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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
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 ...
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 )
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 o...
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
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??
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.
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
Please continue here and refrain from discussing the questions of Mocks till the testing window is over for that particular mock. http://www.pagalguy.com/forum/quantitative-questions-and-answers/82184-official-quant-thread-cat-2012-a.html#post3509369
(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.
((3^1024) -1 ) when divided by 2^x fetches 0 as remainder. Find the highest possible value of x?
Please post the approach... Dont know its OA :(
(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.
Use Euler for this :- E( 2^x ) =2^x ( 1-1/2) = 2^(x-1) Since 3 & 2 are coprime ..using euler we have : 3^ ( 2^(x-1)) mod 2^x =1 in this case 2^(x-1) = 1024 = 2^10 so x = 11 That is max value will be 11 .. & that divisor will be 2^11
Yeah I surmised it the same that aise hi kiya hoga. But now you have got your mistake so :) It is just that when are told to select/choose something out of a lot, we should NOT consider the order of picking up, as the final selection with you will be the same be it picked in any order. The que...
I did it this way: 10c1*10c1 + 10c1*10c1, for white and red being selected first each time.
Yeah I surmised it the same that aise hi kiya hoga. But now you have got your mistake so :) It is just that when are told to select/choose something out of a lot, we should NOT consider the order of picking up, as the final selection with you will be the same be it picked in any order. The question is asking about selecting only and so your job is done by just selecting them. Order won't make any difference.
No fool like an old fool. http://www.pagalguy.com/forums/cat-and-related-bschools/pagalguy-dream-team-2012-t-83955/p-3594200/r-4344097
I think u have done it this way out of 20 balls 1 ball can be selected in 20 ways the ball selected can be either red or white if its red than there are 10 white balls out of which 1 white ball can selected in 10 ways. so 20*10 =200 if you did it this way its fine. you have considered order so we just need to divide it by 2! 200/2! =100
hope it helps.
I did it this way: 10c1*10c1 + 10c1*10c1, for white and red being selected first each time.
GMAT 750. CAT 2012 - 99.90. Final Coverts : ISB, IIM - A, C, L.
I think u have done it this way out of 20 balls 1 ball can be selected in 20 ways the ball selected can be either red or white if its red than there are 10 white balls out of which 1 white ball can selected in 10 ways. so 20*10 =200 if you did it this way its fine. you have cons...
Sir ji pata nahi. Main abhi bhi wahin atka pada hoon. Hope they don't give both choices in the exam. Or else I'll score -1.
I think u have done it this way out of 20 balls 1 ball can be selected in 20 ways the ball selected can be either red or white if its red than there are 10 white balls out of which 1 white ball can selected in 10 ways. so 20*10 =200 if you did it this way its fine. you have considered order so we just need to divide it by 2! 200/2! =100
Sir ji bach gaye mere 4 number. Aap dono sahi ho. :cheers: NCERT class 11. Chapter no 7. P&C.; Example no 19. http://ncertbooks.prashanthellina.com/class_11.Mathematics.Mathematics/Ch-07(Permutation%20and%20Combinations%20FINAL%20%2004.01.06).pdf
well in this case a white ball can be selected in 10c1 ways and a red ball in 10c1 ways so total 10c1*10c1 = 100 ways
think of it this each red ball can make pair with 10 white balls since there are 10 red balls 10*10 = 100
Bhai...See have a look at this Question :
Number of ways to select 7 people comittee consisting of 3men and 4 women out of 5men and 5 women. Here, the answer should be 5C3*5C4 and I guess you would also agree with this one. It wont be 2 times of 5C3*5C4 here.
Like wise, as here those 7 people (3 men and 4 women) are different. In the same way, for selecting 1Red and 1White Ball --> the cases would be 10C1*10C1. Hope that makes it a li'l more clear...
Sir ji bach gaye mere 4 number. Aap dono sahi ho. :cheers:
Bhai...See have a look at this Question : Number of ways to select 7 people committee consisting of 3men and 4 women out of 5men and 5 women. Here, the answer should be 5C3*5C4 and I guess you would also agree with this one. It wont be 2 times of 5C3*5C4 here. As here those 7 people (3 m...
Sir ji pata nahi. Main abhi bhi wahin atka pada hoon. Hope they don't give both choices in the exam. Or else I'll score -1.
Bhai...See have a look at this Question :
Number of ways to select 7 people committee consisting of 3men and 4 women out of 5men and 5 women. Here, the answer should be 5C3*5C4 and I guess you would also agree with this one. It wont be 2 times of 5C3*5C4 here.
As here those 7 people (3 men and 4 women) are different. We are seeing the Committee as a whole and not considering the case that who is selected first (in whichever order they are selected). In the same way, for selecting 1Red and 1White Ball --> the cases would be 10C1*10C1. Hope that makes it a li'l more clear...
No fool like an old fool. http://www.pagalguy.com/forums/cat-and-related-bschools/pagalguy-dream-team-2012-t-83955/p-3594200/r-4344097
well in this case a white ball can be selected in 10c1 ways and a red ball in 10c1 ways so total 10c1*10c1 = 100 ways think of it this each red ball can make pair with 10 white balls since there are 10 red balls 10*10 = 100
Yaar jab irritate hone lago toh bata dena. I am seriously trying to understand this distinction... What is the answer to this question ATY ? No of ways of choosing 1Red and 1White ball from a box containing 10(numbered) red and 10 (numbered)white balls ? Isn't it 200 ?
well in this case a white ball can be selected in 10c1 ways and a red ball in 10c1 ways so total 10c1*10c1 = 100 ways
think of it this each red ball can make pair with 10 white balls since there are 10 red balls 10*10 = 100
My take : 32^(32*(32^31 - 1)) Assuming 32^32^32 = (32)^32^32 = (32)^32*32*...32 times (32)^32*32*...*32 times = (32)^32 * (32)^(32*32*...32 times - 32) so,it's (32)^(32*32*...32 times - 32) or 32^(32^32 - 32) times
It should be 100 according to me. Here, in which order they are picked won't matter Ankit. It's just about choosing two balls => Final outcome in your hand should be a Red and a White Ball choosen in any order.
Yaar jab irritate hone lago toh bata dena. I am seriously trying to understand this distinction... What is the answer to this question ATY ? No of ways of choosing 1Red and 1White ball from a box containing 10 red and 10 white balls ? Isn't it 200 ?
It should be 100 according to me. Here, in which order they are picked won't matter Ankit. It's just about choosing two balls => Final outcome in your hand should be a Red and a White Ball choosen in any order.
No fool like an old fool. http://www.pagalguy.com/forums/cat-and-related-bschools/pagalguy-dream-team-2012-t-83955/p-3594200/r-4344097
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