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: 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.

Answer: C

Correct answer: of, as

“Think of X as Y” is the fixed construction for describing how something is regarded, and both halves are required - “think for” is not English in this sense, and “because” and “since” introduce a reason rather than a description. The sentence says the speakers regarded the thing as a kind of requiem, which is exactly what “of … as” expresses.

Answer: A

Correct answer: cdbda

Filling the gaps gives dcba repeated three times. The five missing positions read in order are c, d, b, d, a. The known letters already show d and b in the first block and a and c in the second, which is enough to identify the four-letter cycle before any option is tested.

Answer: C

Correct answer: LKJM

Filling the gaps gives JKLM, JKLM, JKLM - the same four letters repeated three times. The missing characters are therefore the third of the first group, the second of the second, the first of the third and the fourth of the last, which read in order are L, K, J, M.