In a certain code language, 'BELL' is written as 'MKFA' and 'BOND' is written as 'EMPA'. How will 'BLEW' be written in that language?

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Let's analyze the coding pattern:

First, write the letters and their corresponding codes:

BELL → MKFA
BOND → EMPA

Let's compare each letter position:

BELL:  B  E  L  L
Code:  M  K  F  A

BOND:  B  O  N  D
Code:  E  M  P  A

Let's try to find the pattern for each position:

1st letter:
B (2) → M (13) in BELL
B (2) → E (5) in BOND
So, the first letter is not coded the same way in both words. Let's check the last letter:

4th letter:
L (12) → A (1) in BELL
D (4) → A (1) in BOND
So, the last letter is always coded as 'A'.

Now, let's try the second letter:
E (5) → K (11) in BELL
O (15) → M (13) in BOND

Check the difference:
K (11) - E (5) = 6
M (13) - O (15) = -2
No clear pattern.

Let's try the third letter:
L (12) → F (6) in BELL
N (14) → P (16) in BOND

F (6) - L (12) = -6
P (16) - N (14) = 2
No clear pattern.

Let's try to see if the code is reversed or rearranged. Let's check if the code is an anagram of the original word.

Now, let's look for a pattern in the code words:

BELL: B E L L → MKFA
BOND: B O N D → EMPA

Notice that both code words end with 'A'.

Now, let's check if the code is formed by shifting letters:

BELL: B(2) E(5) L(12) L(12)
MKFA: M(13) K(11) F(6) A(1)

Check the difference for each position:
1st: M(13) - B(2) = 11
2nd: K(11) - E(5) = 6
3rd: F(6) - L(12) = -6
4th: A(1) - L(12) = -11

Similarly for BOND:
B(2) O(15) N(14) D(4)
E(5) M(13) P(16) A(1)

1st: E(5) - B(2) = 3
2nd: M(13) - O(15) = -2
3rd: P(16) - N(14) = 2
4th: A(1) - D(4) = -3

It seems the pattern is not a simple shift. Let's check if the code is formed by rearranging the letters and then shifting.

Notice that in both code words, the last letter is 'A'.

Now, let's check if the code is the reverse of the original word with some shifting:

BELL → MKFA
Reverse of BELL is L L E B
But MKFA is not a direct shift of LLEB.

Let's try mapping the letters:

BELL: B E L L
MKFA: M K F A

B → M
E → K
L → F
L → A

BOND: B O N D
EMPA: E M P A

B → E
O → M
N → P
D → A

So, for each position, let's see if the code is the same for the same position:

First letter: B → M (BELL), B → E (BOND)
Second letter: E → K (BELL), O → M (BOND)
Third letter: L → F (BELL), N → P (BOND)
Fourth letter: L → A (BELL), D → A (BOND)

But the last letter is always 'A'.

Now, let's try to see if the code is formed by mapping the first letter to the last, and so on.

Alternatively, let's try to see if the code is formed by writing the word backwards and shifting each letter by a fixed amount.

Let's try for BELL:
Reverse: L L E B
Now, let's see if L → M, L → K, E → F, B → A
But L(12) → M(13): +1
L(12) → K(11): -1
E(5) → F(6): +1
B(2) → A(1): -1
So, alternately +1, -1, +1, -1

Try for BOND:
Reverse: D N O B
D(4) → E(5): +1
N(14) → M(13): -1
O(15) → P(16): +1
B(2) → A(1): -1
So, again alternately +1, -1, +1, -1

So, the code is formed by reversing the word and then alternately adding 1 and subtracting 1 from the letter's position.

Let's apply this to BLEW:
Original: B L E W
Reverse: W E L B
Now, alternately +1, -1, +1, -1

W(23) + 1 = X(24)
E(5) - 1 = D(4)
L(12) + 1 = M(13)
B(2) - 1 = A(1)

So, the code is: X D M A

Therefore, the answer is XDMA.

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