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

Position-tracking approach (without rewriting the full word at each stage):
The original word \(\texttt{ELECTROCARDIOGRAPH}\) has 18 letters. Reversing its first 9 letters, \(\texttt{ELECTROCA}\), gives the block \(\texttt{ACORTCELE}\), which occupies positions 1 to 9 of the intermediate word; the last 9 letters, \(\texttt{RDIOGRAPH}\), stay at positions 10 to 18.
The last-but-one letter of the original word is \(\texttt{P}\) (position 17), which lies inside this unchanged second block. Pulling it out and placing it first shifts every other letter one slot along, so the new arrangement has \(\texttt{P}\) at position 1, followed by the old positions 1-16, and then the old position 18 at the very end, still 18 letters in all.
Adding \(\texttt{S}\) makes the word 19 letters long, so the middle letter sits at position \(10\). Since position 1 is \(\texttt{P}\), position 10 is simply the 9th letter right after it, that is, the last letter of the reversed block \(\texttt{ACORTCELE}\), which is \(\texttt{E}\).
Hence the middle letter is \(\boxed{E}\), option (D).

Answer 3

The question asks which letter sits exactly in the middle of the word after all three transformations are applied to "ELECTROCARDIOGRAPH." Instead of rewriting the whole word, trace where each candidate letter's occurrences actually end up.

The original word has 18 letters, so its first half is positions 1 to 9, and reversing that half sends the letter originally at position \(i\) (for \(i \le 9\)) to intermediate position \(10 - i\). Moving the last-but-one letter (originally at position 17, which is P) to the front then shifts every letter that was ahead of position 17 forward by one place. So a letter starting at original position \(i \le 9\) ends up at final position \((10-i)+1 = 11-i\).

1. \(L\): L sits at original position 2. Using the formula, its final position is \(11 - 2 = 9\), one short of the middle (position 10), so L is not the answer.
2. \(R\): R appears at original positions 6, 10 and 15. Position 6 falls in the first half, giving final position \(11 - 6 = 5\). Positions 10 and 15 belong to the untouched second half, and after the front-shift caused by moving P, they land at 11 and 16. None of these is position 10.
3. \(D\): D sits at original position 11, in the second half. It shifts to final position 12 after the front-move, not 10.
4. \(E\): E appears at original positions 1 and 3. Position 3 gives final position \(11 - 3 = 8\). Position 1 gives final position \(11 - 1 = 10\), which is exactly the middle of the 19-letter final word.

Only one occurrence, the E that started as the very first letter of "ELECTROCARDIOGRAPH," lands exactly on position 10 after all three steps.

The correct answer is \(E\).

Answer 4

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 5

Vowel-tracking approach (without listing every letter individually):
From the previous question, the transformed word's first 9 letters (positions 1 to 9, all lying to the left of the middle) consist of the moved letter \(\texttt{P}\) followed by the reversed block \(\texttt{ACORTCELE}\) minus its own last letter, since that last letter of the block becomes the middle letter itself.
Removing that last letter (\(\texttt{E}\)) leaves the block's remaining letters \(\texttt{A, C, O, R, T, C, E, L}\) sitting to the left of the middle, alongside the consonant \(\texttt{P}\).
Marking vowels among these: \(\texttt{A}\), \(\texttt{O}\), and \(\texttt{E}\) are vowels, while \(\texttt{P, C, R, T, C, L}\) are consonants. That gives exactly \(\boxed{3}\) vowels to the left of the middle letter, option (D).

Answer 6

The question asks how many vowels appear among the 9 letters to the left of the middle letter in the transformed word. Those 9 letters, from the previous question's result, are P, A, C, O, R, T, C, E, L. Check each answer choice against an actual count.

1. \(2\): this would require exactly two of the nine letters to be vowels. Going through the letters one by one, P, C, R, T, C and L are all consonants, while A, O and E are vowels, three vowels in total, so this option undercounts by one.
2. \(1\): this would mean only one vowel among the nine letters, but the same letter-by-letter check finds three separate vowels (A, O, E) sitting at different positions, so this option is too low.
3. \(4\): this would require a fourth vowel somewhere among the nine letters. No such letter exists, the only vowels present are A, O and E; the rest, P, C, R, T, C and L, are all consonants, so this option overcounts.
4. \(3\): matches the actual tally exactly. Scanning P, A, C, O, R, T, C, E, L gives three vowels (A, O, E) and six consonants (P, C, R, T, C, L), so three is the correct count.

The correct answer is \(3\).

Answer 7

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 8

Alternate Method:
Write the letters of \(\texttt{PACORTCELERDIOGRAH}\) one after another and tag each as vowel (V) or consonant (C):
\[\texttt{P(C) A(V) C(C) O(V) R(C) T(C) C(C) E(V) L(C) E(V) R(C) D(C) I(V) O(V) G(C) R(C) A(V) H(C)}\]
Scanning this V/C string for two V's occurring back to back, the only such occurrence is at the pair \(\texttt{I}\)-\(\texttt{O}\).
So the two vowels that sit next to each other are \(\boxed{\texttt{I, O}}\), matching option (b) \(\texttt{IO}\).

Answer 9

The question asks which pair of vowels from the options actually sits side by side in the coded word \(\texttt{PACORTCELERDIOGRAH}\). Check each option's letter pair directly against the word instead of deriving it from scratch.

1. IE: Look for \(I\) immediately followed by \(E\), or \(E\) immediately followed by \(I\), anywhere in the word. \(I\) appears only once, and its neighbours are \(D\) before it and \(O\) after it, not \(E\), so this pair never occurs together.
2. IO: Check whether \(I\) and \(O\) ever sit side by side. Scanning the word, \(I\) is directly followed by \(O\), so this pair does occur together.
3. AE: \(A\) appears twice and \(E\) appears twice in the word, but none of their positions land next to each other anywhere in \(\texttt{PACORTCELERDIOGRAH}\), so \(A\) and \(E\) never touch.
4. AO: Likewise, every occurrence of \(A\) and every occurrence of \(O\) in the word has a consonant sitting between them, so this pair does not appear side by side either.

Only the pair \(I\) and \(O\) actually appears back to back in the word.

So the correct answer is IO.

Answer 10

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 11

Alternate Method:
Using the vowel/consonant tags for \(\texttt{PACORTCELERDIOGRAH}\):
\[\texttt{C V C V C C C V C V C C V V C C V C}\]
We need a spot with the pattern consonant-vowel-vowel, that is \(\texttt{C V V}\), since the middle vowel then has a consonant on its left and a vowel on its right. The only such block in the tag string is \(\texttt{D I O}\).
So the required vowel is \(\boxed{\texttt{I}}\), option (a).

Answer 12

The question asks which vowel sits with a consonant on its left and a vowel on its right in the coded word \(\texttt{PACORTCELERDIOGRAH}\). Check each option's vowel directly against its actual neighbours in the word.

1. \(I\): The word runs through \(...R\text{-}D\text{-}I\text{-}O\text{-}G...\), so \(I\) has \(D\) (a consonant) on its left and \(O\) (a vowel) on its right. This matches the pattern exactly.
2. \(O\): There are two \(O\)'s. The first, in \(...C\text{-}O\text{-}R...\), has \(C\) on the left and \(R\) on the right, both consonants. The second, in \(...I\text{-}O\text{-}G...\), has \(I\) (a vowel) on its left, so it fails the "consonant on the left" part.
3. \(A\): There are two \(A\)'s. The first, in \(...P\text{-}A\text{-}C...\), has \(P\) on the left and \(C\) on the right, both consonants. The second, in \(...R\text{-}A\text{-}H...\), has \(R\) on the left and \(H\) on the right, again both consonants, so neither \(A\) has a vowel on its right.
4. \(E\): There are two \(E\)'s. The first, in \(...C\text{-}E\text{-}L...\), and the second, in \(...L\text{-}E\text{-}R...\), both have a consonant on the left, but both also have a consonant on the right, not a vowel.

Only \(I\) has a consonant immediately before it and a vowel immediately after it.

So the correct answer is \(I\).

Answer 13

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 14

Alternate Method:
Go through the consonants of \(\texttt{PACORTCELERDIOGRAH}\) one by one and check whether both neighbouring letters are vowels.
The \(\texttt{C}\) at the 3rd position has neighbours \(\texttt{A}\) and \(\texttt{O}\), both vowels, so it qualifies.
The \(\texttt{L}\) at the 9th position has neighbours \(\texttt{E}\) and \(\texttt{E}\), both vowels, so it qualifies too.
Every other consonant in the word has at least one consonant neighbour, so it is excluded.
Hence the letters sandwiched between two vowels are \(\boxed{\texttt{C and L}}\), option (b).

Answer 15

Each option names a pair of consonants; check whether both letters in that pair actually sit with a vowel on each side in \(\texttt{PACORTCELERDIOGRAH}\).

1. R and T: The \(R\) at the 5th position sits between \(O\) and \(T\), and \(T\) at the 6th position sits between \(R\) and \(C\); neither has two vowel neighbours, so this pair fails.
2. C and L: The \(C\) at the 3rd position sits between \(A\) and \(O\), both vowels, and the \(L\) at the 9th position sits between \(E\) and \(E\), both vowels, so both letters in this pair genuinely have a vowel on each side.
3. R and L: \(L\) does have two vowel neighbours as shown above, but every \(R\) in the word sits next to at least one consonant, so this pair fails because of \(R\).
4. D and R: \(D\) sits between \(R\) (a consonant) and \(I\) (a vowel), so it has a consonant neighbour, not two vowels; this pair fails too.

Only the pair \(C\) and \(L\) consists of two letters that each have a vowel immediately before and after them.

So the correct answer is C and L.

Answer 16

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 \(P\).

Answer 17

Alternate Method:
Instead of reversing the first half first, count the required letter directly from the end of \(\texttt{ELECTROCARDIOGRAPH}\). The word has \(18\) letters, and the letter to be prefixed is its second-last letter.
Counting backward from the last letter \(\texttt{H}\): the 1st letter from the end is \(\texttt{H}\), and the 2nd letter from the end is \(\texttt{P}\).
So the letter prefixed to the transformed word is \(\boxed{\texttt{P}}\), option (b), matching what we get by reversing the first half \(\texttt{ELECTROCA} \to \texttt{ACORTCELE}\) and attaching \(\texttt{P}\) in front.

Answer 18

The question asks which single letter got attached to the front of the word once the first half was flipped around. Since the result is the coded word \(\texttt{PACORTCELERDIOGRAH}\), check each option against what actually sits at the very front of that word and against how the rest of it fits together.

1. G: \(G\) does not appear anywhere near the front of \(\texttt{PACORTCELERDIOGRAH}\), it sits deep inside the word, so it was not the letter attached to the front.
2. P: The coded word begins with \(P\), and everything after it, \(\texttt{ACORTCELE}\) followed by \(\texttt{RDIOGRAH}\), is exactly the first nine letters of \(\texttt{ELECTROCARDIOGRAPH}\) written backward, plus the remaining letters left untouched. \(P\) sitting alone at the very front, separate from that reversed block, matches being the letter that was prefixed.
3. H: \(H\) is the very last letter of both the original word and the coded word, not the first, so it was not moved to the front.
4. S: \(S\) does not appear in the original word \(\texttt{ELECTROCARDIOGRAPH}\) at all, so it cannot be the letter that was prefixed.

Only \(P\) sits at the front of the coded word, separate from the reversed block that follows it, exactly as the prefixed letter should.

So the correct answer is P.