# Official Quant Thread for CAT 2012Quantitative Ability & DI Report

Please continue here for all the Quant queries and discussions.
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http://www.pagalguy.com/forum/quantitative-questions-and...
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lemme be the last
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@visionIIM-ACL
It was excellent being with you guys....I hope ki same will be continued(infact i am sure ki same will be continued) in 2013 thread as well...
ATB for other exams
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@chandrakant.k My last comment on the thread (Before Moderator close it)
Thanks everyone for being part of this wonderful journey.
I can't express myself how much I've learned, grown, and enjoyed studying with all of you since the last 5-6 months.
When I saw the post of @chandrakant.k...
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its better if u remember it by the concept24^n
when n is odd it is 24...and even 76..usme n bhi dhoka nhi dega for 1. :P
I followed the concept of 76 * 2^(n) = last 2 digits of 2^nIts valid in all cases except when n=1I calculated 2^21 in the calculator which gives the same result as 2^2001, and realized my mistake :)
As the CAT 2K12 come to a close, lets close this thread for our discussions for quant.. Wishing everyone who was part of this thread and part of CAT 2K12 to have your convert this year itself. All the best for everyone all the best for myslef
For now, this is the thread for discussion
h...
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2^12 = 4096=>2^2001 = (2^9)*(4096)^166 = (2^9)*(4100-4)^166
=>last two digits would be given by (2^9)*(-4)^166 = 2^341
= (2^5)*(4096)^28~(2^5)*(-4)^28 = 2^61 = 2*(4096)^5~ 2*(-4)^5 = -2^11= -2048 hence last two digits = 100-48 = 52.
ATDH.
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2^2001 mod 100 is 52 hence 5.
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2 (24)^2002.76\
52 last two digits.
What is the tens digit of the number 2^2001 ?
Its a very simple question, but I am getting a different answer from one given in the solution, so wanted to verify here . I am getting 0 and the CL solution says 5 :O
if I pick two numbers and take the average.... and if the average is an integer then I go and pick that number and I have found one of the required triplets. Now the average would be integer when the two numbers are both odd or both even. This can be done in number of ways of picking 2 odd nos. ...
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90?
number of AP terms is what is asked for
a, a + d, a + 2d
a = 1 => d = 1,2,3,4,5,6,7,8,9
a = 2 => d = 1,2,3,4,5,6,7,8,9
a = 3 => d = 1 to 8
a = 4 => d = 1 to 8
.
.
a = 18 => d = 1
total = 45 * 2 = 90
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middle no.for 2=>1 pairfor 3=>2 pairs....for 10=>9for 11=>9for 12=>8..for 19=>1 pairtotal=90
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From the first 20 natural numbers, a combination of 3 distinct numbers is selected and then arranged in ascending order. How many such combinations can be picked such that the average of the extreme numbers gives the middle element?
1) 1022) 903) 944) 86
if C(n,0)*C(n,n-2) + C(n,1)*C(n,n-3) + C(n,2)*C(n,n-4)............C(n,n-2)*C(n,0) =C(x,y)
given n = 100. find the values of (x,y)
ATDH.
@stellarPG bro please the tricks in http://www.pagalguy.com/forums/quantitative-ability-and-di/quant-tricks-short-cuts-t-87923/p-3609086?page=7 it will be helpful for all.. please post all these tricks there.
Hogwarts School of Witchcraft and Wizardry
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i dont knw why my post aint coming properly! the angle is greater than 0 and less than pi
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Arey yaa, sumthing wrong with my typing today!
its 0<*
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muggle should not interfere
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hw??? that means d angle is in 2nd quadrant !! if that case cot would be -ve
coz he was in slytherin first year n u were in griffindor's final year
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Hahaha I would give anything to go to that place, I plan to take a tour of the warner bros hogwarts studio someday! :P
And yeah Cot^-1(cot *) = * for 0<*
hence the subtraction from pi
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Haha i was in slytherine..u were in griffindoor i guess.. arch rivalry :D
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how come i never met u over there?
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@Vascent said:sorry i got cot-1(cot a/2)i delted it my post to correct my typo, how come u got to quote it
hes faster
@ScareCrow28 said:cot^-1[ 2cos^2(a/2)/ 2sin(a/2)cos(a/2)] = cot^-1 [cot(a/2)]The answer should be pie + a/2 ...right?? But the answer given is pie-a/2@krum@Brooklyn
IMO ...
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igjaaactlyyyy i guess the answer is wrong thenn ..
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I went for 3 months cpurse in The Hogwarts School of Wizardry and Witchcraft :D
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hes faster :D
IMO shld be ans in either 1st or 3rd quadrent so : pi + a/2 or only a/2
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yups buddy...m getng the same..
sorry i got cot-1(cot a/2)
i delted it my post to correct my typo, how come u got to quote it :(
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cot^-1[ 2cos^2(a/2)/ 2sin(a/2)cos(a/2)] = cot^-1 [cot(a/2)] The answer should be pie + a/2 ...right?? But the answer given is pie-a/2@krum @Brooklyn
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156-3pie
how yar ??
dost m getng as 156-36pi....wats ur logic?
ur soln is correct
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(2) 169 – 36∏
Ans given is: pie - a/2
ek baat batao...agar cot-1(a/2) hai to y not the ans is NOT.???...means shudn't we have the range for a...yaa simply options dekh k karoge??..coz its tricky