[OFFICIAL] AIMCAT 1310 Discussion Thread (Please DON'T OPEN if you have not taken the test)

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Invigilated.7th to 13th August. All the best. :)
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@PriyankaMitra said: explain qstn 20 sumone pls
Okay, here goes...
I guess you should have arrived at the equation given in the solution given at TIME's website... (n+4)(n-4)^2
Now, in the inequality above, since n is a positive integer (given in the question), (n+4) will always be positive, i.e. >0.
And, (n-4)^2 cannot be negative, i.e.
Now the only possible solution for the above inequality is being equal to zero. For that (n-4)=0, which implies n=4.
OVERRIDE the "Luck Factor".
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explain qstn 20 sumone pls

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CAn any temme %ile for dese scores:


Sec 1 -> 14
Sec 2 -> 38
Why do we fall??? So that we can learn to pick ourselves up again.
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hi everyone can u plz send me all the aimcats till date ..plz pm me the aimcats as soon as possible ,,i need those urgently

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@global87 said:
Ahh yes, got it. Don't know why I assumed all to be 90 degrees. Thanks for your help and for the drawing of course.
@CatAspirant742 said:
got it... couldn't imagine that figure at first...
well Data Sufficiency does that...we have to look at all possible conditions. I always skip the DS questions for trying at last......
|DT '12| |CAT'12: 99.93| |SNAP'12: 99.93| |XAT'13: 99.87| |NMAT'13: 99.86| |IIFT'12: 67.25|
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@sivaniim said:
try it, it definitely is possible. Draw a 90 degree angle. From one of the vertices draw a 30 degree and then drop a perpendicular from the original angle to the new line. It is possible.
just drew a crude drawing for you...
got it...:) couldn't imagine that figure at first...:D
screwed up CAT '12... resolved for CAT '13... 'YOU ARE NOT DEFEATED UNTIL YOU GIVE UP'
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@sivaniim said:
try it, it definitely is possible. Draw a 90 degree angle. From one of the vertices draw a 30 degree and then drop a perpendicular to the original angle from the new line. It is possible.
just drew a crude drawing for you...

Ahh yes, got it. Don't know why I assumed all to be 90 degrees. Thanks for your help and for the drawing of course.

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@global87 said:
Is that possible? If two opposite sides of any quadrilateral are 90 deg each, the other two have to be 90. Try drawing a quadrilateral on a piece of paper with two 'OPPOSITE ANGLES' as 90 and the other two not 90 ( like (120,60), (150,30)). I don't think its possible.

try it, it definitely is possible. Draw a 90 degree angle. From one of the vertices draw a 30 degree and then drop a perpendicular from the original angle to the new line. It is possible.

just drew a crude drawing for you...
|DT '12| |CAT'12: 99.93| |SNAP'12: 99.93| |XAT'13: 99.87| |NMAT'13: 99.86| |IIFT'12: 67.25|
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@sivaniim said:
you did not get my point. Sum of the opposite angles of a cyclic quadrilateral is 180. So, if one angle is 90 degree, the angle opposite it is 90 degree. But it does not say anything about the other two angles, all we know is that their sum is 180. It can be (120,60),(150,30) or any other pair, never it has been mentioned they are equal.
Is that possible? If two opposite sides of any quadrilateral are 90 deg each, the other two have to be 90. Try drawing a quadrilateral on a piece of paper with two 'OPPOSITE ANGLES' as 90 and the other two not 90 ( like (120,60), (150,30)). I don't think its possible.



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@global87 said:
Exactly my point! When any angle of a cyclic quadrilateral is 90 degrees, all become 90 right? Since opp sides of a cyclic quadrilateral add up to 180 degrees. And adjacent sides are given equal, not opposite sides, which makes it a square. If opposite sides were equal, it would have been a Rectangle or a square.
you did not get my point. Sum of the opposite angles of a cyclic quadrilateral is 180. So, if one angle is 90 degree, the angle opposite it is 90 degree. But it does not say anything about the other two angles, all we know is that their sum is 180. It can be (120,60),(150,30) or any other pair, never it has been mentioned they are equal.
|DT '12| |CAT'12: 99.93| |SNAP'12: 99.93| |XAT'13: 99.87| |NMAT'13: 99.86| |IIFT'12: 67.25|
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