADC areas are in ratio 4/5 because area on same bases FD and DC of triangle FDE and EDC have ratio 4/5 => \(\frac{x}{(y+5)} = \frac{4}{5}\)

=> 5x = 4y+20 ---------------------------------2

using 1 and 2 we get x= 12 and y =10

our requirement = x+y = 22 = answer D

PS: The solution looks complicated but if you try to understand it is very easy question and I think is probable option for 750+ question in GMAT. There is nothing un usual and it only require 1 min of work ..but it took me 10 mins because m out of touch _________________

thanks a lot gurupreet. This was a refresher, but not of GMAT type.

I just recovered from mild heart-attack, as I took 5 mins per question on an average. GMAT math is not that hard, even for 750 level, it is the verbal score that takes you there.

And regarding Indian CAT, these questions are from paper-pen test. Now these are out in the current format. The strategy were different then and one was supposed to skip such questions. --- exp. from a 99 percentile-er in Indian CAT _________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! shit-happens-my-journey-to-172475.html#p1372807

Question 4 was a nightmare. Took me 15 mins. have to use the formula for are of a triangle when three sides are known. Then using that find the height has to be found. Then similarity of triangles has to be used. Dont think I'l have the time to solve this on test day. _________________

Did you find this post helpful?... Please let me know through the Kudos button.

2. Triangle ABC is divided into four parts by straight lines from two of its vertices. Area of three triangular parts are 8 , 5 and 10. what is the area of remaining part?

a. 32 b. 40 c. 54 d. 22 e. None of these.

Donot have OA, open for discussion

Attachment:

2.jpg

Simple application of Ladder's Theorem: 1/18 + 1/15= 1/10 +1/A where A is the Area of the triangle. Required Area=22

I took a shot at no.3; given r=10. So, AC=QC=10. Now QC=PR=10(as diagonals of a rectangle are equal). Triangles QPR and RPC are right triangles with hypotenuse as 10. This means that QR=8 and PQ=6 as both the triangles form a 3,4,5 Pythagorean triplet. So, AP=10-8=2. BR=10-6=4. Perimeter= 2+10+4+5pi(perimeter of arc AB). So, answer is 16+5pi. I believe this is the closest to GMAT level.

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