Very Nice Question.... I don't think the answer would have been derivable without the options..
Anyway, The solution can be arrived as:

Condition 1:
Notice the pattern here:
4=2*2,(2,2)is a pair of prime numbers.
14=2*7,(2,7)is a pair of prime numbers.
11=1*11,(1,11)is a pair of prime numbers.
34=(2*17),(2,17)is a pair of prime numbers.
35=5*7,(5,7)is a pair of prime numbers.
26=2*13,(2,13)is a pair of prime numbers.
73=1*73,(1,73)is a pair of prime numbers.

Condition 2:
Again notice if we reverse both the series and write each term individually:
3 5 2 6 7 3 - - ...........(1)
0 4 1 4 1 1 3 4 ...........(2)

Now if we subtract we get
3 1 1 2 6 2 (-) (-)

If any case (-) and (-) would be positive so, the missing number has to greater than 34 in both the tenth's place and unit's place.

So, the last missing number would also have to satisfy condition (1) and condition(2)...

Let's check the option:
51=3*17,(3,17)is a pair of prime numbers.
56=8*7,(8,7)is not a pair of prime numbers.
45=5*9,(5,9)is not a pair of prime numbers.
55=5*11,(5,11)is a pair of prime numbers.
52=4*13,(4,13)is not a pair of prime numbers.

We get two numbers i.e. 51 and 55 which satisfy condition 1.
Let's Check Condition 2.
51 does not satisfy condition 2, as 1 in the unit's place of 51 is less than 4 in the unit's place of 34.
55 also satisfies this condition.

Hence,Answer is D)55.

Note: Some of you may argue that we how can I write 11=1*11, which would also mean that I can write 14=1*2*7.In that case, i would like to say that see in case of 11 and 73, they are whole prime numbers whereas others in the list are partition prime numbers...

Lastly, in case of condition 2, we need not write both the series upside down, but that is just shown for convenience.

[ 4, 14] [11,31]
[35,26] [73, ?]

In the 1st set the 1st number is 4.just note that the double of 4 plus 3= 11 which is the 1st number of 2nd set. and the 2nd number of 1st set is 14.the double of 14 plus 3=31 which is the 2nd number of 2nd set.in the 3rd set in the same way, 1st number is 31. the double of 35 plus 3=73 which is the 1st number of 4th set.so,2nd number of 3rd set is 28.so the double of 26 plus 3=55

so,the missing number is 55.

Mr. Sourangsu Bandyopadhayay
I got to tell you something . (1,11) & (1,73) you have indicated them as a pair of prime Number .. ri8?? As far as I know 1 is neither a PRIME no a composite . So will you please explain this point ?? uhm waiting for ur answer.

Barnisa Chottopadhyay : how do u think so... :-(
kindly give me a tip to solve series...
And also in interview u wont find time to think like this.rite??

2nd digit 14(7*2)
5th digit 35(7*5)
8th digit will 56(7*8)
ans 56

exact question is:
Identify the missing number.
4 14 11 34
35 26 73 ?
options:
A B C D E
51 56 45 55 52

and ans is (D)55

In the 1st set the 1st number is 4.just note that the double of 4 plus 3= 11 which is the 1st number of 2nd set. and the 2nd number of 1st set is 14.the double of 14 plus 3=31 which is the 2nd number of 2nd set.in the 3rd set in the same way, 1st number is 31. the double of 35 plus 3=73 which is the 1st number of 4th set.so,2nd number of 3rd set is 28.so the double of 26 plus 3=55

so,the missing number is 55.

In the 1st set the 1st number is 4.just note that the double of 4 plus 3= 11 which is the 1st number of 2nd set. and the 2nd number of 1st set is 14.the double of 14 plus 3=31 which is the 2nd number of 2nd set.in the 3rd set in the same way, 1st number is 31. the double of 35 plus 3=73 which is the 1st number of 4th set.so,2nd number of 3rd set is 26.so the double of 26 plus 3=55

so,the missing number is 58

11*2+4=26
34*2+5=73
4*2+6=14
14*2+7=35
the final answer is 14