Q&A: practice questions with answers

Concept: A word present in both sentences gets the code present in both codes.

1. Sentence 1: cup, of, coffee → lh, em, ro.
2. Sentence 2: he, loves, coffee → em, tf, za.
3. Common word = coffee; common code = em.

Trick: Circle the repeated word and the repeated code – match them directly, no need to decode the rest.

Answer: (C) em

Concept: Track the 1st, 2nd and 3rd letters of each group separately.

1. 1st letters: L, M, N, O, P, Q (+1).
2. 2nd letters: M, K, I, G, E, C (−2).
3. 3rd letters: D, G, J, M, P, S (+3).
4. Missing group = PEP.

Trick: First letter must be P (L→Q runs +1) – this alone removes QEP; then 2nd letter E removes PGP and PFR.

Answer: (C) PEP

Concept: 20 places with 5 blanks → split into 5 blocks of 4 letters each.

1. Blocks: _HMG / D_NH / EJ_I / FKP_ / GLQ_.
2. Each column rises by +1: C D E F G, H I J K L, M N O P Q, G H I J K.
3. Blanks in order: C, I, O, J, K.

Trick: Look at the last block GLQ_ – the 4th column G H I J → K, so the answer must end in K: only CIOJK and CGOLK; 2nd blank (column H→I) gives I.

Answer: (A) CIOJK

Concept: If no class is a part of another and they share no member, draw three separate circles.

1. Fruit – something we eat; Light – a form of energy; Shirt – clothing.
2. No fruit is light or a shirt; no light is a shirt.
3. So three disjoint circles.

Trick: Three totally different categories (food, energy, clothes) = three separate circles.

Answer: (D) Three separate circles

Concept: Find figure (X) inside an option in the same orientation (no rotation).

1. (X) = a vertical line with a hollow triangle on its lower part, apex pointing left.
2. Scan each option for that exact vertical line + left-pointing triangle.
3. Only option (d) contains it; others show the triangle on the wrong side or no such line.

Trick: Fix one unique feature (here the triangle's apex direction) and reject options where it is reversed.

Answer: (D) Option (d)

Concept: Number seats 1–6 from the left (all face north) and place fixed persons first.

1. Two to the left of Y → Y = seat 3.
2. Two between Y and Z → Z = seat 6.
3. One between Z and X → X = seat 4.
4. R immediately left of S → R = 1, S = 2; T = 5.
5. S (2) and Y (3) are neighbours → 0 people between.

Trick: Start with the clue that fixes an exact seat ('only two to the left') – it locks the rest quickly.

Answer: (C) Zero

Concept: Code = sum of reverse alphabet positions (reverse = $27-\text{position}$).

1. FWYO: $21+4+2+12=39$ ✓
2. SIXR: $8+18+3+9=38$ ✓
3. LARZ: L=15, A=26, R=9, Z=1 → $15+26+9+1=51$.

Trick: If the forward sum is too big (FWYO forward = 69), try reverse: forward + reverse = $27\times 4=108$, so reverse = $108-69=39$. For LARZ: forward 57 → $108-57=51$.

Answer: (A) 51

Concept: Compare the gaps between consecutive letters in each cluster.

1. KLQ: K→L +1, L→Q +5.
2. TUZ: +1, +5; OPU: +1, +5.
3. ORV: O→R +3, R→V +4 – different.

Trick: In three options the first two letters are neighbours (KL, TU, OP); ORV is the only one where they are not.

Answer: (B) ORV

Concept: Write the word and its alphabetical arrangement one below the other; count matching places.

1. Original: Q U A R T O S
2. Sorted: A O Q R S T U
3. Place 4 has R in both; no other place matches.
4. Unchanged letters = 1.

Trick: Check the middle letter first – in a 7-letter word, the letter that is 4th in alphabetical rank often stays (R here).

Answer: (A) One

Concept: Row 1 faces south, row 2 faces north, so their 'right' sides are opposite.

1. No one right of K → K at the right end of row 2; two between R and K → R at the other end.
2. Person facing O is 2nd right of the one facing K → O sits next to R; M fills the last seat.
3. Row 1: S, A, N consecutive (A right of S, N right of A); N must not face O → N faces R, A faces O, S faces M, D faces K.
4. Matching pair: A – O.

Trick: Draw both rows in one table with seat columns; each column is a facing pair – then just read the option.

Answer: (A) A and O

Concept: Count how many symbols each box has and in what order they are filled.

1. Boxes hold 1, 2, 3, 4 symbols → next box holds 5.
2. Symbols fill the first column top to bottom, then start the second column.
3. Shapes alternate T, V, T, V → the 5th is T at the top of column 2.

Trick: Count first: options without exactly 5 symbols are rejected instantly.

Answer: (A) Option (a)

Concept: Build the family tree from T upward.

1. C is wife of Q and T is C's daughter → Q is T's father.
2. R is Q's father → R is T's father's father (grandfather).
3. P is R's brother → P = father's father's brother.

Trick: Go generation-wise: P and R are on the same level, two generations above T → answer must contain 'father's father'.

Answer: (A) Father's father's brother

Concept: Find the shift of each letter in the given pairs and apply it.

1. FROM → CPLK: −3, −2, −3, −2.
2. TJCE → QHZC: −3, −2, −3, −2 ✓
3. Apply to KDTY: K→H, D→B, T→Q, Y→W, giving HBQW.

Trick: Only the first two letters H, B are enough – options HBQW and HBWQ remain; T−3 = Q picks HBQW.

Answer: (A) HBQW

Concept: Differences between terms form a pattern.

1. Differences: 1, 3, 5, 7 (odd numbers).
2. Next difference = 9.
3. $18+9=27$.

Trick: Terms are $n^2+2$ ($0+2$, $1+2$, $4+2$, $9+2$, $16+2$) → $25+2=27$.

Answer: (D) 27

Concept: Terms decrease by consecutive odd numbers.

1. 451 − 1 = 450, 450 − 3 = 447, 447 − 5 = 442.
2. Next: 442 − 7 = 435.
3. Check: 435 − 9 = 426 ✓

Trick: Work backwards from the last term: 426 + 9 = 435 – confirms in one step.

Answer: (A) 435

Concept: Faces seen together in any view are adjacent, so they cannot be opposite.

1. View I: P with E and U.
2. View II: P with C and O.
3. P is adjacent to E, U, C, O → the only remaining face M is opposite P.

Trick: The letter that appears in both views and touches four faces – its opposite is the one face never seen beside it.

Answer: (D) P

Concept: Swap the signs as told, then solve by BODMAS.

1. After swapping: $40+8\times 1-72÷8$.
2. $8\times 1=8$; $72÷8=9$.
3. $40+8-9=39$.

Trick: Rewrite the whole expression with new signs first – most errors come from swapping in the head.

Answer: (D) 39

Concept: A face that touches four other faces is opposite the fifth one.

1. View I: S with E and A.
2. View II: S with I and D.
3. S touches E, A, I, D → S is opposite Y.

Trick: Find the letter common to both views; the face never seen with it is its opposite.

Answer: (A) S

Concept: Each letter moves forward by the same number of places.

1. GRSN → JUVQ: every letter +3.
2. KDEA → NGHD: every letter +3 ✓
3. BSTG: E, V, W, J → EVWJ.

Trick: Check just the first letter: B + 3 = E leaves EVWJ and EVUH; T + 3 = W fixes EVWJ.

Answer: (A) EVWJ

Concept: All A are B + No B is C ⇒ No A is C; 'All A are B' also gives 'Some B are A'.

1. Inks ⊂ Pens ⊂ Rubbers; Sharpener lies outside Rubbers.
2. Inks sit wholly inside Rubbers, and Sharpeners touch no Rubber → Sharpeners touch no Ink: (I) follows.
3. All inks are rubbers → Some rubbers are inks: (II) follows.

Trick: 'All + No = No' and 'All' can always be reversed as 'Some' – both rules apply here.

Answer: (A) Both conclusions (I) and (II) follow