Q&A: practice questions with answers

Concept: Replace the letters by the given operators and apply BODMAS (division and multiplication before addition and subtraction).

1. $81\text{A}3\text{D}7\text{B}2\text{C}16\Rightarrow 81÷3-7\times 2+16$
2. $=27-14+16$
3. $=29$

Trick: Write the substituted expression fully before computing – mixing substitution and calculation causes sign slips.

Answer: (C) ग

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

1. Letters: G→E→C→A (each $-2$), A→Y→W→U ($-2$), O→M→K→I ($-2$). Next: A$-2=$Y, U$-2=$S, I$-2=$G → YSG.
2. Numbers: $81,75,69,63$ – each $-6$, so next $=57$.
3. Combined: YSG 57.

Trick: Numbers alone eliminate three options ($63-6=57$); only option D has 57.

Answer: (D) क

Concept: Substitute the operators and follow BODMAS.

1. $25\text{A}51\text{D}3\text{B}6\text{C}5\Rightarrow 25+51÷3-6\times 5$
2. $=25+17-30$
3. $=12$

Trick: Do all $÷$ and $\times$ first ($51÷3=17$, $6\times 5=30$), then run left to right.

Answer: (A) प

Concept: Decode from the end (X) backwards so that each person’s generation relative to X is clear.

1. P = X → P is the father of X. O € P → O is the son of P, so O is X’s brother.
2. C ? O → C is the daughter of O. B % C → B is the sister of C, so B is also a daughter of O.
3. Therefore B is the daughter of X’s brother.

Trick: “Son of X’s father” = X’s brother (gender known!). The child of a brother is a niece/nephew – here a daughter, so “brother’s daughter”.

Answer: (D) “

Concept: Draw Venn circles: “All A are B” puts circle A inside circle B.

1. All tigers are chairs and all chairs are monks → Tigers ⊂ Chairs ⊂ Monks.
2. Conclusion (I) “Some monks are chairs” is the valid conversion of “All chairs are monks” – follows.
3. Conclusion (II) “All tigers are monks” follows by the chain (two universal affirmatives give a universal affirmative).

Trick: All + All = All (A→B, B→C gives A→C), and every “All B are C” also yields “Some C are B”.

Answer: (D) A

Concept: Look for an odd digit whose left neighbour is a perfect cube (1 or 8) and whose right neighbour is even.

1. Series: 3 5 8 9 7 4 1 5 9 6 2 3 6 5 8 7 4 1 5 9 8 6 3 2 1 5 4 7 8 4 1 2 5 9 8 7 4 1 5 6 2 3 9 8 7 8 9 7 5 1.
2. Matches: 8 7 4, 1 5 4, 8 7 4, 1 5 6, 8 7 8.
3. Near-misses: 8 9 7 (followed by odd) ✗, 1 5 9 ✗, 1 2 5 (2 is even) ✗. Total = 5.

Trick: Circle every 1 and 8, then read the two digits after each – odd then even is a hit.

Answer: (B) ह

Concept: Find an odd digit with an odd digit immediately before it and an even digit immediately after it.

1. Series: 7 5 9 2 7 9 4 1 5 1 7 8 3 2 8 4 1 2 3 8 2 7 9.
2. Matches: 5 9 2, 7 9 4, 1 7 8.
3. Others fail: 7 5 9 (followed by odd), 1 5 1 ✗, 7 8 → 8 is even itself, 1 2 3 → 1 not odd-preceded... So three.

Trick: Scan for the pattern “odd, odd, even” as a sliding window of three – the middle digit is the one you count.

Answer: (A) त

Concept: Find the rule from the first pair and verify with the second.

1. FR 6 → KN $-6$: F$+5=$K, R$-4=$N, $6-12=-6$.
2. LU 9 → QQ $-3$: L$+5=$Q, U$-4=$Q, $9-12=-3$ ✓.
3. LS 14 → L$+5=$Q, S$-4=$O, $14-12=2$ → QO 2.

Trick: The first letter rule ($+5$: L→Q) alone leaves only option A starting with Q.

Answer: (A) क

Concept: Take A as the origin and track the position after every turn (facing south, a left turn faces east; facing east, left faces north, and so on).

1. A → 23 km south → (0, −23); left (east) 19 km → (19, −23); left (north) 36 km → (19, 13).
2. Left (west) 19 km → (0, 13); left (south) 9 km → (0, 4); right (west) 5 km → (−5, 4); left (south) 4 km → P (−5, 0).
3. P is 5 km due west of A, so he must drive 5 km to the east to reach A.

Trick: Add up the east–west moves and north–south moves separately: E $19-19-5=-5$ (5 west); N $-23+36-9-4=0$ – P is 5 km west of A.

Answer: (D) प

Concept: Common letters of the two words take the common digits; the remaining letter takes the remaining digit.

1. VIBE → 4 3 9 5 and BITE → 5 9 1 4.
2. Common letters I, B, E ↔ common digits 4, 9, 5.
3. Extra letter in BITE is T, extra digit is 1 → T = 1 (and V = 3).

Trick: Cancel what is common on both sides; the leftovers pair up instantly.

Answer: (A) द

Concept: Find a rule linking the three numbers of each given set and test the options with the same rule.

1. (14, 7, 21): $7=14÷2$ and $21=7+14$. (56, 28, 42): $28=56÷2$ and $42=28+14$ ✓.
2. Rule: 2nd $=$ 1st $÷2$, 3rd $=$ 2nd $+14$.
3. (62, 31, 45): $31=62÷2$ ✓, $45=31+14$ ✓. Others fail: $57+14=71\ne 72$, $121+14=135\ne 134$, $173+14=187\ne 188$.

Trick: All options have 2nd $=$ 1st $÷2$, so the 3rd number decides: just check “2nd $+14$”.

Answer: (D) स

Concept: Position from the left $=$ Total $-$ position from the right $+1$; moving left reduces it.

1. Yusuf from the left $=15-10+1=6$th.
2. Suman is 4th to the left of Yusuf: $6-4=2$nd from the left.

Trick: Convert to one end first, then add for “right of” and subtract for “left of”.

Answer: (B) प

Concept: Compare the shift from the first cluster to the second for both letters.

1. RF–NB: R(18)→N(14) $-4$, F(6)→B(2) $-4$. MJ–IF: M(13)→I(9) $-4$, J(10)→F(6) $-4$. PL–LH: P(16)→L(12) $-4$, L(12)→H(8) $-4$.
2. XA–TU: X(24)→T(20) $-4$, but A(1)→U(21) is $+20$ (or $-6$ with wrap-around), not $-4$.
3. So XA–TU is the odd one.

Trick: Three pairs show a clean $(-4,-4)$; the pair that needs “wrapping past A” to explain is the odd one.

Answer: (A) त

Concept: Evaluate the equation as given, then try each interchange and recompute with BODMAS until the LHS equals 67.

1. Given: $(28÷4)\times 6+14\times 3-7\times 9+(66÷2)+38=42+42-63+33+38=92\ne 67$.
2. Interchange 2 and 3: $(28÷4)\times 6+14\times 2-7\times 9+(66÷3)+38$
3. $=42+28-63+22+38=67$ ✓.

Trick: LHS is 25 too big; swapping 3 and 2 lowers $14\times 3$ by 14 and $66÷2$ by 11 – exactly 25.

Answer: (C) L

Concept: Draw Venn circles: “Some poets are writers” makes the Poets and Writers circles overlap; “All writers are dreamers” puts Writers inside Dreamers.

1. The overlap of Poets and Writers lies inside Dreamers, so some poets are dreamers – hence “Some dreamers are poets” (I) follows.
2. Poets outside the Writers circle may or may not be dreamers, so “All poets are dreamers” (II) does not follow.

Trick: Some + All = Some (particular + universal gives only a particular conclusion); “Some A are B” always converts to “Some B are A”.

Answer: (D) S

Concept: Greek ambassador Megasthenes and his book Indica.

1. Seleucus I Nicator, after his treaty with Chandragupta Maurya (c. 305 BCE), sent Megasthenes as ambassador to Pataliputra.
2. Megasthenes' Indica describes Chandragupta's court and administration.
3. Deimachus was sent later to Bindusara's court – the usual confusion.

Yaad rakhein: Megasthenes → Chandragupta; Deimachus → Bindusara.

Answer: (D) Chandragupta Maurya

Concept: Facts about the Great Indian (Thar) Desert from NCERT physiography.

1. It lies north-west of the Aravalis, has an arid climate, scanty vegetation, and is believed to have been under the sea in the Mesozoic era – statements A, B, D are correct.
2. Its rainfall is low but below about 150 mm per year, not 15 mm; so statement C is incorrect.

Yaad rakhein: Thar rainfall figure = 150 mm (NCERT); "15 mm" is a trap.

Answer: (C) This region receives low rainfall below 15 mm per year

Concept: India–Spain cooperation agreements of October 2025.

1. During the October 2025 engagements between India and Spain, an MoU was signed for developing a world-class convention centre in Maharashtra.
2. Maharashtra (Mumbai region) has been the focus of such large-scale convention and tourism infrastructure tie-ups.

Yaad rakhein: Spain + convention centre (2025) → Maharashtra.

Answer: (A) Maharashtra

Concept: Rashtriya Avishkar Abhiyan (RAA), launched 9 July 2015 by the Ministry of HRD.

1. Its aim: make science, mathematics and technology learning joyful for children aged 6–18 through observation, experiments and inquiry.
2. It was inspired by Dr. A. P. J. Abdul Kalam and runs under Samagra Shiksha with mentoring by IITs/IISERs.

Yaad rakhein: "Avishkar" = invention → science & maths.

Answer: (B) Science and mathematics learning

Concept: Host of the 2030 Centenary Commonwealth Games.

1. At the Commonwealth Sport General Assembly in Glasgow (November 2025), Ahmedabad (Gujarat) was confirmed as host of the 2030 Centenary Commonwealth Games.
2. India last hosted the Games in New Delhi in 2010; 2030 marks 100 years since the first Games (Hamilton, 1930).

Yaad rakhein: CWG 2026 Glasgow → CWG 2030 Ahmedabad.

Answer: (A) Ahmedabad