Inserting The Missing Character Questions and Answers

Answer: B

The missing number is 23.

In each figure, the number inside is the top number divided by the bottom one, plus the left number divided by the right one:

\[ \begin{aligned} 81 \div 9 + 28 \div 7 &= 9 + 4 = 13 \\ 56 \div 8 + 63 \div 3 &= 7 + 21 = 28 \\ 48 \div 4 + 55 \div 5 &= 12 + 11 = 23 \end{aligned} \]

The rule fits both complete figures, so the third gives 23. 21, 25 and 27 do not come from this rule.

Answer: C

The missing number is 16.

In each row, the product of the first two numbers divided by the third gives the fourth:

\[ \begin{aligned} \frac{4 \times 6}{3} &= 8 \\ \frac{6 \times 10}{5} &= 12 \\ \frac{4 \times 8}{2} &= 16 \end{aligned} \]

The rule works for both complete rows, so it gives the missing value in the third. The option 8 repeats the first row's answer, and 12 the second's; 20 does not fit the rule at all.

Answer: B

The missing number is 27.

In each row, the middle number is the sum of the products of the digits of the two outer numbers:

• Row 1: 51 gives 5 × 1 = 5 and 61 gives 6 × 1 = 6; 5 + 6 = 11.
• Row 2: 64 gives 24 and 32 gives 6; 24 + 6 = 30.
• Row 3: 35 gives 15 and 43 gives 12; 15 + 12 = 27.

The rule fits the first two rows exactly, and only 27 completes the third, so 25, 32 and 37 are ruled out.

Answer: B

The missing number is 25.

In each figure, subtract the bottom number from the left number and the right number from the top number, then add the squares of the two differences:

\[ \begin{aligned} (16 - 6)^{2} + (5 - 2)^{2} &= 100 + 9 = 109 \\ (22 - 15)^{2} + (21 - 19)^{2} &= 49 + 4 = 53 \\ (17 - 13)^{2} + (51 - 48)^{2} &= 16 + 9 = 25 \end{aligned} \]

The rule gives the centre numbers of the first two figures exactly, so the third centre is 25. 7 is just the sum of the two differences, and 49 and 129 do not fit the rule.

Answer: C

The missing number is 32.

In each triangle, the number inside is the product of the three outer numbers divided by 10:

\[ \begin{aligned} \frac{5 \times 6 \times 4}{10} &= \frac{120}{10} = 12 \\ \frac{6 \times 7 \times 5}{10} &= \frac{210}{10} = 21 \\ \frac{4 \times 8 \times 10}{10} &= \frac{320}{10} = 32 \end{aligned} \]

The rule matches both given triangles, so the third is 32. 22 is the sum of the outer numbers, and 30 and 42 do not follow the pattern.

Answer: A

The missing number is 7.

Each row is a Pythagorean triple: the first number squared equals the sum of the squares of the other two.

\[ \begin{aligned} 13^{2} &= 12^{2} + 5^{2},\quad 17^{2} = 15^{2} + 8^{2} \\ 29^{2} &= 21^{2} + 20^{2},\quad 25^{2} = 24^{2} + 7^{2} \end{aligned} \]

Since 625 - 576 = 49, the missing number is \(\sqrt{49} = 7\). 9, 11 and 15 give 657, 697 and 801, which are not 625.

Answer: C

Correct answer: ÷, x and =

\[ 15 \div 3 \times 4 = 5 \times 4 = 20, \] so the expression closes with an equals sign. Working left to right is correct here since division and multiplication have equal precedence. The alternative with multiplication first, \( 15\times3\div4 \), gives 11.25, which is neither greater than nor less than 20 in a way that would make one of the inequality options work.

Answer: D

Correct answer: UUTSU

Filling the gaps gives TUS repeated four times: \[ \mathrm{TUS\ TUS\ TUS\ TUS}. \] Reading off the five missing positions in order gives U, U, T, S, U. Once a short repeating block is suspected, the fastest check is to see whether the known letters are consistent with it before testing any option.

Answer: B

Correct answer: 343

The series is the cubes: \[ 1^3, 2^3, 3^3, 4^3, 5^3, 6^3, 7^3 = 1, 8, 27, 64, 125, 216, 343. \] Recognising 216 as \( 6^3 \) settles it at once, and the differences would take much longer to work through.

Answer: C

Correct answer: jks

Each position advances by two letters across the terms. The first letters run d, f, h, then j; the second run e, g, i, then k; and the third run m, o, q, then s. So the missing term is jks. Checking all three positions rather than only the first is what separates it from the jkt offered beside it.