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Geometry for CAT 2011

Hi Guys, Geometry, Algebra and Number system form the major chunk of our QA section for CAT. Proficiency in these three sections would definitely boost our Quants scores. Contents of Geometry 1\. Plane Geometry - Basics and Tri...
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If x, y, z are the angles of a triangle where x, y, z are integers, what is the number if values that x-y-z can take?
1. 90
2. 355
3. 178
4. 177


Please provide me with a solution.....

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dpiict Says
my take is option b.



calculation error, 1/2
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ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED = FD. Find the area of the shaded figure.


Choose one answer.
a. 1/2 b. 5/9 c. 1/3 d. 13/16






my take is option b.
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In an acute angle triangle all the angles are positive integers and 13 times one angle is equal to 17 times another angle then what could be the minimum angle??

Please solve it ..i dint get it after trying for 3 hrs


ans can also be 30 ,65,85..
30+65+85=180...all are positive integers....and 17*65=13*85 also..
so the ans should be 30..

That's correct!
The possible triplets are - (17,13,150),(34,26,120),(51,39,90),(68,52,60),(85,65,30).
Only last 2 satisfy the acute angled tr. criteria...so you have ur answer.
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ADE is the triangle here...not a straight line. Plz refer to original detailed solution by ravitoons!

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How 2 wrongs make up for one right answer...
If ADF is x then BDF = 2x. => DBE = 2x=> EDC = 4x. Total unshaded = 9x.
Then BEA = 6x and ABC = 18x. Since 18x = 1, 9x = 1/2!


Can I ask a doubt here how you reach that ADE is a straight Line??
Can you please clarify this?
Thanks with a :) , AB
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Ques in the attached image. Options are:

a. 152/3
b. 32
c. 12
d. 41


wats the ans?????
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In an acute angle triangle all the angles are positive integers and 13 times one angle is equal to 17 times another angle then what could be the minimum angle??

Please solve it ..i dint get it after trying for 3 hrs


ans can also be 30 ,65,85..
30+65+85=180...all are positive integers....and 17*65=13*85 also..
so the ans should be 30..

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Correct. My bad

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Hi tukku,

Let area of ADF be x then BDF will be 3x. { traingles on same line segment and common vertex have their areas in ratio of their bases }

As BDF is 3x, DBE will also be 3x. And thus EDC will be 9x.

Total unshaded portion is 16x.

Now draw a line EA. DFA is x so EDA is x.

Then BEA is 8x. and so ECA is 24x. Total triangle area is 32 x.

So shaded portion area = unshaded portion = 1/2

How 2 wrongs make up for one right answer...
If ADF is x then BDF = 2x. => DBE = 2x=> EDC = 4x. Total unshaded = 9x.
Then BEA = 6x and ABC = 18x. Since 18x = 1, 9x = 1/2!
  • 1 Like  
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yes u r right ... i too get 1/2

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ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED = FD. Find the area of the shaded figure.


Choose one answer.
a. 1/2 b. 5/9 c. 1/3 d. 13/16


Hi tukku,

Let area of ADF be x then BDF will be 3x. { traingles on same line segment and common vertex have their areas in ratio of their bases }

As BDF is 3x, DBE will also be 3x. And thus EDC will be 9x.

Total unshaded portion is 16x.

Now draw a line EA. DFA is x so EDA is x.

Then BEA is 8x. and so ECA is 24x. Total triangle area is 32 x.

So shaded portion area = unshaded portion = 1/2
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dagattani Says
hi can sm1 pls solve this question for me.. i m unable to solve it.. at least give an approach.. i will at least try.. thanks...


Hi,

The approach is simple.. avoid all confusions of what height has come down, etc.. just stick to the difference of volume ' V ' which is proportional to surface area at a given time.

2/3V-1/2V = K*pi*R2
1/2V- xV = K*pi*r2

Here R is initial surface radius, which can be found by cone equations. r is final surface radius, which also can be expressed in terms of R, as the final volume at the end of 1st day is given.

Now get x.
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SET - 21

A conical vessel, with a circular base, is filled with water to two-thirds of its volume. The pointed end of the cone is snipped off and replaced with a lid. The lid is kept open for 10 hours every day during which some water evaporates. The volume of the water that evaporates on a day is directly proportional to the area of the water surface at the beginning of the day. The volume of the water left in the container after evaporation on the 1st day is half the volume of the original cone. If V is the volume of the original cone, then what is the volume of the water that evaporates on the 2nd day?





a.

b.

c.

d.




hi can sm1 pls solve this question for me.. i m unable to solve it.. at least give an approach.. i will at least try.. thanks...
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ye derivation ki thi sirjjiii

(_/4 + _/9)^2 = 25


hi cud sm1 please help me with this.. i m unable to solve this question.. thanks..!!
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