Odd Man Out & Series Questions and Answers

Answer: D

Three of the four are anagrams of cereals.

• Hewat → Wheat
• Ceri → Rice
• Rowaj → Jowar

Cereals is the name of the whole class, not a scrambled member of it, so it does not belong.

Answer: A

Each term is roughly double the previous, with the correction alternating between \( +1 \) and \( -1 \).

\[ 16 \times 2 + 1 = 33 \]

\[ 33 \times 2 - 1 = 65 \]

\[ 65 \times 2 + 1 = 131 \]

\[ 131 \times 2 - 1 = 261 \]

\[ 261 \times 2 + 1 = \mathbf{523} \]

Doubling alone would give 522, so the alternating sign is what separates 523 from the 521 on offer.

Answer: B

Three of the four letters are vowels.

\[ A,\; E,\; I \;\text{are vowels}; \qquad R \;\text{is a consonant} \]

So R does not belong.

Answer: A

Convert each group to letter positions and look at the gaps.

• PRUZ: 16, 18, 21, 26 → \( +2,\, +3,\, +5 \)
• JLOT: 10, 12, 15, 20 → \( +2,\, +3,\, +5 \)
• EGJO: 5, 7, 10, 15 → \( +2,\, +3,\, +5 \)
• BDFL: 2, 4, 6, 12 → \( +2,\, +2,\, +6 \)

So BDFL does not belong - the only group whose gaps are not \( 2,\, 3,\, 5 \).

Answer: B

The series alternates between multiplying by 1.5 and dividing by 3.

\[ 36 \xrightarrow{\times 1.5} 54 \xrightarrow{\div 3} 18 \xrightarrow{\times 1.5} 27 \xrightarrow{\div 3} 9 \xrightarrow{\times 1.5} 13.5 \xrightarrow{\div 3} 4.5 \]

The sixth term should be \( 9 \times 1.5 = 13.5 \), not 18.5.

So 18.5 is wrong. The final 4.5 confirms it, since \( 13.5 \div 3 = 4.5 \) while \( 18.5 \div 3 \) is not a clean value.

Answer: C

The series is the perfect squares in order.

\[ 1^{2},\; 2^{2},\; 3^{2},\; 4^{2},\; 5^{2},\; 6^{2},\; 7^{2} \]

\[ \text{next} = 8^{2} = \mathbf{64} \]

The differences are the odd numbers 3, 5, 7, 9, 11, 13, so the next gap is 15 and \( 49 + 15 = 64 \) - the same answer from the other direction.

Answer: A

The differences should grow by 3 each time.

\[ 8 \xrightarrow{+5} 13 \xrightarrow{+8} 21 \xrightarrow{+11} 32 \xrightarrow{+14} 46 \xrightarrow{+17} 63 \xrightarrow{+20} 83 \]

The fifth term should therefore be \( 32 + 14 = 46 \), not 47.

So 47 is the wrong number. Every later term is consistent with 46, which is what confirms the error sits at that single position.

Answer: B

All four are made from the digits 1, 2 and 3, so look at the last digit.

\[ 123,\; 231,\; 321 \;\text{end in an odd digit and are therefore odd} \]

\[ 132 \;\text{ends in 2 and is even} \]

So 132 does not belong. All four share a digit sum of 6, so divisibility by 3 cannot separate them - parity can.

Answer: C

Three of the four are products of two consecutive integers.

\[ 20 = 4 \times 5, \quad 42 = 6 \times 7, \quad 72 = 8 \times 9 \]

\[ 58 = 2 \times 29 \text{, and } 7 \times 8 = 56, \; 8 \times 9 = 72 \]

So 58 does not belong - it falls between two such products without being one.

Answer: D

The differences should be the prime numbers in order.

\[ 125 \xrightarrow{+2} 127 \xrightarrow{+3} 130 \xrightarrow{+5} 135 \xrightarrow{+7} 142 \xrightarrow{+11} 153 \xrightarrow{+13} 166 \]

The last term should be \( 153 + 13 = 166 \), not 165.

So 165 is wrong. The actual last gap is 12, which is not prime.