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Official Quant Thread for CAT 2012 [part 3]

  • Created by @pkaman
*All the quant enthusiasts [image] please continue here. here is the link to the previous thread: http://www.pagalguy.com/forum/quantitative-questions-and-answers/78107-official-quant-thread-cat-2012-a.html*
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Can anyone help me with this question

lagaan is levied on the 60% of the cultivated land .the revenue department collected total of RS 384000 through the lagaan from the village of sukhaya ,sukhaiya a rich farmer paid only RS 480 as lagaan .the percentage of the total land of sukhaiya over the total taxable land of the village is:
a)0.15% b)15% c)0.125% d)none of these


Hi ..
is it 0.125% ?
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nil1989 Says
what are the last 2 digits of the nos 169^31?



for finding the last 2 digits,always try and find out whether the number can be made to end with 1,which makes it very easy thereafter as final digit in this case will always be 1.
In this case 169 if squared will end up with number ending with 1 :
so (169^2)^29 = (...61)^29 [its enough to find out the last 2 digits)

therefore last 2 digits of the above power number = 51 (5 from 6 *9 = 54 )
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Can anyone help me with this question

lagaan is levied on the 60% of the cultivated land .the revenue department collected total of RS 384000 through the lagaan from the village of sukhaya ,sukhaiya a rich farmer paid only RS 480 as lagaan .the percentage of the total land of sukhaiya over the total taxable land of the village is:
a)0.15% b)15% c)0.125% d)none of these
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ananthraghavalu Says
page not found??? also still 5 more pages we can spend in this thread itself!!

are sorry use the link of Aizen.ya i know u can use.but ek jagah pe discussion hota he to its good for everyone i guess
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Corporate Communications Cell - MDI Gurgaon
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Find the remainder when
1^5+2^5+3^5+... ...+100^5 is
divided by 4? (explain plz)


my take is also 0
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bro, in the binomial expansion of (2+1)^1024= 2^1024*1^0+2^1023*1^1.....2^0*1^1024 -1
thus the highest term will be 2^1024-1=2^1023..
am i wrong anywhere??

(3^1024-1)can be written as (2+1)^1024-1 which means in the expansion of (2+1)^1024 last term will be 1 nad will be cancelled by -1....ao the whole expansion will be C(1024,0)2^1024+C(1024,1)2^1023+.......C(1024,1)2^1
LAST TERM IS 1024*2=2048 IE 2^11=2^X
HENCE x=11.
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it is 12.. break the above into a^2 - b^2 form.. it will reduce like (3^512 +1) (3^512-1) .. at last (3^2-1)(3^2+1) will be left..
CAT 2012 : 99.41
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Q) If the graph of the function f(x)= (a^x)-1/x^n*((a^x)+1) is symmetric about y-axis then what is the value of n?

ans -1/3

please explain

TIA


Symmetric about Y axis So ,this is an even function.
so use f(x)=f(-x) to find the value of n
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Q) If the graph of the function f(x)= (a^x)-1/x^n*((a^x)+1) is symmetric about y-axis then what is the value of n?

ans -1/3

please explain

TIA
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both are same bro,10/(3)^1/2((3^1/2)+1) = 10/(3 + rt(3)) :)

My mistake sorry brother...
kindly share your approach....

and please will someone explain on which thread we are supposed to post question???????only open thread after this one is closed
SBT-2012 VC|| one last hurray
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thanks fapsian19... here's another 1...
*1!+2*2!+....+12*12! when divided by 13 leaves remainder??

12 remainder ... but i find it manually no gud ways
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