My scores:

OA______45A____34C____91
Sec1____20A____16C____44
Sec2____25A____18C____47

PS: Can anyone tell me how to view the poll result, as in names of guys with >120 score ??

The guy who started the thread should have made it public while creating a poll, then its possible to see the names of guys who polled.
I think, now a moderator can change it.

How many 3 digit numbers are divisible by 3 or 5 but not by 7?

1. 51
2. 300
3. 360
4. 180
5. 128

We need to find numbers divisible by 3,5 and eliminate those divisible by 21,35,15 and add those divisible by 105.

divisible by 3 => 102 to 999 => 897/3 + 1 => 300
divisible by 5 => 105 to 995 => 895/5 + 1 => 180
Divisible by 21 => 21*47 is near 999 so , 47 - 4 = 43 such numbers
Divisible by 35 => 35*28 is near 999 so, 28 - 2 = 26 such numbers
Divisible by 15 => 105 to 990 => 885/15 + 1 => 60
Divisible by 105 are 9 numbers

so, Answer is 300 + 180 - 129 + 9 => 360

emanresu Says

P is a group of four numbers 1, 2, 3 and 1. In every step, 1 is added to any two numbers in group P. In how many such steps is it possible to make all the four numbers in group P equal?

Its not possible to make all of them equal.

Find the sum of the last three digits of S=871x873x875x878x881x883?

129*127*125*122*119*117
Last digit is 0
now, Last two digits of
29*27*25*61*19*17 is 25

h.kagda Says

How many ordered pairs (x,y) of integers are possible for xy/(x+y) = 196?

1/x + 1/y = 1/196
(x+196)(y+196) = 196^2
RHS takes all the divisors of 196^2 = 14^4 => 7^4 * 2^4 => 25
(Oops Ordered pairs i ignored)

@RC89 : Wish you a Happy Birthday

yeh wala try kariye zara bhai log .. mera frnd planet of the apes dekh ke aaya aur mujhe pucha .. he said usi movie me dekha tha ...

there are 4 wts . a,b,c,d
wts = a>b>c>d
there are 3 pans . the wts are kept in the first pan such that the heaviest is at the bottom.
u need to re-shuffle them and place them back again using all the 3 pans such that the same formation is obtained ( each wt. needs to be moved atleast once ).
rule = at no point can u keep the heavier wt on a lighter one

d
c
b
a
_ _ _
1 2 3

question is , minimum how many moves are needed ?

Its 2^n - 1, n is number of weights
(For more info search for towers of Hanoi)

not sure if am right .. jus approximating the calculation

number of 5 will be 2*67 = 134

number of 5^2 will be 1*13 = 13

number of 5^3 will be 1*6 = 6

number of 5^4 will be 1*3 = 1

so i think its around 154

It will be 168

Ten tickets are numbered 1,2,3,...,10. Six tickets are selected at random one at a time with replacement. The probability that the largest number appearing on the selected ticket is 7 is???

7^6 is the number of cases in which the 6 tickets chosen are numbered between 1 to 7 i.e., where there is no 8, 9 or 10.
6^6 is the number of cases in which the 6 tickets are numbered between 1 to 6 i.e. where there is no 7, 8, 9 or 10.
7^6 - 6^6 is the no of cases in which 7 is there and all are numbered between 1 to 7.

So, Probability is (7^6 - 6^6)/10^6

can u post the solution for this prob of why it is 5 ?

i am getting just two solutions

y takes value 44 and 60

Here are the solutions
(19,684)
(21,252)
(27,108 )
(45,60)
(99,44)

There are two numbers A and B. A can be expressed as a product of 13 and a two-digit prime number and B can be expressed as a product of 17 and a two-digit prime number. If the unit's digit of the product of A and B is 7, then how many distinct products of A and B are possible?

Two digit prime numbers can be
(11,31,41,61,71) & (17,37,47,67,97)
or
(13,23,43,53,73,83)&(19,29,59,79,89)
So, total distinct products are 5*5 + 6*5 = 55

we have 8 digits------2,2,3,3,4,4,4,4

out of these how many 4digits are possible greater than 3000?

Case 1: 3_ _ _ => 27 - 2 => 25
Because, we do not get 222 & 333 case
Case 2: 4_ _ _ => 27 - 1 => 26
Because we do not get 222 case

So, 51 numbers