official quant thread for cat08

[nigam jha]

Originally Posted by rahul_4096

how u can say that bced is cyclic quadrilateral ????

made an asumption and solved the question and after that checked if
assumption is violated anywhere.
i didnt find any violation in my solution for concyclic assumption

[hi.shivani]

Q1) LCM of two numbers A and B =P^x*Q^y,where P and Q are prime numbers and x and y are positive whole numbers. How many set of values are positive whole numbers.How many set of values are possible for A and B?

a) xy(x+y) b) xy(x-y) c) x^2*y^2(x+y) d) None of these


Q2) When 7179 and 9699 are divided by another natural number N,remainder obtained is same.How many values of n will be ending with one or more than one zeroes?

a) 24 b) 124 c) 46 d) None of these

[seba_catrpillar]

Originally Posted by shashank3012

Hi seba,
The n in the condition belongs to the set of real nos. and not integers,right.
1)If the question is,
2(n!)*2(n!)-in this case,answer comes to be 4x.
2)If n belongs to real nos.
[(2n)!]^2=x
So,
(2n)!=sqrt(x)
Here,(2n)!=6
So,x=36.
Btw whats the solution.

why is 2n! = 6???
btw the sum is (2n!)^2
I also didn't got the total logic behind your solution...... can you please elaborate it a bit

[prakharc]

Originally Posted by hi.shivani

Q1) LCM of two numbers A and B =P^x*Q^y,where P and Q are prime numbers and x and y are positive whole numbers. How many set of values are positive whole numbers.How many set of values are possible for A and B?

a) xy(x+y) b) xy(x-y) c) x^2*y^2(x+y) d) None of these


Q2) When 7179 and 9699 are divided by another natural number N,remainder obtained is same.How many values of n will be ending with one or more than one zeroes?

a) 24 b) 124 c) 46 d) None of these

Ans 1) d) None of these
One number has P^x then other number should have Q^y
So we can have 1st number in y ways
We can have 2nd number in x ways
So possible set of values= xy

One number has both P^x and Q^y ......2nd number we can have in (x+1)(y+1) ways
So total number of sets = xy+(x+1)(y+1).........and if it's ordered pair...then 2{xy+(x+1)(y+1)}

Ans 2)9699= Na+r
7179=Nb+r
=> N(a-b)= 2520=2*5* 2^2 * 7 * 3^2

So N must be a factor of 2520.Excluding one 2 and 5 for getting 0 at the end....possible values of N are= 3*2*3=18

So answer d) None of these

[poojadatta]

How many set of the form Sn={n-3,n-2,n-2,n,n+1,n+2,n+3} where n is a natural number such that n<=100 do not contain 8 or any integral multiple of 8.
1) 13 2)14 3)15 4)16

Please post your approach also.

[hi.shivani]

Q) Let S={1,2,3,...,n} be a set of n natural numbers. Let T be a subset of S such that the sum of any three natural number is not less than N .Find the maximumnumber of elements in any such subset T for N=40?
a) 26 b) 27 c) 28 d) None of these

I am getting 28 but the answer is d) None of these..
Can some one point out where I m wrong??

I am taking T={13,14,...40}

[ankaj]

Originally Posted by poojadatta

How many set of the form Sn={n-3,n-2,n-2,n,n+1,n+2,n+3} where n is a natural number such that n<=100 do not contain 8 or any integral multiple of 8.
1) 13 2)14 3)15 4)16

Please post your approach also.

Is the answer 13?
if we consider both n-3 and n+3 non multiples of 8...we can get these sets...
1-7; 9-17; .....; 97-103. total 13 sets...

[HYgoldking]

2) 7^ 50 % 1001
1001=7*11*13
7^ 50 % 7 = remainder is zero
7^ 50 % 11 = remainder is 1 (Eulers Method)
7^ 50 % 13 = 7^2 % 13 (Eulers Method) = remainder is 10
7x, 11y+1, 13z+10

3) 244^ 1500 % 1001
1001=7*11*13

244^ 1500 / 7 = (243+1)^500 remainder is 1
244^ 1500 / 11 = remainder is 1 (Eulers Method)
244^ 1500 / 13 = remainder is 1 (Eulers Method)
7x+1, 11y+1, 13z+1

[Fuzon]

How many set of the form Sn={n-3,n-2,n-2,n,n+1,n+2,n+3} where n is a natural number such that n<=100 do not contain 8 or any integral multiple of 8.
1) 13 2)14 3)15 4)16

Please post your approach also.

Ans option 1. 13

Logic :

Starting from 4 , the next time that the condition is satisfied is at 12 , then again at 20

We get a series of 4,12,20,......

Now 4+8x=100 --> x=12 ,
thus the total number of sets is 13....

[poojadatta]

@ankaj yes u are right can u elaborate ur answer.