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Just attempted 1314(had missed out due to some ofc work )

QA 23A 18C 5W - 49

VA 23A 17C 6W - 45

OA 46A 35C 11W-94

Hardest mock of the season(havn't given 1301)

QA was difficult(not one question was a sitter)

LR was out of this world...i took 42 min to solve 3 sets and was left with 28 min for VA in which i attempted all grammar/vocab questions and 1 RC( 4 questions wala- attempted 3)

so it tested my patience as i have never taken more than 25 min in LR section...!!!

Can anyone tell me what %ile would these scores have gotten ?

QA 23A 18C 5W - 49

VA 23A 17C 6W - 45

OA 46A 35C 11W-94

Hardest mock of the season(havn't given 1301)

QA was difficult(not one question was a sitter)

LR was out of this world...i took 42 min to solve 3 sets and was left with 28 min for VA in which i attempted all grammar/vocab questions and 1 RC( 4 questions wala- attempted 3)

so it tested my patience as i have never taken more than 25 min in LR section...!!!

Can anyone tell me what %ile would these scores have gotten ?

@edgy here odd multiples of 25 are 25,75 and 125 [correct me if i am wrong] , wouldnt they be covered in odd multiples of 5 ??

@Dipika1234 said: Manoj multiplied all the odd numbers for 1 to 150 and all even numbers from 151 to 175.Find the number of zeroes at the end of this product.

Number of 5s before 150 = odd multiples of 5 till 150 + odd multiples of 25 till 150 + odd multiples of 125 till 150 = 15 + 3 + 1 = 19.

The number of 5s after 150 and upto 175 are found only in 160 and 170. So total 5s = 21.

Now u can either calculate the number of 2s or can safely assume that 2s outnumber 5s ( because of the presence of multiples of 2, 4, 8 etc etc between 150 and 175)Hence the answer is 21 zeroes.

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@VijayRawat very well said! I've ben going through the same! These aimcats never fail to demoralize me! But who cares!

@[549423:ashish1909] Please send me the aimcat papers1318,1316 and 1314 guys..Just started preparing for cat 2012...Tension has gripped my nerves! E Mail: pravatnanda@yahoo.com

Thanks

Can someone mail me the PDFs of this Exam Please?

@[267494:priyaghosh] : please give me your mail id.. I will fwd you !

someone please mail me the pdf of 1314..it will be very helpful

@ashish1909 said: please send me the aimcat ques papers 1318,1316and 1314 ........will be very thankfull to you ....my mail id is ash_sharma45@yahoo.com

check your inbox.

please send me the aimcat ques papers 1318,1316and 1314 ........will be very thankfull to you ....my mail id is ash_sharma45@yahoo.com

Hi Puys,

Doing some question of Maxima and Minima

Don't have answere for some question

I am writitng question and my answere

Please check

Ques.To find the dimension of the cubical box of maximum vol that has one croner at the origin, three sides that contains the corner lying in the 3 cordiante planes and the opposited corner lying on the plane 2x+3y+4z=12 in the first octant ?

My answere 64/81 cubic unit.

Q2.Find The Maximum Value of x^2y if x and y are restricred to positive real number satisfying 6x+5y=45

My Answer 225

Q3.For any +ve constant a, what is the maximum value of x/(x^2 +a) for +ve x ?

My Answer 1/(2 squroot a)

Q4.Find the Least value of xy+2xz+3yz for +ve number x,y,z saisfying xyz=48.

My ANswer 72

Q5.Find the minimum value of the product (x+y^2+z^3)( (1/x) +(1/y^2) + (1/z^3)) for +va value of x,y,z

My answer is 9

Q6.Find the smallest value of 5x+(16/x)+21 for +ve value of x

My Answer 8*squrt5+21

Please reply ur answer with explanation.

Thanks and Regards

Another Puy

@[346223:yesarpit] PM me also my mail id is mpsgrover@gmail.com

:- Plz forward it to me as well.. Its jaiswal.bit@gmail.com.

Thanks in advance

@shyguy21 said: Can someone send em the PDFs of this mock?thanks in advanceInbox me I'll message you my email ID!.thanks in Advance

PM me your email id

@shyguy21 said: Can someone send em the PDFs of this mock?thanks in advanceInbox me I'll message you my email ID!.thanks in Advance

PM me your email id

Can someone send em the PDFs of this mock?

thanks in advance :)

Inbox me I'll message you my email ID!.

thanks in Advance :)

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