Previous Year Topic Wise Questions With Solutions: Coding & Decoding | CSAT Preparation - UPSC PDF Download

Ans: (a)
Here, the code is being calculated by multiplying the position values of the alphabet. 
ABCD = 24 1 × 2 × 3 × 4 = 24
EFGH = 1680
5 × 6 × 7 × 8 = 1680
IJKL = 9 × 10 × 11 × 12 = 11880

Ans: (d)
In a certain code, ‘POT’ is written as ‘ATOP’ and ‘TRAP’ is written as ‘APART’, Logic: Here the word is being reversed and prefix ‘A’ is being added to the front. Reverse order of POT = TOP
Thus, code of POT will be ’ATOP’. Reverse order of TRAP = PART
Thus, code of TRAP will be ’APART’.
Similarly, Reverse order of ARENA = ANERA Thus, code of ARENA will be ’AANERA’.

Ans: (d)
Here, we are required to decipher or decode the pattern.
ZERO is written as CHUR. We can see that the underlying pattern is very simple, as shown below:
Z + 3 = C
E + 3 = H
R + 3 = U
O + 3 = R
We will follow a similar pattern to code PLAYER.
P + 3 = S
L + 3 = O
A + 3 = D
Y + 3 = B
E + 3 = H
R + 3 = U
So, the required code is SODBHU.
Hence, option (d) is correct.

Ans: (d)
Method I: The given sequence: ABC_ _ABC_DABBCD_ABCD
We can break this sequence in five sets of 4 elements each.
On analysing it in this manner, we can see that the last element moves to the first position in a cyclic manner.
The complete sequence is: ABCD DABC CDAB BCDA ABCD
Hence, option (d) is correct.
Method II: The given sequence: ABC_ _ABC_DABBCD_ABCD
There are 20 elements in this series. Let’s break this sequence in four sets of 5 elements each.
ABC_ _
ABC_D
ABBCD
_ABCD
The third set ABBCD gives us a clue that one letter is being repeated. Now, we can check from the options to further confirm our presumption.
Using option (d), we get the complete sequence: ABCDD ABCCD ABBCD AABCD
We can see that in the first set 4^th letter D is getting repeated, in the second set 3^rd letter is getting repeated, and so on.
Hence, option (d) is correct.

Ans: (d)
We basically need to find the opposite letter of the letters in the given word.
Opposite Letter Position = 27 – Letter Position
So, in case of LUCKNOW, we get:
27 – 12 = 15 = O
27 – 21 = 6 = F
27 – 3 = 24 = X
27 – 11 = 16 = P
27 – 14 = 13 = M
27 – 15 = 12 = L
27 – 23 = 4 = D
The word that we get is OFXPMLD.

Ans: (b)
In a code language 'MATHEMATICS' is written as 'LBSIDNZUHDR'.
The pattern used in the above coding is as follows:
M (13) → L (12), i.e. reduction of 1
A (1) → B (2), i.e. increase of 1
T (20) → S (19), i.e. reduction of 1
H (8) → I (9), i.e. increase of 1
And so on.
So, the code of CHEMISTRY will be: BIDNHTSSX