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Symbol Coding - Intermediate Level: arrow coding INTERMEDIATE

Exam-focused quick response training ★ worksheet: 20 intermediate-level symbol coding questions. Worksheet 13 of 30 targets arrow coding. Build proficiency in symbol mapping, graphic codes, symbol sequences with detailed solutions. Ideal for mid-level competitive exam preparation.

📝 Worksheet 13 of 30 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Intermediate level

What you'll learn in this worksheet:
Your progress through Symbol Coding
Worksheet 13 of 30 (43% complete)

Question 1

If +=*, -=&, *=@, /=#, then evaluate: 8 & 8 # 3
Decode symbols: 8 & 8 # 3 → 8 - 8 / 3 = 6
arithmetic, operations

Question 2

Using the symbol matrix: E → & | G → = | A → * | D → ! | B → @ Code the word: 'ABGD'
Using matrix: A→* → B→@ → G→= → D→! = *@=!
matrix, mapping

Question 3

If +=@, -=#, *=%, /=*, then evaluate: 8 * 11 # 11
Decode symbols: 8 * 11 # 11 → 8 / 11 - 11 = -11
arithmetic, operations

Question 4

If Layer 1: Vowels=@, Consonants=#, Layer 2: @=1, #=2, then code 'DATA'
Layer 1: DATA → #@#@
Layer 2: #@#@ → 2121
multilayer, complex

Question 5

If @=C, #=L, $=O, ... then decode: '@#$%&'
Reverse mapping: @#$%& → CLOUD = CLOUD
reverse, decoding

Question 6

If vowels at even positions (1-based) = @, vowels at odd positions = #, consonants remain same, then 'DEVELOP' = ?
Vowels at positions: D, E(pos2,even→@), V, E(pos4,even→@), L, O(pos6,even→@), P = D#V#L#P
conditional, rules

Question 7

If I=$, G=5, H=6, L=9, T==, then 'LIGHT' is coded as?
Applying symbol substitution: L→9 → I→$ → G→5 → H→6 → T→= = 9$56=
substitution, basic

Question 8

If letters repeat pattern @, #, $ in order, then 'ALGORITHM' = ?
Triplet pattern: positions 1,4,7...=@, 2,5,8...=#, 3,6,9...=$ → @#$@#$@#$
pattern, position

Question 9

If vowels at even positions (1-based) = @, vowels at odd positions = #, consonants remain same, then 'HARDWARE' = ?
Vowels at positions: H, A(pos2,even→@), R, D, W, A(pos6,even→@), R, E(pos8,even→@) = H#RDW#R#
conditional, rules

Question 10

If vowels = $, consonants = %, then 'ALGORITHM' = ?
Vowel-consonant coding: vowels (A, O, I)→$, consonants (L, G, R, T, H, M)→% → $%%$%$%%%
pattern, position

Question 11

If Layer 1: Letters → position numbers (A=1, B=2...), Layer 2: Numbers → symbols (1=@,2=#,3=$,4=%,5=&,6=*,7=+,8==,9=!,0=?), then code 'COPY'
Layer 1: COPY → C=3, O=15, P=16, Y=25 = 3151625
Layer 2: 3151625 → $@&@*#&
multilayer, complex

Question 12

If +=*, -=$, *=&, /=@, then evaluate: 4 @ 7 * 8
Decode symbols: 4 @ 7 * 8 → 4 / 7 + 8 = 8
arithmetic, operations

Question 13

If vowels = @, even position consonants = #, odd position consonants = *, then 'KEYBOARD' = ?
Position-wise coding: K(pos 1, even)→#, E(vowel)→@, Y(pos 3, even)→#, B(pos 4, odd)→*, O(vowel)→@, A(vowel)→@, R(pos 7, even)→#, D(pos 8, odd)→* = #@#*@@#*
conditional, rules

Question 14

If Layer 1: Letters → position numbers (A=1, B=2...), Layer 2: Numbers → symbols (1=@,2=#,3=$,4=%,5=&,6=*,7=+,8==,9=!,0=?), then code 'DATA'
Layer 1: DATA → D=4, A=1, T=20, A=1 = 41201
Layer 2: 41201 → %@#?@
multilayer, complex

Question 15

If vowels = @, even position consonants = #, odd position consonants = *, then 'MONITOR' = ?
Position-wise coding: M(pos 1, even)→#, O(vowel)→@, N(pos 3, even)→#, I(vowel)→@, T(pos 5, even)→#, O(vowel)→@, R(pos 7, even)→# = #@#@#@#
conditional, rules

Question 16

If first and last letters = &, others = *, then 'ALGORITHM' = ?
First-last special: A(first)=&, M(last)=&, middle letters=* → &***&
pattern, position

Question 17

If I=$, B=1, D=3, R=+, then 'BIRD' is coded as?
Applying symbol substitution: B→1 → I→$ → R→+ → D→3 = 1$+3
substitution, basic

Question 18

If vowels = @, even position consonants = #, odd position consonants = *, then 'PRINTER' = ?
Position-wise coding: P(pos 1, even)→#, R(pos 2, odd)→*, I(vowel)→@, N(pos 4, odd)→*, T(pos 5, even)→#, E(vowel)→@, R(pos 7, even)→# = #*@*#@#
conditional, rules

Question 19

If +=*, -=$, *=~, /=@, then evaluate: 6 @ 4 * 12
Decode symbols: 6 @ 4 * 12 → 6 / 4 + 12 = 13
arithmetic, operations

Question 20

If first and last letters = &, others = *, then 'INTELLIGENCE' = ?
First-last special: I(first)=&, E(last)=&, middle letters=* → &**&
pattern, position
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