Odd Man Out & Series Questions and Answers

Answer: B

The multiplier grows by one at each step.

\[ 2 \times 2 = 4,\quad 4 \times 3 = 12,\quad 12 \times 4 = 48,\quad 48 \times 5 = 240 \]

\[ 240 \times 6 = \mathbf{1440} \]

Assuming a constant multiplier of 5 would give 1200, and of 4 would give 960 - the first option.

Answer: B

Two series are interleaved.

\[ \text{positions 1, 3, 5, 7, 9: } 8,\; 11,\; 14,\; 17,\; \mathbf{20} \quad (+3) \]

\[ \text{positions 2, 4, 6, 8: } 7,\; 12,\; 17,\; 22 \quad (+5) \]

The ninth term belongs to the first series:

\[ 17 + 3 = 20 \]

Continuing the second series instead would give 27, which is offered as the first option.

Answer: A

The series lists the prime numbers from 11 upward.

\[ 11,\; 13,\; 17,\; 19,\; 23,\; 29,\; 31,\; 37,\; 41 \]

The next prime after 41 is 43.

42 is even and 51 is \( 3 \times 17 \), so 43 is the only candidate; 47 is the prime after that, not the next one.

Answer: A

The operations alternate between multiplying by 3 and halving.

\[ 8 \times 3 = 24,\quad 24 \div 2 = 12,\quad 12 \times 3 = 36,\quad 36 \div 2 = 18,\quad 18 \times 3 = 54 \]

\[ 54 \div 2 = \mathbf{27} \]

Continuing with a multiplication instead would give 162; the alternation is what makes the next step a division.

Answer: C

Each term is a product of two consecutive integers.

\[ 1 \times 2 = 2,\quad 2 \times 3 = 6,\quad 3 \times 4 = 12,\quad 4 \times 5 = 20,\quad 5 \times 6 = 30 \]

\[ 6 \times 7 = 42,\quad 7 \times 8 = 56,\quad 8 \times 9 = \mathbf{72} \]

Equivalently the differences are the even numbers 4, 6, 8, 10, 12, 14, so the next gap is 16 and \( 56 + 16 = 72 \).

Answer: B

Each term is the previous one multiplied by \( -2 \).

\[ 4 \times (-2) = -8,\quad -8 \times (-2) = 16,\quad 16 \times (-2) = -32 \]

\[ -32 \times (-2) = 64,\quad 64 \times (-2) = \mathbf{-128} \]

The signs alternate, and the sixth term must be negative - which by itself rules out the positive options.

Answer: B

Each term is one less than a perfect cube.

\[ 2^{3}-1 = 7,\quad 3^{3}-1 = 26,\quad 4^{3}-1 = 63,\quad 5^{3}-1 = 124 \]

\[ 6^{3}-1 = 215,\quad 7^{3}-1 = 342,\quad 8^{3}-1 = \mathbf{511} \]

The near miss 513 would be \( 8^{3} + 1 \) - the same cube with the sign of the correction flipped.

Answer: A

The operations alternate between doubling and adding 3.

\[ 5 \times 2 = 10,\quad 10 + 3 = 13,\quad 13 \times 2 = 26,\quad 26 + 3 = 29 \]

\[ 29 \times 2 = 58,\quad 58 + 3 = 61,\quad 61 \times 2 = \mathbf{122} \]

The next operation is a doubling because the previous one was an addition; 64 would come from adding 3 again.

Answer: B

Each term is double the previous plus one.

\[ 15 \times 2 + 1 = 31,\quad 31 \times 2 + 1 = 63,\quad 63 \times 2 + 1 = 127 \]

\[ 127 \times 2 + 1 = 255,\quad 255 \times 2 + 1 = \mathbf{511} \]

Equivalently every term is \( 2^{n} - 1 \): 15, 31, 63, 127, 255, 511. The 513 on offer is \( 2^{9} + 1 \), one on the wrong side.

Answer: B

The series is the cubes of the natural numbers.

\[ 1^{3},\; 2^{3},\; 3^{3},\; 4^{3},\; 5^{3},\; 6^{3} = 1,\; 8,\; 27,\; 64,\; 125,\; 216 \]

\[ \text{next} = 7^{3} = \mathbf{343} \]

The distractor 392 is \( 216 + 176 \), continuing the differences by eye rather than recognising the cubes.