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The Pahadi Mod
@raghav507
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Can anyone help me with this questionlagaan 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 Sayswhat 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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*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 Sayspage not found??? also still 5 more pages we can spend in this thread itself!!

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page not found??? also still 5 more pages we can spend in this thread itself!!

%29"]http://www.pagalguy.com/forum/quanti...ml#post3509369 (Official Quant Thread for CAT 2012 )

bhai please continue here.request

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mba4mheart Saysmy take is also 0

)"]http://www.pagalguy.com/forum/quanti...ml#post3509369 (Official Quant Thread for CAT 2012 )

bhai please continue here.request

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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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Please continue your discussions here.

http://www.pagalguy.com/discussions/official-quant-thread-for-cat-2012-part-4-25082184/3509369

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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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