Q&A: practice questions with answers

Concept: Identify the required numbers, then pick the asked digit.

1. Numbers: 547, 589, 652, 598, 637. Highest = 652, lowest = 547.
2. Second digit of 652 = 5; second digit of 547 = 4.
3. $5+4=9$.

Trick: Compare hundreds digits first (6 > 5) to spot the highest/lowest instantly.

Answer: (A) 9

Concept: Count-based clues ("only four after/before") fix exact days; then place the rest.

1. Only four after V → V = Wednesday. Only four before S → S = Friday.
2. R before P before V → R = Monday, P = Tuesday.
3. Left: Thu, Sat, Sun for Q, T, U. T after Q and not Sunday → Q = Thursday, T = Saturday, U = Sunday.
4. Between Q (Thu) and U (Sun): Fri, Sat → Two.

Trick: "Only four after X" in a 7-day week means X is on the 3rd day (Wednesday).

Answer: (D) Two

Concept: Start from the fixed floor (L = 1) and chain the "between" clues.

1. L = 1; one person between L and K → K = 3.
2. Two between S and K → S = 6 (floor 0 doesn't exist).
3. Two between J and T with J above K: J = 5, T = 2 (J = 4 would need T = 1 or 7).
4. R takes floor 4. Between R (4) and S (6): only J → One.

Trick: "Two people between A and B" = gap of 3 floors; with one end fixed, test both directions and discard the impossible one.

Answer: (D) One

Concept: Find the chain of operations in the given sets and apply it to the options.

1. $6\stackrel{+5}{\to }11\stackrel{\times 2}{\to }22\stackrel{+10}{\to }32$; $9\stackrel{+5}{\to }14\stackrel{\times 2}{\to }28\stackrel{+10}{\to }38$.
2. Option C: $13+5=18$, $18\times 2=36$, $36+10=46$ ✓.
3. A ends in 14 (should be 34), B in 58 (should be 48), D in 54 (should be 44).

Trick: The first two steps match in all options – check only the last step ($+10$).

Answer: (C) 13-18-36-46

Concept: Try the sum of alphabetical positions and look for a multiplier.

1. VAJR: $22+1+10+18=51$; $51\times 2=102$ ✓.
2. DOTG: $4+15+20+7=46$; $46\times 2=92$ ✓.
3. ZELW: $26+5+12+23=66$; $66\times 2=132$.

Trick: Even codes for 4-letter words → suspect (sum of positions) $\times 2$ first.

Answer: (B) 132

Concept: Compare corresponding letters of the two clusters in each pair.

1. QU–TX: Q→T $+3$, U→X $+3$.
2. ZD–CG: Z→C $+3$ (wrapping), D→G $+3$. KO–NR: K→N $+3$, O→R $+3$.
3. FI–JM: F→J $+4$, I→M $+4$ → different.

Trick: Write positions quickly: F=6, J=10 gives $+4$ while every other pair gives $+3$.

Answer: (D) FI-JM

Concept: For each number, check: symbol on the left AND letter on the right.

1. Series: G 6 E * @ # 3 D 2 % 7 4 Y & 5 + C R K S T.
2. 6: G before (letter) ✗. 3: # before, D after ✓.
3. 2: D before ✗. 7: % before but 4 after ✗. 4: 7 before ✗. 5: & before but + after ✗.
4. Only one such number.

Trick: Write the type under each character (L/N/S) and look for the pattern S-N-L.

Answer: (B) 1

Concept: Swap the signs as instructed, then apply BODMAS.

1. Given: $156\times 12-13÷8+71$. After swapping: $156÷12+13\times 8-71$.
2. $156÷12=13$; $13\times 8=104$.
3. $13+104-71=46$.

Trick: Rewrite the whole expression with new signs first – never swap mentally mid-calculation.

Answer: (C) 46

Concept: Trace the path on N–E–S–W axes; a left turn from North faces West, a left turn from West faces South, a right turn from South faces West.

1. Start → 11 km North → turn left (West) 7 km → turn left (South) 11 km: now she is back on the starting line, 7 km West of the start.
2. Turn right (West) 2 km → total $7+2=9$ km West of the start, at point R.
3. The two 11 km legs (North and South) cancel each other.

Trick: Equal and opposite legs cancel – just add the net East–West distances.

Answer: (A) 9 km, West

Concept: Treat the letters and the numbers as two separate series and find each pattern.

1. Letters: J→M→P→S (+3), M→P→S→V (+3), O→R→U→X (+3). Next: S+3 = V, V+3 = Y, X+3 = A (wrap after Z) → VYA.
2. Numbers: 50, 61, 72, 83 → +11 each → next $83+11=94$.
3. So the next cluster is VYA 94.

Trick: Within each cluster letters are at gap 3 (J-M-O is +3, +2) – check the first letter column only, then build the rest.

Answer: (C) VYA 94

Concept: Draw the sets: Tea lies completely inside Snack and completely inside Brunch; Brunch lies completely inside Dinner.

1. Conclusion (I): All tea ⊂ Brunch ⊂ Dinner, so every tea is a dinner – follows.
2. Conclusion (II): Tea is part of both Brunch and Snack, so the Tea region is common to them – "Some brunches are snacks" follows.
3. Both conclusions are definite, so both follow.

Trick: Two "All A is B" + "All A is C" statements always give "Some B is C" (through A).

Answer: (C) Both conclusions (I) and (II) follow.

Concept: The same letter-shift rule must connect # → EGC and QSO → %. Test the options.

1. Option (A): HJF → EGC: H−3 = E, J−3 = G, F−3 = C ✓ (each letter −3).
2. Apply −3 to QSO: Q−3 = N, S−3 = P, O−3 = L → NPL ✓, which matches option (A).
3. Other options do not follow one uniform rule for both pairs.

Trick: Check the first letter of each pair only (H→E, Q→N: both −3) to eliminate fast.

Answer: (A) # = HJF; % = NPL

Concept: Find one operation linking the first and third numbers to the middle one.

1. (44, 968, 22): $44\times 22=968$ ✓. (26, 650, 25): $26\times 25=650$ ✓. Rule: middle = product of the outer two.
2. Option (B): $18\times 6=108$ ✓.
3. (A) $34\times 5=170\ne 173$; (C) $15\times 23=345\ne 348$; (D) $23\times 9=207\ne 210$.

Trick: When the middle number is much larger, try the product of the outer numbers first.

Answer: (B) (18, 108, 6)

Concept: "All X are wolves" only places X inside Wolves; it says nothing about how the different sub-groups relate to each other.

1. Tigers, Jackals and Animals are all inside Wolves, but they may be three separate circles (as drawn) – no overlap is forced.
2. Conclusion (I) "Some animals are tigers" – possible but not definite → does not follow.
3. Conclusion (II) "Some jackals are animals" – also not definite → does not follow.

Trick: Three "All … are B" statements with the same predicate B never give any definite relation between the subjects.

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

Concept: Apply the rule digit by digit: even digit → +3, odd digit → −2, then count repeated digits in the new number.

1. 5723946 → 5−2 = 3, 7−2 = 5, 2+3 = 5, 3−2 = 1, 9−2 = 7, 4+3 = 7, 6+3 = 9.
2. New number: 3 5 5 1 7 7 9.
3. Digits appearing more than once: 5 and 7 → two digits.

Trick: Write the new digits under the old ones in a row – repeats become visible instantly.

Answer: (B) Two

Concept: One uniform letter shift must turn # into OIM and ICG into %. Test the options.

1. Option (C): QKO → OIM: Q−2 = O, K−2 = I, O−2 = M ✓ (each letter −2).
2. ICG with −2: I−2 = G, C−2 = A, G−2 = E → GAE ✓, which is given in option (C).
3. Option (A)/(B) MKO → OIM would need +2, −2, −2 – not uniform; (D) fails too.

Trick: Check the % part first – only (C) gives GAE from ICG with a single rule.

Answer: (C) # = QKO, % = GAE

Concept: Find the shift from the first cluster to the second in each pair.

1. LN → MO: +1, +1. HJ → IK: +1, +1. DF → EG: +1, +1.
2. KM → NP: K+3 = N, M+3 = P → +3, +3 – different.
3. So KM-NP is the odd one out.

Trick: Compare only the first letters: L→M, H→I, D→E are +1; K→N is +3.

Answer: (D) KM-NP

Concept: The same letter shift must give # → ZJR and PZH → %. Work backwards from the given clusters.

1. Options for # are TDL / UDL / TDM. TDL → ZJR: T+6 = Z, D+6 = J, L+6 = R ✓ (uniform +6).
2. Apply +6 to PZH: P+6 = V, Z+6 = F (wrap after Z), H+6 = N → VFN ✓.
3. Only option (A) has both TDL and VFN.

Trick: Z+6: count A(1)…F(6) after Z → F; options with VCN are instantly wrong.

Answer: (A) # = TDL % = VFN

Concept: Each letter has a fixed digit code; common letters in two words share common digits.

1. ALIT → 2159 and HILT → 6912.
2. Common letters L, I, T ↔ common digits 1, 9, 2.
3. The uncommon letter in ALIT is A → leftover digit 5; the uncommon letter in HILT is H → leftover digit 6.

Trick: Cancel the common letters and common digits – whatever remains is the code of the odd letter.

Answer: (C) 6

Concept: All face the centre, so "left" is clockwise and "right" is anticlockwise. Fix P and place the others step by step.

1. Two people between P and B counted from P's right → B is 3^rd to the right of P (equivalently 4^th to P's left).
2. Q is immediately left of B. Only S sits between O and P: the side P–S–O must be P's left (the right side is needed for the two seats before B), so S is immediate left of P and O next.
3. Seats left: the one immediately right of P and the one between O and B. C is not P's neighbour → C sits between O and B; R sits immediately right of P.
4. Counting left (clockwise) from R: P (1), S (2), O (3).

Trick: "Only X sits between A and B" = A and B are 2 seats apart – place such pairs first.

Answer: (B) O