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% ?

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 )

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

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

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

mba4mheart Says

my take is also 0

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


bhai please continue here.request

Find the remainder when
1^5+2^5+3^5+... ...+100^5 is
divided by 4? (explain plz)

my take is also 0

Please continue your discussions here.

http://www.pagalguy.com/forum/quantitative-questions-and-answers/82184-official-quant-thread-cat-2012-a.html#post3509369

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.

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..

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

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

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

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

yup im also getting 12 !!!!

thanks fapsian19... here's another 1...
*1!+2*2!+....+12*12! when divided by 13 leaves remainder??

the expression 1*1!+2*2!+....+12*12! = (2! - 1) + (3! - 2!) + (4! - 3!) + ... + (12! - 11!) + (13! - 12!) = 13! - 1

13! - 1 mod 13 = -1 = 12

gudda1122 Says

Bhai answer is 10/(3)^1/2((3^1/2)+1)

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

thanks fapsian19... here's another 1...
*1!+2*2!+....+12*12! when divided by 13 leaves remainder??

SIMPLIFIED BECOMES 13!-1/13=-1 REMAINDER I.E 12 remainder as ans

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

mani0303 Says

I guess it's 10/(3 + rt(3)) ; ll share the approach if the answer's right by any chance:)

Bhai answer is 10/(3)^1/2((3^1/2)+1)

((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.

thanks fapsian19... here's another 1...
1*1!+2*2!+....+12*12! when divided by 13 leaves remainder??

1*1!+2*2!+....+12*12! = 1!*(2-1) + 2!(3-1) + 3!(4-1) + ... + 12!(13-1)

= 2! - 1! + 3*2! - 2! + 4*3! - 3! + ... + 13*12! - 12!

Every term will be cancelled except 13! - 1!

=> 13! - 1! mod 13 = -1 or 12

thanks fapsian19... here's another 1...
*1!+2*2!+....+12*12! when divided by 13 leaves remainder??

3^1024-1^1024

(3^512-1^512)(3^512+1^512)=only one power of 2

similarly u will get 2 for every step till 3^1-1=2

hence answer will be 2^11

Please continue here :-

http://www.pagalguy.com/forum/quantitative-questions-and-answers/82184-official-quant-thread-cat-2012-a.html#post3509369

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

((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

in the triangle ABC,Angle c=30.O is the center.if AM:MB=(3)^1/2:1.find the radius of the circle given that BC=10.

please post approach

I guess it's 10/(3 + rt(3)) ; ll share the approach if the answer's right by any chance:)

in the triangle ABC,Angle c=30.O is the center.if AM:MB=(3)^1/2:1.find the radius of the circle given that BC=10.

please post approach

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.

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.

AnkitCat2012 Says

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

hope it helps.

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:

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

AnkitCat2012 Says

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...

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

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.

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

singhjaskiran81 Says

In 32^32^32 (not (32^32)^32 ), How many times is 32^32 multiplied ?

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

Estallar12 Says

Should be 32^32 /32 = 32^31 times Bhai... :)

.............................................

singhjaskiran81 Says

In 32^32^32 (not (32^32)^32 ), How many times is 32^32 multiplied ?

Should be 32^32 /32 = 32^31 times Bhai...

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.