46. There is a circular table and 60 people can sit in that. If there are N number of people sitting and a monster comes and want to occupy a seat such that he has someone on his side. What is the minimum value of N?
a) 15
b) 20
c) 29
d) 30

Say, one person is sitting in seat no. 2 then d monster sits in seat no. 1 or 3 (monster will find a person beside it)
Again there is a person sitting in seat no. 5, then d monster can sit in seat no 4 and 6...
again there is a person sitting in seat no. 8
Thus d pattern of ppl sitting is-
sit no. 2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59
thus min no. of n is 20

47. How many 3 digits numbers less than 1000 are there in which, if there is a 3 it is followed by 7? Given that no two digit of the number should be same.

suppose d 3 digit no. starts with 1
10__(This blank space can b filled in 7 ways 2,4,5,6,7,8,9)
in d above case we cannot take 3 since it should be followed by 7 which is not possible here.
now we take-
11__ this space can also be filled in 7 ways
12__ 7 ways
13__ this space can be filled only by 1 way and dat is 7.
14__ again 7 ways
15__ 7 ways
16__ 7 ways
17__ 7 ways
18__ 7 ways
19__ 7 ways
that gives us 57 ways in total

We will follow d same procedure for 3 digits starting with 2,4,5,6,8,9(but not in case of 3 and 7)

thus we get 57*7= 399 ways

now lets check condition for 3. 3 is always followed by 7 and nothing else
thus
37__ this space can be filled in 8 ways(0,1,2,4,5,6,8,9)
no other condition is possible in case of 3.
that makes our total to be 399+8=407

now lets check condition for 7
70__ this can be filled in 7 ways
71__ 7 ways
72__ 7 ways
73__ this condition is not possible since 3 should be followed by 7 but we cannot repeat 7.
74__ 7 ways
75__ 7 ways
76__ 7 ways
78__ 7 ways
79__ 7 ways
this gives us 56 ways
therefore our total becomes 407+56=463 and this is the answer.

48. How many numbers are there between 1100 - 1300 which are divisible by and that all 4-digits of number should be odd (ex:1331)

1300-1100=200 now divide it by 2 we have 100 even number hence 100 odd no's
now we see 50 are in each 1100-1200 and 1200-1300 now we see in 1200 to 1299 2 will always be there
hence no number frm here in 1100-1199 we have 1113,1119,1131,1137,1155,1173,1179,1191,1197
hence
9 no's

49. Three dice are rolled simultaneously. Find the probability of getting at least one six.
a) 1/6*5/6*5/6
b) 5/6*5/6*5/6
c) 3*1/6*5/6*5/6
d) none of the above

condition for no sixes is-
(5/6)*(5/6)*(5/6)
i.e. 125/216
therefore for atleast 1 six-
=1-(125/216)
=91/216
therefore option d)

50. If there are n numbers of square cubes and you are given two colors, black and white. How many unique square cubes can u color?

Let B= Black color W=white color
If all 6 face colured with B and none with W -> 1 unique cube
5 B and 1 W->1
4 B and 2 W->2
3 B and 3 W->2
2 B and 4 W->2
1 B and 5 W->1
0 B and 6 W->1
----------
Answer=10

how many numbers can be formed using digits(1,2,3,4,5,6,7,8,9) such that they are in increasing order (e.g. 12345,345,6789,123456789)

ReplyDeletenumbers may be 1 or 2 or 3 or.........9 digit

ReplyDelete9c1+9c2+9c3+9c4+...........+9c9

=2^9-1

=511

ReplyDeleteBest place to prepare elitmus exam...

Click Here to know more ant elitmus pattern

Best place to prepare elitmus exam...

Click Here to know more ant elitmus pattern

Best place to prepare elitmus exam...

Click Here to know more ant elitmus pattern

Best place to prepare elitmus exam...

Click Here to know more ant elitmus pattern