Q&A: practice questions with answers

Concept: Sort the word's letters A→Z, pick the asked positions, then count letters lying between them in the alphabet.

1. DUMPERS sorted: D E M P R S U.
2. 4th from left = P; 2nd from right = S.
3. P(16) and S(19): letters between = Q, R → 19 − 16 − 1 = 2.

Trick: Letters between two positions = difference of ranks − 1.

Answer: (A) 2

Concept: The same figure keeps rotating by a fixed angle; continue the rotation one more step.

1. The pentagon, inner circle and forked stalk never change – only the direction of the stalk does.
2. Frame to frame the stalk moves up → up-right → right → down-right, i.e. 45° clockwise each time.
3. Next step: stalk points straight down with the pentagon on top, and the fork must remain open (two spreading arms).

Trick: Track one distinctive part (the stalk) like a clock hand – 45° per step means the 5th frame is exactly opposite the 1st.

Answer: (A) Figure (A)

Concept: A fixed group of letters repeats; find the group from the given letters.

1. Given: C _ _ _ B _ _ A X B _ Z (12 places).
2. C at 1, B at 5, and A X B at 8–10 suggest the block CZAXB of length 5.
3. Full series: CZAXB CZAXB CZ – place 12 is Z ✓.
4. Blanks (places 2,3,4,6,7,11): Z A X C Z C.

Trick: Try option (c) directly in the blanks – the whole line splits into equal repeating blocks only for ZAXCZC.

Answer: (C) ZAXCZC

Concept: Each letter position of the clusters follows its own +3 pattern.

1. 1st letters: B, E, H, K → +3 → N.
2. 2nd letters: C, F, I, L → +3 → O.
3. 3rd letters: G, J, M, P → +3 → S.
4. Next cluster = NOS.

Trick: Only the first column matters to cut options: next is N (removes Q-options); third column P+3 = S picks NOS.

Answer: (A) NOS

Concept: Every fold doubles the layers; on opening, each crease reflects the punch like a mirror.

1. Fold 1: top half onto bottom (horizontal crease). Fold 2: right half onto left (vertical crease).
2. Two folds = 4 layers, so the single punch appears 4 times – once in each quarter.
3. Neighbouring cuts are mirror images across the creases: the flat bar stays towards the outer edge and the curved side faces the centre.

Trick: Count first (4 cuts), then check mirror direction – options with 2 cuts or wrongly turned cuts are out in seconds.

Answer: (B) Figure (B)

Concept: Letters common to two words take the digits common to their codes.

1. TIDE → 2986, KITE → 8216.
2. Common letters: T, I, E; common digits: 2, 8, 6.
3. TIDE's leftover letter D gets the leftover digit 9.

Trick: Strike out common letters and common digits – whatever remains pairs up.

Answer: (A) 9

Concept: Facing the centre, a person's left is clockwise and right is anticlockwise.

1. Fix F; L is just right of F (one seat anticlockwise).
2. H is 4 seats from L going left (clockwise); J is 4 seats from F going left.
3. Only one adjacent empty pair remains for K and E with E to K's right; G takes the last seat.
4. From K moving right: E, then F → one person between.

Trick: Number seats 0–6 clockwise: F=0, L=6, H=3, J=4, K=2, E=1, G=5 – count directly.

Answer: (C) One

Concept: Apply the same letter shifts and number change found in the example pairs.

1. AR 18 → HO 31: 1st letter +7, 2nd letter −3, number +13.
2. Same rule works for the second pair (K→R, N→K, 05→18).
3. For QL 15: Q moves 7 ahead to X, L moves 3 back to I, and 15 becomes 28 → XI 28.

Trick: Number part alone decides: 15+13 = 28 leaves XI 28 or WJ 28; first letter Q+7 = X picks XI 28.

Answer: (B) XI 28

Concept: Swap the signs first, then solve by BODMAS.

1. After swapping: 41 − 8 × 6 + 24 ÷ 2.
2. Multiplication gives 48; division gives 12.
3. So the value is 41 + 12 − 48 = 53 − 48 = 5.

Trick: Write the new expression on paper before calculating – most mistakes come from swapping mentally.

Answer: (A) 5

Concept: Facing north, right = right end of the row; place fixed people first.

1. Seats 1–6 from left. Nobody to F's right → F = 6.
2. Three between F and J → J = 2.
3. Two between J and G → G = 5.
4. H second to the left of I with seats 1, 3, 4 free → H = 1, I = 3; S = 4.
5. Between H(1) and S(4): J, I → 2.

Trick: People between seats a and b = |a − b| − 1 = 4 − 1 − 1 = 2.

Answer: (A) Two

Concept: A vertical mirror swaps left and right; top and bottom stay the same.

1. MN stands vertically beside the figure, so only left–right reverses.
2. The figure is symmetric about its own vertical centre line, so its image looks the same as the original.
3. So pick the option that reproduces the original completely – no missing line at the base, neither the outer uprights nor the small middle one under the V.

Trick: For a symmetric figure, just look for the option that is an exact copy with no line missing.

Answer: (D) Figure (D)

Concept: Find the gap pattern between consecutive letters of each cluster.

1. LOI: +3, −6. PSM: +3, −6. RUO: +3, −6.
2. TWS: T→W = +3 but W→S = −4.
3. TWS breaks the pattern.

Trick: Check only the 1st and 3rd letters: L→I, P→M, R→O are all −3, but T→S is −1.

Answer: (D) TWS

Concept: The code depends only on how many letters the word has.

1. TNRA: 4 letters → 4 × 7 = 28.
2. LOPGVO: 6 letters → 6 × 7 = 42.
3. JDSHKQB: 7 letters → 7 × 7 = 49.

Trick: When codes are multiples of one number and words look random, divide code by length first.

Answer: (D) 49

Concept: Differences between terms are consecutive odd numbers.

1. Differences: 1, 3, 5, 7.
2. Next difference = 9.
3. 70 + 9 = 79.

Trick: Terms = 54 + n²: 54, 55, 58, 63, 70, 79 (adding 0², 1², 2², 3², 4², 5²).

Answer: (D) 79

Concept: Show subset as a circle inside another, and unrelated classes as separate circles.

1. All cheetahs are wild animals → Cheetah circle inside Wild animals.
2. Cow is a domestic animal – it is neither a cheetah nor wild → separate circle.
3. Required: one circle within another + one independent circle.

Trick: Ask 'is every X a Y?' for each pair – only Cheetah→Wild is 'yes', so only one nesting.

Answer: (B) Figure (B)

Concept: A conclusion follows only if it is true in every possible diagram.

1. Sacks, mugs and pugs are each fully inside rugs.
2. No link among sacks, mugs and pugs is given.
3. Draw them as three separate circles inside rugs → both 'Some pugs are sacks' and 'Some mugs are pugs' fail.

Trick: When all statements share only a common big class, sub-classes have no definite relation with each other.

Answer: (C) Neither conclusion (I) nor (II) follows.

Concept: The differences themselves increase by a constant.

1. Differences: 10, 13, 16, 19 (each +3).
2. Next difference = 22.
3. 111 + 22 = 133.

Trick: Last digit check: differences end 0,3,6,9,2 → 111 + …2 ends in 3; only 133 fits.

Answer: (B) 133

Concept: Decode each symbol and build the family tree link by link.

1. P − Q: P is Q's brother. Q + R: Q is R's daughter.
2. R × S: R is S's wife. S ÷ T: S is T's father.
3. So P and T are both children of S and R; P is male.

Trick: A male who shares both parents with T is T's brother – no need to draw the full tree.

Answer: (B) Brother

Concept: Find the change in both letters and the number, then repeat it.

1. MH 45 → KF 40: each letter −2, number −5.
2. TN 35 → RL 30 confirms the same rule.
3. ZI 25: Z−2 = X, I−2 = G, 25−5 = 20 → XG 20.

Trick: Number alone: 25 − 5 = 20 is in only one option.

Answer: (B) XG 20

Concept: Each cut is repeated in every layer; on opening, creases act as mirrors.

1. Two folds (horizontal then vertical) make 4 layers, so each cut appears 4 times.
2. The square was cut at the outer corner of the folded quarter → squares go to the 4 corners of the sheet.
3. The circle was near the inner corner (where creases meet) → 4 circles cluster round the centre.

Trick: Outer corner of the folded piece → sheet corners; inner corner → sheet centre.

Answer: (A) Figure (A)