Alphabet Test Questions and Answers

Answer: C

There are 8 such pairs.

A pair qualifies when the gap between two letters in the word equals their gap in the alphabet. Checking every pair in I N C R E D I B L E (counting both forward and backward order in the alphabet) gives:

I-E, I-D, I-B (the first I with E, D and B), N-I, C-E, E-D, E-B and D-B.

For example, I and E are four places apart in the word (I N C R E) and four apart in the alphabet (E F G H I). That makes 8 pairs; the smaller options miss some of them.

Answer: D

The correct pair is I R.

When the 1st letter is paired with the 26th, the 2nd with the 25th and so on, the positions in every pair add up to 27. I is the 9th letter and R the 18th, and \(9 + 18 = 27\), so I pairs with R.

The other pairs fail the test: H (8) and T (20) add to 28, J (10) and P (16) to 26, and F (6) and V (22) to 28. H actually pairs with S, J with Q, and F with U.

Answer: D

Look at each number in the series: F G + @ - T U V @ A C G 1 M N 3 4 K L 5 6 @ # S 1 # 2. The number 1 in 'G 1 M' is followed by the letter M, so it does not count. The number 3 is followed by 4, a number, and 4 is followed by K, a letter. The number 5 is followed by 6, a number, and 6 is followed by the symbol @, so 6 counts. The number 1 after S is followed by the symbol #, so it counts. The last number, 2, has nothing after it. So two numbers are immediately followed by a symbol, and the answer is 2, option D.

Answer: D

Arrange the letters of PALINDRONE alphabetically: A, D, E, I, L, N, N, O, P, R. Compare each place with the original word P, A, L, I, N, D, R, O, N, E. The 4th place is I in both, and the 8th place is O in both. No other place matches, because the 1st is P against A, the 2nd A against D, the 3rd L against E, the 5th N against L, the 6th D against N, the 7th R against N, the 9th N against P and the 10th E against R. So 2 letters keep their positions, option (D). The options 1, 3 and 0 miss or add a match.

Answer: B

Read the string E U V Y F S C B A B C W U E A X F I M D K T from the left and look at each vowel with the letter just before it. The first E has nothing before it, U follows E, which is a vowel, and so does not count. The A after B counts, because B is a consonant. The U after W counts. The E after U does not count, and the A after E does not count. The I after F counts. So three vowels are immediately preceded by a consonant, option (B). The counts 2, 4 and 5 miss one of them or include a vowel that follows another vowel.

Answer: B

Number the letters of RAPIDLY as R1, A2, P3, I4, D5, L6, Y7 and test pairs for the same number of letters between them in the word as in the alphabet. R and P have one letter between them in the word (A), and in the alphabet only Q lies between P and R, so they fit. A and D have two letters between them in the word (P and I), and in the alphabet B and C lie between A and D, so they fit. Every other pair fails, for instance R and Y have five letters between them in the word but six in the alphabet. So 2 pairs fit, option (B); 0, 1 and 3 miscount them.

Answer: D

Writing the first half of the alphabet in reverse order and leaving the second half unchanged gives the series: M L K J I H G F E D C B A N O P Q R S T U V W X Y Z. In this series, counting from the end, the 7th letter is T (Z, Y, X, W, V, U, T).

T is at position 20 from the left. The 11th letter to its left is at position 20 - 11 = 9, and the 9th letter of the new series is E. So the answer is E. The letters R, F and I are the letters at other positions in the series, which come from miscounting the 7th letter from the end or the 11 places.

Answer: A

Only one word has no vowel after the change.

Move each letter to the next letter: END becomes FOE (contains O and E), MAT becomes NBU (contains U), JAR becomes KBS (no vowel) and BED becomes CFE (contains E). So only KBS has no vowel.

Two, three and four would need more of the words to lose all their vowels, which does not happen.

It is easiest to check each word separately. END has the vowel E in the original, and after shifting it becomes FOE, which has O and E. MAT becomes NBU, and U is a vowel. BED becomes CFE, which ends with E. Only JAR, which becomes KBS, has no vowel left.

Answer: A

END only forms a meaningful word.

Adding B in front gives BEND, BMAT, BJAR and BBED. Only BEND is an English word. The others are not words.

So MAT only, both END and MAT, and both JAR and BED are all wrong.

Test each word one at a time: BEND is a common English word meaning to curve, BMAT, BJAR and BBED are not words at all. Since just one of the four forms a word, the answer is END only, and any option naming MAT, JAR or BED fails.

Answer: A

The letter exactly between them is N. The 15th letter from the beginning is O. The 14th letter from the end is the \(27 - 14 = 13\)th from the beginning, which is M. M is the 13th and O the 15th, so the letter exactly between them is the 14th, N. L is before M and so lies outside the pair, and O is one of the two end letters, so neither can be between them. M likewise is itself one of the two letters. In alphabetical order the 13th, 14th and 15th letters are M, N and O, so the letter between the 13th (M) and the 15th (O) is the 14th, N.