Question

Which letter will be exactly in the middle?

• A. \(L\)
• B. \(R\)
• C. \(D\)
• D. \(E\)
• A. \(2\)
• B. \(1\)
• C. \(4\)
• D. \(3\)
• A. IE
• B. IO
• C. AE
• D. AO
• A. \(I\)
• B. \(O\)
• C. \(A\)
• D. \(E\)
• A. R and T
• B. C and L
• C. R and L
• D. D and R
• A. G
• B. P
• C. H
• D. S

Answer 1

The Correct Option is D

Step 1: Start with the given word: \[ \texttt{ELECTROCARDIOGRAPH} \] This word has 18 letters. The first half (9 letters) is: \[ \texttt{ELECTROCA} \] Reversing this half gives: \[ \texttt{ACORTCELE} \] Step 2: The second half of the word (remaining 9 letters) is: \[ \texttt{RDIOGRAPH} \] After Step 1, the intermediate word is: \[ \texttt{ACORTCELERDIOGRAPH} \] Step 3: The last but one letter of the {original} word \texttt{ELECTROCARDIOGRAPH} is \(\texttt{P}\). Move this \(\texttt{P}\) to the beginning of the intermediate word (removing it from its original position there): \[ \texttt{PACORTCELERDIOGR A H} \;\Rightarrow\; \texttt{PACORTCELERDIOGRAH} \] Now the word still has 18 letters. Step 4: Add the letter \(\texttt{S}\) at the end: \[ \texttt{PACORTCELERDIOGRAHS} \] This final word has 19 letters, so the middle letter is the \(10^{\text{th}}\) letter. Counting carefully: \[ \texttt{P(1)\,A(2)\,C(3)\,O(4)\,R(5)\,T(6)\,C(7)\,E(8)\,L(9)\,\boxed{E(10)}\,R(11)\,D(12)\,I(13)\,O(14)\,G(15)\,R(16)\,A(17)\,H(18)\,S(19)} \] Thus, the letter exactly in the middle is \(\boxed{E}\).

Answer 2

From Question 1, the final transformed word is \[ \texttt{PACORTCELERDIOGRAHS}, \] which has 19 letters, so the middle letter is the \(10^{\text{th}}\) one, \(E\). Listing the letters to the left of the middle letter: \[ \texttt{P, A, C, O, R, T, C, E, L} \] Among these, the vowels are \(A, O, E\). Hence, there are \(3\) vowels to the left of the middle letter.

Answer 3

Using the final word: \[ \texttt{PACORTCELERDIOGRAHS} \] the vowels are at the positions: \[ A(2),\, O(4),\, E(8),\, E(10),\, I(13),\, O(14),\, A(17). \] The only pair of vowels that are side by side is at positions \(13\) and \(14\): \(\texttt{I}\) and \(\texttt{O}\). Thus, the adjoining vowels are \(\boxed{\texttt{IO}}\).

Answer 4

Again, consider: \[ \texttt{PACORTCELERDIOGRAHS} \] Marking each letter with neighbours: \[ \begin{array}{c|c|c} \text{Position} & \text{Letter} & \text{Left / Right neighbour}
\hline 13 & I & D\ (\text{consonant}),\ O\ (\text{vowel}) \end{array} \] At position 13, the letter \(I\) has \(D\) (a consonant) on its left and \(O\) (a vowel) on its right, fitting the required pattern. No other vowel satisfies “consonant on the left and vowel on the right”. Hence, the vowel is \(\boxed{I}\).

Answer 5

We look for patterns of the form \(\text{vowel} - \text{consonant} - \text{vowel}\) in \[ \texttt{PACORTCELERDIOGRAHS}. \] Checking triplets: \[ \texttt{A(2)C(3)O(4)} \quad\Rightarrow\quad C \text{ is between } A \text{ and } O, \] \[ \texttt{E(8)L(9)E(10)} \quad\Rightarrow\quad L \text{ is between } E \text{ and } E. \] Thus the consonants sandwiched between two vowels are \(C\) and \(L\).

Answer 6

The original word is: \[ \texttt{ELECTROCARDIOGRAPH}. \] The first half \(\texttt{ELECTROCA}\) is reversed to \(\texttt{ACORTCELE}\), and the second half is left as it is, giving: \[ \texttt{ACORTCELERDIOGRAPH}. \] According to the instructions, the {last but one} letter of the original word is identified and moved to the beginning. The last but one letter of \(\texttt{ELECTROCARDIOGRAPH}\) is \(\boxed{P}\), so the prefixed word becomes: \[ \texttt{PACORTCELERDIOGRAH}. \] Therefore, the letter prefixed after the first half is reversed is \(P\).