Q&A: practice questions with answers

Concept: The letter common to both words takes the digit common to both codes.

1. NICE → 8 2 4 6 and COST → 3 8 5 1.
2. Only letter common to both words is C; only digit common to both codes is 8.
3. So C = 8.

Trick: One common letter ↔ one common digit – no need to decode the rest.

Answer: (B) 8

Concept: Track the position of each letter. In CLUB → CUBL the letters move as 1-2-3-4 → 1-3-4-2; in CUBL → ULCB they move as 1-2-3-4 → 2-4-1-3. HANG → HNGA → NAHG follows exactly the same two shuffles.

1. Apply 1-3-4-2 to MASK: M, S, K, A → MSKA.
2. Apply 2-4-1-3 to MSKA: S, A, M, K → SAMK.
3. So MASK-MSKA-SAMK matches. In option B, EARN → ARNE is 2-3-4-1, a different shuffle; C and D also break the pattern at the first step.

Trick: Write the position numbers under the first pair, then just test the first step of each option — most options fail there.

Answer: (A) MASK-MSKA-SAMK

Concept: Take A as origin, East = $+x$, North = $+y$, and track the position after each leg (every left turn rotates the heading anticlockwise).

1. West 53: $(-53,0)$. Left → South 50: $(-53,-50)$.
2. Left → East 54: $(1,-50)$. Left → North 60: $(1,10)$.
3. Left → West 1: $(0,10)$ = P. So P is exactly 10 km north of A.
4. To return to A he must drive $10$ km towards the South.

Trick: Net East–West: $-53+54-1=0$; net North–South: $-50+60=+10$. No Pythagoras needed when one of them is zero.

Answer: (D) 10 km to the South

Concept: Replace the signs as told, then apply BODMAS: division and multiplication before addition and subtraction.

1. Swap: $95÷5+378\times 3-161$ becomes $95\times 5-378÷3+161$.
2. $95\times 5=475$ and $378÷3=126$.
3. $475-126+161=510$.

Trick: Rewrite the whole expression with new signs first, then solve — never swap and solve in the same breath.

Answer: (C) 510

Concept: Compare letter positions place by place: DRU → BWS is $-2,+5,-2$ (D→B, R→W, U→S). ZBQ → XGO is also $-2,+5,-2$.

1. Apply to VLM: V$-2$ = T, L$+5$ = Q, M$-2$ = K → TQK.
2. Check others: BCG → WGB is $-5,+4,-5$; CLU → AOS is $-2,+3,-2$; HKD → DNZ is $-4,+3,-4$. None match.

Trick: Write the pattern as a signed triple like $(-2,+5,-2)$ and test only the middle letter of each option first — it eliminates fastest.

Answer: (A) VLM : TQK

Concept: Fix the definite clue first (V = Thursday), then chain the 'after/before' clues from it.

1. R is after V, Q after R, S after Q: that needs three days after Thursday → R = Fri, Q = Sat, S = Sun.
2. Remaining Mon–Wed go to P, T, U with P before T before U: P = Mon, T = Tue, U = Wed.
3. Sunday → S.

Trick: Count how many people must come after the fixed person; if it equals the days left, the order is forced.

Answer: (C) S

Concept: Find the differences between consecutive terms; both given sets must share them.

1. 12 → 26 → 46 → 70: $+14,+20,+24$. 20 → 34 → 54 → 78: $+14,+20,+24$.
2. Option A: 25 → 39 ($+14$) → 59 ($+20$) → 83 ($+24$). Matches.
3. B: $+14,+10,+24$; C: $+14,+30,+24$; D: $+15,+20,+24$ – all fail.

Trick: When the first differences are the same in both sets, just check the same differences in options – no need to hunt for a formula.

Answer: (A) 25 – 39 – 59 – 83

Concept: Use the product and sum clues to limit the choices, then test each case against the 'immediately above' clue.

1. D × B = 15 → {D, B} = {3, 5}. C + B = 7 → if B = 3, C = 4; if B = 5, C = 2.
2. Case B = 5, D = 3, C = 2: floors left 1, 4, 6 – A cannot be immediately above F. Rejected.
3. Case B = 3, D = 5, C = 4: floors left 1, 2, 6 → F = 1, A = 2, E = 6. Works.
4. B + F = 3 + 1 = 4.

Trick: With 6 floors, a product of 15 can only be $3\times 5$ – start from the most restrictive clue.

Answer: (B) 4

Concept: Look for a multiplying pattern when the numbers grow fast.

1. 5 → 20 → 60 → 120: $\times 4,\times 3,\times 2$. 8 → 32 → 96 → 192: $\times 4,\times 3,\times 2$.
2. Option B: $10\times 4=40$, $40\times 3=120$, $120\times 2=240$. Matches.
3. A: $24\times 2=48$ (not $\times 4$); C: 60 → 240 → 80 breaks; D: $144\times 2=288\ne 248$.

Trick: Ratios $4,3,2$ falling by one each step – once seen in the first set, verify only the last step of each option.

Answer: (B) 10 – 40 – 120 – 240

Concept: Drop every symbol (non-letter, non-digit), then count from the left in the new series.

1. Symbols in the series: £, $,\copyright ,*,\#,$.
2. New series: 5 C T R 6 B D 9 6 9 Z 7 2 R O K.
3. Fourth from the left: 5, C, T, R.

Trick: Strike the symbols with a pencil stroke and count the survivors – don't rewrite the whole series.

Answer: (D) R

Concept: Check the gap between corresponding letters of the two clusters in each pair.

1. GI–JL: G→J $+3$, I→L $+3$. MO–PR: $+3,+3$. JL–MO: $+3,+3$.
2. BD–FH: B→F $+4$, D→H $+4$ – the odd one.

Trick: Only the first letters need checking: G→J, M→P, J→M are all $+3$; B→F is $+4$.

Answer: (D) BD–FH

Concept: Scan each symbol and look at its immediate left and right neighbours; we need number–symbol–number.

1. Symbols with their neighbours: % (start), 3 < $,\&lt;$ 3, 4 ^ <, ^ < &, < & #, & # 5, 5 < 3, 3 ^ %, ^ % >, % > \, > \ &, \ & (end).
2. Only '5 < 3' has a number on both sides.
3. Count = one.

Trick: Underline every number first; a qualifying symbol must sit squeezed between two underlined items.

Answer: (D) One

Concept: Draw the Venn diagram from the statements and test each conclusion against every possible drawing.

1. All hands are legs + some hands are ears → the hands that are ears lie inside legs, so some legs are ears. Conclusion I follows.
2. Some ears are arms: these arms may be ears that are not hands at all, so 'some hands are arms' is not certain. Conclusion II does not follow.

Trick: Two 'some' statements in a chain (hands–ears, ears–arms) never give a definite 'some' conclusion.

Answer: (B) Only conclusion I follows.

Concept: Facing the centre, 'right' means anticlockwise. Fix one person and place the rest from the definite clues.

1. Put U at the top. G is third to the right of U; H is to the immediate left of G (since G is to H's immediate right).
2. Counting from H's right, three people sit between F and H → F is the fifth seat to H's right, which lands F just to the left of U.
3. Remaining seats: W with V to his immediate right fills two of them, I takes the last one (between U and H).
4. Second to the right of I is G.

Trick: Draw 7 dots, mark arrows for 'right', and place the pair with a fixed gap (U–G) first.

Answer: (B) G

Concept: Decode the chain from left to right and build the family tree.

1. T @ S: T is the son of S. S % G: S is the brother of G.
2. G + H: G is the wife of H. H & K: H is the father of K → G is K's mother.
3. S is K's mother's brother (maternal uncle); T, the son of S, is K's mother's brother's son.

Trick: Decode the symbols nearest the target (K) first, then walk back along the chain.

Answer: (D) Mother's brother's son

Concept: The question asks for the shortest of five people; all five must be placed, or at least the bottom position must be certain.

1. Statement I: Ketan > Mayank > Harsh. Rohit and Sidharth unknown – not sufficient.
2. Statement II: Mayank > Rohit > Harsh. Ketan and Sidharth unknown – not sufficient.
3. Together: Ketan > Mayank > Rohit > Harsh, but Sidharth is never mentioned – he could be shortest. Still not sufficient.

Trick: Count the names in the statements; if one person is missing from both, the shortest/tallest can't be fixed.

Answer: (B) Both statements I and II together are NOT sufficient to answer the question

Concept: Number Monday–Sunday as 1–7 and start with the clue that pins positions.

1. 'Only G has exam before K' → G = Mon (1), K = Tue (2).
2. Three people between H and K → H = Sat (6).
3. Three between L and J → the free pair with gap 4 is (Wed, Sun). F < E < J forces J = Sun (7), L = Wed (3); F = Thu, E = Fri.
4. Before L (Wed): G and K → two people.

Trick: 'Only X before Y' fixes two slots at once – always the first clue to use.

Answer: (D) Two

Concept: Tughlaqabad Fort in Delhi was built by the founder of the Tughlaq dynasty.

1. Ghiyasuddin Tughlaq (Ghazi Malik) founded the Tughlaq dynasty in 1320 and built the massive Tughlaqabad Fort in 1321 as his new capital city.
2. Muhammad bin Tughlaq is known for shifting the capital to Daulatabad; Firoz Shah built Firozabad (Firoz Shah Kotla); Alauddin Khalji built Siri Fort.

Yaad rakhein: Delhi's forts by ruler: Siri – Alauddin Khalji; Tughlaqabad – Ghiyasuddin; Firoz Shah Kotla – Firoz Shah.

Answer: (C) Ghiyasuddin Tughlaq

Concept: Both Indian classical systems rest on the same two pillars – Raga (melodic framework) and Tala (rhythmic cycle).

1. Hindustani (North) and Carnatic (South) music differ in style, instruments and compositions, but both are built on Raga and Tala and both accept the concept of Shruti (microtones).
2. Compositions are not identical, and both systems are primarily vocal, not exclusively instrumental – so the other options are wrong.

Yaad rakhein: Hindustani vs Carnatic: different flavours, same grammar – Raga + Tala.

Answer: (D) Foundation on Raga and Tala

Concept: The gold foil (alpha-particle scattering) experiment of 1909–11 revealed the tiny, dense, positively charged nucleus.

1. Ernest Rutherford interpreted the experiment: most alpha particles passed straight through, a few bounced back → atom is mostly empty space with a central nucleus – the nuclear model (1911).
2. Dalton gave the indivisible-atom theory, Thomson the 'plum pudding' model (electron discovery), and Bohr the quantised-orbit model (1913) that refined Rutherford's.

Yaad rakhein: Dalton → Thomson (plum pudding) → Rutherford (nucleus) → Bohr (orbits).

Answer: (B) Ernest Rutherford