Q&A: practice questions with answers

Concept: $\sin \theta =\frac{\text{perpendicular}}{\text{hypotenuse}}$; find the base by Pythagoras, then $\cos \theta =\frac{\text{base}}{\text{hypotenuse}}$.

1. Perpendicular $=p^2-1$, hypotenuse $=p^2+1$
2. Base${}^2=(p^2+1{)}^2-(p^2-1{)}^2=4p^2$, so base $=2p$
3. $\cos \theta =\frac{2p}{p^2+1}$

Trick: $(p^2-1,2p,p^2+1)$ is a standard Pythagorean triplet — learn it.

Answer: (C) $\frac{2p}{p^2+1}$

Concept: Doubling speed halves the time. Compare Anmol’s two times with Nikhil’s time $t$.

1. At speed $a$: $\frac{52}{a}=t+6$
2. At speed $2a$: $\frac{52}{2a}=\frac{26}{a}=t-7$
3. Subtract: $\frac{26}{a}=13$
4. $a=2\text{km/h}$

Trick: Halved time saves $6+7=13$ h, so the full time is 26 h; speed $=\frac{52}{26}=2$.

Answer: (C) 2 km/hr

Concept: Convert each average into a total, then combine the totals.

1. Salil + Mukund + Sachin $=3\times 45=135$
2. Salil + Mukund $=88$ ⇒ Sachin $=135-88=47$
3. Mukund + Sachin $=96$ ⇒ Mukund $=96-47=49$

Trick: Middle person $=(88+96)-135=49$ directly.

Answer: (A) 49

Concept: Expand each bracket, then collect like terms.

1. $x(6x-9)=6x^2-9x$
2. $6(x^2-3)=6x^2-18$
3. Sum $+19$: $12x^2-9x-18+19=12x^2-9x+1$

Trick: Put $x=1$: $-3+(-12)+19=4$; only $12-9+1=4$ matches.

Answer: (B) $12x^2-9x+1$

Concept: For equal time periods, profit is shared in the ratio of investments.

1. $1120:1320:1160=28:33:29$ (divide by 40)
2. Total $=90$ parts
3. Madhur $=\frac{29}{90}\times 1080=29\times 12=348$

Trick: $1080÷90=12$ per part makes the multiplication easy.

Answer: (A) ₹348

Concept: Write the alphabet positions and check the gaps between the letters.

1. HPJ: H(8) → P(16) is +8, P → J(10) is −6
2. KSM: 11 → 19 (+8) → 13 (−6); NVP: 14 → 22 (+8) → 16 (−6)
3. QYR: 17 → 25 (+8) → 18 (−7) — breaks the pattern

Trick: Pattern +8, −6 also means 3rd letter = 1st + 2; Q + 2 = S, not R.

Answer: (B) QYR

Concept: Facing the centre, a person’s right is anticlockwise. Fix Q and count seats to Q’s right as positions 1–6.

1. Two between Q and K from Q’s right ⇒ K at 3. Three between Q and P ⇒ P at 4 (3 is taken).
2. Two between K and O ⇒ O at 6.
3. Left seats 1, 2, 5; I immediately left of L ⇒ I = 1, L = 2, J = 5.
4. Third to the left of J: P, K, L ⇒ L.

Trick: In a 7-seat circle, “n between” on one side = “5 − n between” on the other side.

Answer: (B) L

Concept: Facing north, the right end is the east end. Number positions 1–7 from the left.

1. Two people to the right of K ⇒ K = 5; Q immediately left ⇒ Q = 4.
2. Two between K and R ⇒ R = 2.
3. Two between I and Q ⇒ I = 1 or 7. P–L must be adjacent: only I = 1 leaves 6, 7 free ⇒ P = 6, L = 7.
4. Remaining J = 3.

Trick: Test both cases for I quickly — the adjacent-pair condition (P, L) usually rejects one.

Answer: (A) J

Concept: Swap the signs first, then apply BODMAS.

1. $100\times 4+11÷2-18$ becomes $100÷4-11\times 2+18$
2. $=25-22+18=21$

Trick: Rewrite the whole expression with new signs before calculating — never swap mentally.

Answer: (C) 21

Concept: Check the differences between consecutive terms.

1. 22 → 25 (+3) → 36 (+11) → 39 (+3) → 50 (+11)
2. Next: 50 + 3 = 53

Trick: Alternating differences (+3, +11) repeat — continue the cycle.

Answer: (B) 53

Concept: Track each letter position of the cluster separately.

1. 1st: E, H, K, N → +3 → Q
2. 2nd: L, O, R, U → +3 → X
3. 3rd: J, M, P, S → +3 → V
4. Answer: QXV

Trick: If one letter shows a constant jump, check the others first with the same jump.

Answer: (C) QXV

Concept: Find the gaps between consecutive letters.

1. FIM: F → I (+3) → M (+4)
2. ORV: +3, +4; PSW: +3, +4
3. ZBF: Z → B (+2) → F (+4) — different

Trick: For wrap-around letters (Z → A …) count carefully: Z + 3 = C, not B.

Answer: (C) ZBF

Concept: Letters and numbers follow separate rules.

1. Each letter moves −4: N→J→F→B→X→T, U→Q→M→I→E→A, B→X→T→P→L→H
2. Numbers: 1, 4, 10, 22, 46 — each = previous × 2 + 2
3. Next number $=46imes2+2=94$; cluster = TAH 94

Trick: Options share 94, so only the letters decide: −4 from X, E, L gives T, A, H.

Answer: (B) TAH 94

Concept: Draw the least-overlap Venn diagram and test each conclusion.

1. All Melons lie inside Grapefruits; some Blueberries are Melons.
2. So those Blueberries are also Grapefruits ⇒ “Some Grapefruits are Blueberries” is definite — II follows.
3. I (“No Blueberries are Grapefruits”) contradicts this — does not follow.

Trick: Some A are B + All B are C ⇒ Some A are C (and its converse Some C are A).

Answer: (D) Only conclusion (II) follows.

Concept: Sort the numbers and compare each position with the original.

1. Original: 869, 509, 779, 491, 863
2. Ascending: 491, 509, 779, 863, 869
3. Same position: 509 (2nd) and 779 (3rd) ⇒ 2 numbers

Trick: Write both rows one under the other and tick matching columns.

Answer: (B) 2

Concept: Differences double every time.

1. Differences: 8, 16, 32, 64
2. Next difference = 128
3. 203 + 128 = 331

Trick: When differences are 8, 16, 32 … they form a ×2 chain.

Answer: (D) 331

Concept: Interchange the signs, then solve with BODMAS.

1. $25\times 5-18÷3+15$ becomes $25÷5+18\times 3-15$
2. $=5+54-15=44$

Trick: Do ÷ and × first after swapping: 5 and 54.

Answer: (C) 44

Concept: Compare letter positions of each pair.

1. QMJ → GCZ: Q(17)→G(7), M(13)→C(3), J(10)→Z(26): each −10 (wrapping after A)
2. CYV → SOL: C−10 = S, Y−10 = O, V−10 = L
3. VRO: V(22)−10 = L, R(18)−10 = H, O(15)−10 = E ⇒ LHE

Trick: −10 is the same as +16; using +16 makes wrap-arounds easy.

Answer: (C) VRO−LHE

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

1. RAID → 3527, DOGS → 1346
2. Common letter: D; common digit: 3
3. So D = 3

Trick: Common letter ↔ common digit (order of digits may be jumbled).

Answer: (B) 3

Concept: Plot each move on a grid with A at (0, 0).

1. 7 km N → (0, 7); right ⇒ 4 km E → (4, 7)
2. Right ⇒ 11 km S → (4, −4); right ⇒ 9 km W → (−5, −4)
3. Right ⇒ 4 km N → P(−5, 0)
4. P is 5 km west of A ⇒ drive 5 km east

Trick: Net N–S: 7 − 11 + 4 = 0; net E–W: 4 − 9 = −5 (west). Go back 5 km east.

Answer: (D) 5 km to the East