official quant thread for cat08

[KIMSTER]

Originally Posted by amitkrsingh

Is answer b)61

Kindly post your approach...Thanks in advance
@all the puys on this thread online :Is this thread always active at this time of the day...or are u guys in a different time-zone or wat :P

[nikijpr]

Originally Posted by KIMSTER

I think we can use reminder theoreum for this equating divisor to zero
(3)^20 + 1 =0; 3^20 =-1

we can substitute for 3^84+3^63+3^42+3^21+1

(-1)^4+ (-1)^3+ (-1)^2+(-1)^1 +1= 1

Is the answer option c) ? really rusty with reminder theoreum...so not so sure,we can do like this...

Using this approach (-1)^4 * 3^4 + (-1)^3 * 3^3 + (-1)^2 * 3^2 + (-1)^1 * 3^1 + 1
It comes out to be 61

[amitkrsingh]

Originally Posted by KIMSTER

Kindly post your approach...Thanks in advance
@all the puys on this thread online :Is this thread always active at this time of the day...or are u guys in a different time-zone or wat :P

Previous post has been edited for approach.... well was seeing movie at this time...... just got over and saw few question which could be solved..... so.....

[KIMSTER]

Originally Posted by nikijpr

Using this approach (-1)^4 * 3^4 + (-1)^3 * 3^3 + (-1)^2 * 3^2 + (-1)^1 * 3^1 + 1
It comes out to be 61

Yes you are rite...Both are appoach are the same,but i left out multiplying the remaining powers of 3.... Ans=81-27+9-3+1=61..

[implex]

Originally Posted by naga25french

is the answer b) 4 ?
solution:

55x^2+2xy+y^2=2007..

54x^2 + (x + y )^2 = 2007

when x= 3 (x+y)^2 = 1521 =======> 39^2


the ordered pairs can be formed by writing 39^2 in 4 different ways...


ordered pairs are (3,36) ,(-3,-36), (-3 , 42) and (3,-39)

nicely done 4 pairs are there !!

[implex]

Originally Posted by selebratinglife

Ans is 2/3....Assuming 'a' to be the side of hexagon, We have BE=2a and BE bisects AC for M and N to be collinear the ratio AM/MC should be greater than 0.5, else N and E would coincide. So checked for 2/3 and it clicked..Else would be none of these. This method can be applied during exam to save time. It can be solved by proper geometry though but is a bit time consuming process.

answer is correct , approach isn't that bad either! but can be done better!

[selebratinglife]

Originally Posted by implex

New Problem!!
Let X={1,2,3...k} and let S be any non-empty collection of subsets of X. Then Define S' to be collection of all subsets of X that are subsets of an odd number of elements in S.

a) S'=S
b) (S')'=S
c) S'=X
d) (S')'=X
e) none of the above!!

If we take for example X={1,2,3}
Total number of subsets are 7(2^3-1){non-empty)=S
S' would have all the elements of X only when it has odd number of elements and when it has even number of elements this is negated..So it should be none of these. Right?

[naga25french]

Originally Posted by amitkrsingh

Is answer b)61

Using reaminder theorem would be:- 81 - 27 +9 -3 +1= 61

yeah its right...

i posted this question for one funda..


here is that funda

The point to be remembered is whenever x^n is divided by (x^(n-1)) + 1 the remainder will always be -x...


in our problem
3^n is divided by (3^n-1) + 1 the remainder will always be -3.


hope its useful...

[KIMSTER]

New Question:
1)How many 3 digit natural numbers have their cubes ending with 24?
a)20
b)18
c)16
d)22


Ps: Kindly post your approach along with the answer,as i got the answer but its a lil time consuming.

[naga25french]

Originally Posted by KIMSTER

New Question:
1)How many 3 digit natural numbers have their cubes ending with 24?
a)20
b)18
c)16
d)22


Ps: Kindly post your approach along with the answer,as i got the answer but its a lil time consuming.

answer is b) 18