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: Set up the segments.
After reversing the first 9 letters of "ELECTROCARDIOGRAPH" and moving the original word's 17th letter (P) to the front, the word splits into three clear blocks: position 1 is the moved letter P, positions 2 to 10 are the reversed first half, and positions 11 to 18 are the rest of the original second half with P taken out.

Step 2: Fill in the middle block.
The first half of the original word is "ELECTROCA" (positions 1 to 9). Reversed, this reads "ACORTCELE," and this exact 9-letter string fills positions 2 to 10 of the word after the P-move.

Step 3: Add the final letter and locate the middle.
Adding S at the end makes the word 19 letters long, so the middle sits at position 10. From Step 2, position 10 is the last letter of the block "ACORTCELE," which is E.
\[ \boxed{E} \]

Answer 2

Since the final word has 19 letters, an odd number, its middle letter is equally distant from both ends. Rather than counting forward from the front, count backward from the end and see which candidate letter lands exactly 10th from the right.

The last 9 letters of the final word come from the original second half of "ELECTROCARDIOGRAPH" (which was "RDIOGRAPH") with its P removed and pulled to the front, and then S added at the very end. That gives the tail "RDIOGRAH" followed by "S," occupying the last 9 positions of the 19-letter word.

1. \(L\): L does not appear anywhere in that 9-letter tail "RDIOGRAHS," so it cannot be the 10th letter from the right either.
2. \(R\): R does appear inside the tail, but only within the last 9 positions, meaning any R lands closer to the end than the middle, not at the 10th position from the right.
3. \(D\): D likewise sits inside that same 9-letter tail, one position short of the exact middle.
4. \(E\): E does not appear in the tail at all, since none of the letters in "RDIOGRAHS" is E. That means the middle letter must come from the block just before the tail, which is exactly the reversed first-half block ending in E.

Counting 9 letters in from the right accounts for the whole tail block, so the 10th letter from the right is the last letter of the reversed first half, E, confirming the middle letter from the other direction too.

The correct answer is \(E\).

Answer 3

Step 1: Locate the original vowel positions.
In the untouched first half of "ELECTROCARDIOGRAPH," the letters "ELECTROCA" sit at original positions 1 to 9. Among these, the vowels are E (position 1), E (position 3), O (position 7) and A (position 9).

Step 2: Map each position to its place after the transformations.
As shown for the middle-letter question, a letter starting at original position \(i\) (for \(i \le 9\)) ends up at final position \(11 - i\) once the first half is reversed and the letter P is moved to the front. Applying this: position 1 maps to \(11-1=10\), position 3 maps to \(11-3=8\), position 7 maps to \(11-7=4\), and position 9 maps to \(11-9=2\).

Step 3: Keep only the positions to the left of the middle.
Position 10 is the middle letter itself, so it is excluded from the "left of middle" count. The remaining mapped positions, 8, 4 and 2, all fall between positions 1 and 9, meaning three of the four original vowels land to the left of the middle letter.
\[ \boxed{3} \]

Answer 4

Here is a check that does not need the exact left-hand letters at all: count vowels across the whole 19-letter transformed word, then subtract the ones that are not on the left side.

None of the three operations, reversing part of the word, moving a letter to the front, or adding "S," changes which letters are vowels, they only move letters around or add a consonant. So the total number of vowels in the transformed word is the same as in the original "ELECTROCARDIOGRAPH": E, E, O, A, I, O, A, seven vowels in all.

The middle letter itself, found earlier to be E, is one of those seven vowels, and it sits in the middle position, not to the left. The letters to the right of the middle are R, D, I, O, G, R, A, H, S, among which I, O and A are vowels, three more of the seven.

1. \(2\): would leave \(7 - 1 - 2 = 4\) vowels unaccounted for elsewhere, which does not match, so this option is wrong.
2. \(1\): would leave \(7 - 1 - 1 = 5\) vowels unaccounted for, also too many, so this is wrong too.
3. \(4\): would need \(7 - 1 - 4 = 2\) vowels on the right, but there are actually three (I, O, A), so this overshoots.
4. \(3\): leaves \(7 - 1 - 3 = 3\) vowels needed on the right side, and there are exactly three there (I, O, A), so the numbers balance perfectly.

The correct answer is \(3\).

Answer 5

Step 1: Write out the coded word \(\texttt{PACORTCELERDIOGRAH}\) and list the vowels in the order they appear: \(A, O, E, E, I, O, A\).

Step 2: For each vowel, count how many letters sit between it and the next vowel in the word. Between \(A\) and \(O\) there are two consonants, between \(O\) and \(E\) there are three, between the two \(E\)'s there is one, between \(E\) and \(I\) there are two, between \(I\) and \(O\) there are zero letters, and between \(O\) and the last \(A\) there are two.

Step 3: A gap of zero means the two vowels are touching, and that happens only for the \(I\) and \(O\) pair.
\[ \boxed{\texttt{IO}} \]

Answer 6

Another way to check this is to mark every vowel in the coded word \(\texttt{PACORTCELERDIOGRAH}\) and see which pair from the choices actually sits with nothing between the two letters.

1. IE: Marking the word letter by letter, \(I\) sits between \(D\) and \(O\), and neither of its neighbours is \(E\), so \(I\) and \(E\) are never written next to each other.
2. IO: The same marking shows \(I\) sitting immediately before \(O\), with no letter in between, so this pair does sit side by side.
3. AE: The two \(A\)'s in the word sit next to consonants on both sides, and the two \(E\)'s do too, so an \(A\) is never written beside an \(E\) anywhere in the word.
4. AO: Likewise, every \(A\) in the word has a consonant on both sides, so \(A\) and \(O\) never touch either.

Marking the word this way confirms that \(I\) and \(O\) are the only vowels written with nothing between them.

So the correct answer is IO.

Answer 7

Step 1: Write out the coded word \(\texttt{PACORTCELERDIOGRAH}\) and pick out every vowel along with the letters sitting right before and right after it.

Step 2: Going through them: \(A\) is flanked by consonants on both sides, \(O\) is flanked by two consonants the first time and by a vowel then a consonant the second time, \(E\) is flanked by consonants on both sides both times it appears, and \(I\) is flanked by \(D\) and \(O\).

Step 3: We need a consonant just before the vowel and a vowel just after it. Checking each one, only \(I\) fits, since \(D\) before it is a consonant and \(O\) after it is a vowel; every other vowel either has a consonant on both sides or a vowel on the left instead of the right.
\[ \boxed{I} \]

Answer 8

A quicker way to think about this is to first work out what shape of triplet we're hunting for, then search the word for that exact shape.

1. \(I\): The needed shape is consonant, then vowel, then vowel, because the middle letter of that triplet is a vowel with a consonant before it and a vowel after it. Scanning \(\texttt{PACORTCELERDIOGRAH}\) for this shape turns up exactly one spot, \(D\text{-}I\text{-}O\), with \(I\) in the middle.
2. \(O\): For \(O\) to fit this shape as the middle letter, it would need a consonant right before it and a vowel right after it. The only \(O\)'s in the word sit in \(C\text{-}O\text{-}R\) and \(I\text{-}O\text{-}G\), neither of which has this shape, so \(O\) does not fit.
3. \(A\): Both occurrences of \(A\) sit inside consonant-vowel-consonant blocks, not consonant-vowel-vowel, so \(A\) never fits the shape either.
4. \(E\): Both occurrences of \(E\) also sit inside consonant-vowel-consonant blocks, so \(E\) doesn't fit the shape.

The only vowel that lands as the middle letter of a consonant-vowel-vowel triplet is \(I\), in the block \(D\text{-}I\text{-}O\).

So the correct answer is \(I\).

Answer 9

Step 1: Break the coded word \(\texttt{PACORTCELERDIOGRAH}\) into every group of three consecutive letters and look at the middle letter of each group.

Step 2: A consonant is "sandwiched between two vowels" only when its group has the shape vowel, consonant, vowel.

Step 3: Checking the groups in order, \(A\text{-}C\text{-}O\) gives this shape with \(C\) in the middle, and \(E\text{-}L\text{-}E\) gives the same shape with \(L\) in the middle. No other three-letter group in the word has a vowel on both outer letters.
\[ \boxed{\text{C and L}} \]

Answer 10

Another route is to look at the vowels first, find any two vowels that are exactly one letter apart, and then read off whichever consonant sits in that gap.

1. R and T: Neither \(R\) nor \(T\) sits in a one-letter gap between two vowels anywhere in \(\texttt{PACORTCELERDIOGRAH}\).
2. C and L: The vowels \(A\) and \(O\) are exactly one letter apart, with \(C\) in between, and the vowels \(E\) and \(E\) are exactly one letter apart, with \(L\) in between. Both letters in this pair are confirmed this way.
3. R and L: \(L\) is confirmed as shown above, but no vowel pair one letter apart surrounds any occurrence of \(R\).
4. D and R: No vowel pair one letter apart surrounds \(D\) either. The letters flanking \(D\) are \(R\) and \(I\), and \(R\) is a consonant, not a vowel, so this gap does not qualify.

Scanning for vowels that are one letter apart turns up exactly two such gaps, and they contain \(C\) and \(L\).

So the correct answer is C and L.

Answer 11

Step 1: Count the letters of \(\texttt{ELECTROCARDIOGRAPH}\) from the end: \(H\) is 1st from the end, \(P\) is 2nd from the end.

Step 2: The instruction is to take this 2nd-last letter out of the word and place it at the front, before reversing the first half of what remains.

Step 3: That 2nd-last letter is \(P\), and checking the coded word \(\texttt{PACORTCELERDIOGRAH}\) confirms \(P\) sitting alone at the start, ahead of the reversed block \(\texttt{ACORTCELE}\).
\[ \boxed{P} \]

Answer 12

Another way to find the prefixed letter is to line up the original 18-letter word against the 18-letter coded word position by position and see what changed at the front.

1. G: In \(\texttt{ELECTROCARDIOGRAPH}\), \(G\) sits well inside the word, not at either edge, so it can't be the letter moved to the front.
2. P: \(\texttt{ELECTROCARDIOGRAPH}\) has 18 letters, and its 17th letter is \(P\). Removing that \(P\) and sliding it to the very front, then reversing the first nine of the remaining letters, produces \(\texttt{PACORTCELERDIOGRAH}\), which matches the coded word exactly.
3. H: \(H\) is the 18th and final letter of the original word, and it stays at the end of the coded word too, so it never moved to the front.
4. S: There is no \(S\) in \(\texttt{ELECTROCARDIOGRAPH}\) to begin with, so it cannot be the answer.

Lining up both 18-letter words this way shows \(P\) is the one letter pulled out and placed at the front.

So the correct answer is P.