Inserting The Missing Character Questions and Answers

Answer: A

The value of x is 15.

In each circle, the sum of the two upper numbers divided by 7 gives the lower number:

\[ \begin{aligned} (38 + 18) \div 7 &= 8 \\ (25 + 17) \div 7 &= 6 \\ (89 + 16) \div 7 &= 15 \end{aligned} \]

The rule works for both complete circles, so it fixes \(x = 105 \div 7 = 15\). 16, 17 and 18 do not fit it, since 112, 119 and 126 would be needed as sums.

Answer: C

The missing number is 8.

In each row, the middle number is the sum of the squares of the outer two:

\[ \begin{aligned} 5^{2} + 9^{2} &= 25 + 81 = 106 \\ 6^{2} + 7^{2} &= 36 + 49 = 85 \\ 4^{2} + ?^{2} &= 80 \Rightarrow ?^{2} = 64 \end{aligned} \]

So the missing number is 8. 11 and 15 would make the middle number 137 and 241, and 3 would make it 25.

Answer: A

The value of x is 1.

In each circle, the product of the two lower numbers minus the sum of the two upper numbers gives the centre:

\[ \begin{aligned} 9 \times 8 - (15 + 17) &= 72 - 32 = 40 \\ 3 \times 12 - (6 + 2) &= 36 - 8 = 28 \\ 13 \times 4 - (23 + 28) &= 52 - 51 = 1 \end{aligned} \]

The rule fits both known circles, so x = 1. The other options (8, 12, 27) do not satisfy any rule that also produces 40 and 28 in the first two circles.

Answer: D

The missing number is 15.

The diagonals and midlines cut the square into eight triangles, which pair up along each side of the square: the two triangles touching the same side add up to the same total.

• Left side: 21 + 35 = 56
• Bottom: 14 + 42 = 56
• Right side: 16 + 40 = 56
• Top: 41 + ? = 56

So the missing number is 56 - 41 = 15. The other options give top totals of 53, 54 and 55, which break the pattern.

Answer: B

The missing number is 6.

In each cross the four outer numbers add up to 8 times the centre:

• First: 4 + 5 + 1 + 6 = 16 = 8 × 2
• Second: 9 + 7 + 2 + 6 = 24 = 8 × 3
• Third: 19 + 13 + 5 + 11 = 48 = 8 × 6

So the centre of the third cross is 48/8 = 6. The other options would need outer totals of 32, 64 or 80.

Answer: B

The missing number is 64.

In the first circle, the product of the top two numbers minus the product of the bottom two gives the centre:

\[ \begin{aligned} 10 \times 11 - 6 \times 9 &= 110 - 54 = 56 \\ 12 \times 10 - 7 \times 8 &= 120 - 56 = 64 \end{aligned} \]

Applying the same rule to the second circle gives 64. Simple sums do not work for the first circle (10 + 11 + 6 + 9 = 36), so a sum-based rule cannot be the pattern, and 44, 60 and 54 do not follow from the product rule.

Answer: A

The missing number is 17.

In each column, the first number minus the second plus the third gives the fourth:

\[ \begin{aligned} 8 - 6 + 4 &= 6 \\ 10 - 5 + 6 &= 11 \\ 15 - 7 + 8 &= 16 \\ 13 - 4 + 8 &= 17 \end{aligned} \]

The rule holds for the three complete columns, so it fixes the fourth at 17. The other options would need a different rule in the last column alone; 19, 21 and 23 fit none of the patterns of the first three.

Answer: D

The missing number is 32.

Reading clockwise from the sector holding 1, the numbers are 1, 2, 4, 8, 16 and then the blank: each number is double the previous one. So the blank is

\[ 16 \times 2 = 32 \]

which also completes the circle back to 1, since the six sectors hold the powers of 2 from \(2^{0}\) to \(2^{5}\). 24 is 16 + 8 (adding instead of doubling), and 8 and 30 do not continue the pattern.

Answer: C

The missing number is 9.

In each figure the bottom numbers are fixed multiples of the top ones: in the first, \(80 = 2 \times 40\) and \(24 = 6 \times 4\). The second follows the same rule: \(120 = 3 \times 40\), so the right-hand column must satisfy \(36 = ? \times 4\), giving 9.

The other options do not give 36 when multiplied by 4.

The same reading also works column-wise: 2 and 3 are each multiplied by 40, and 6 and the missing number are each multiplied by 4.

Answer: A

The missing number is 5.

Each number in the second row is the top number times one less than itself, \(n(n - 1)\):

\[ \begin{aligned} 3 \times 2 &= 6,\quad 8 \times 7 = 56,\quad 10 \times 9 = 90 \\ 2 \times 1 &= 2,\quad 1 \times 0 = 0,\quad 5 \times 4 = 20 \end{aligned} \]

So the top number over 20 is 5. 0 gives 0, 7 gives 42, and 3 gives 6.

Recognising products of consecutive integers (2, 6, 12, 20, 30, 42, 56, 72, 90) makes such matrices quick to solve.