Q&A: practice questions with answers

Concept: Each cluster shifts by a fixed number of places.

1. CEGI → GIKM → KMOQ: every letter moves +4.
2. KMOQ + 4 → O, Q, S, U.
3. Next = OQSU.

Trick: The last two letters of a cluster open the next one (CEGI → GI.., GIKM → KM..): so after KMOQ comes OQ.. → keep +2 → OQSU.

Answer: (C) OQSU

Concept: Apply the defined meanings of the symbols, left to right.

1. 8 $ 6 $=(8-6{)}^2=2^2=4$.
2. 4 % 2 $=(4+2)\times 3=18$.
3. Value = 18.

Trick: Answer must be a multiple of 3 (last step ×3): only 12 and 18 – and 12 would need sum 4, impossible here.

Answer: (B) 18

Concept: Causation: one thing produces the other. Correlation: two things simply occur together.

1. Contaminated water → cholera, insulin deficiency → diabetes, bacteria → food poisoning: proven causes.
2. Front-row seating and higher marks only go together; sincere students tend to sit in front – the seat does not create marks.
3. So (A) shows only correlation.

Trick: The options using "causes" state causation; the one without a cause word is usually the correlation.

Answer: (A) Students who sit in the front rows score higher marks.

Concept: Track each letter position separately.

1. 1st: B, E, _, K, N (+3) → H.
2. 2nd: S, P, _, J, G (−3) → M.
3. 3rd: F, I, _, O, R (+3) → L → HML.

Trick: Only (A) and (B) start with H; middle letter must be P − 3 = M → HML.

Answer: (A) HML

Concept: Check each position's rule; the term that breaks it is wrong.

1. 1st: N, J, F, B, X (−4, wrapping B → X) ✓.
2. 2nd: V, R, N, J, F (−4) ✓.
3. 3rd: K, G, C, Y, ? → Y − 4 = U, but given W → XFW is wrong (should be XFU).

Trick: Backward wrap: B − 4 → (A, Z, Y) X; Y − 4 → U.

Answer: (B) XFW

Concept: Different positions follow different steps.

1. 1st: H, J, _, N, P (+2) → L.
2. 2nd: Y, U, _, M, I (−4) → Q.
3. 3rd: V, B, _, N, T (+6, V → B wraps past Z) → H → LQH.

Trick: Only (A) and (D) start with L; 2nd letter U − 4 = Q → LQH.

Answer: (A) LQH

Concept: Find the step for each of the three positions.

1. 1st: G, K, _, S, W (+4) → O.
2. 2nd: P, L, _, D, Z (−4) → H.
3. 3rd: Q, M, _, E, A (−4) → I → OHI.

Trick: In KLM and SDE the last two letters are consecutive; so the missing one also ends in consecutive letters: H, I → OHI.

Answer: (C) OHI

Concept: Here the rule works inside each cluster: 1st +9 = 2nd, 2nd +5 = 3rd.

1. DMR: 4+9=13 (M), 13+5=18 (R) ✓; FOT ✓; BKP ✓.
2. ZIN: Z+9 → I (wrap), I+5 → N ✓.
3. HQW: H+9 = Q ✓ but Q(17)+5 = 22 = V, not W → HQW is wrong.

Trick: First and last letters differ by 14 in every correct term (D→R, F→T); H→W is 15 – odd one out.

Answer: (B) HQW

Concept: A word can be formed only if each letter is available as many times as needed.

1. ADMINISTRATION has A2, D1, M1, I3, N2, S1, T2, R1, O1.
2. RATION – all once ✓; DOMINANT needs N×2 ✓; TRANSMIT needs T×2 ✓.
3. All three can be formed.

Trick: Check only the repeated letters of each word (N in DOMINANT, T in TRANSMIT) – that is where it usually fails.

Answer: (D) All the words can be formed using the letters only once.

Concept: Look at the differences of consecutive terms.

1. Differences: 4, 9, 25, 49, 121 = $2^2,3^2,5^2,7^2,{11}^2$.
2. Next prime is 13 → ${13}^2=169$.
3. $210+169=$ 379.

Trick: Differences that are perfect squares but skip $4^2$, $6^2$ → think of prime squares.

Answer: (C) 379

Concept: Work with totals: total = average × count.

1. Old total = $20\times 55=1100$; correction adds $70-40=30$ → 1130.
2. Corrected average = $1130÷20=56.5$; new student = $56.5\times 1.2=67.8$.
3. Final average = $(1130+67.8)÷21=1197.8÷21\approx$ 57.04.

Trick: Newcomer is $67.8-56.5=11.3$ above average → average rises by $\frac{11.3}{21}\approx 0.54$ → $56.5+0.54=57.04$.

Answer: (B) 57.04

Concept: Overall return % = weighted average of the separate returns (weights = share invested).

1. Shares: MF 32%, FD 28%, RD $100-60=40\%$.
2. Return = $0.32\times 25+0.28\times 10+0.40\times 20=8+2.8+8=$ 18.8%.
3. (In rupees: gain ₹33,840 on ₹1,80,000 = 18.8%.)

Trick: The income figure is not needed – percentage weights alone give the answer.

Answer: (A) 18.8%

Concept: Net effect of +a% then −b% = $a-b-\frac{ab}{100}$.

1. +25% then −20%: $25-20-\frac{25\times 20}{100}=0\%$ net change.
2. So final price = original price.
3. Original = ₹84,000.

Trick: $+25\%=\times \frac{5}{4}$ and $-20\%=\times \frac{4}{5}$ – they cancel exactly.

Answer: (D) ₹84,000

Concept: A number ending in 3 has a square ending in 9, and the option must also be a perfect square.

1. All options end in 9, so check which is a perfect square.
2. ${12}^2=144$, ${13}^2=169$, ${14}^2=196$ → only 169 is a square.
3. $169={13}^2$, and 13 ends in 3 ✓.

Yaad rakhein: Squares end only in 0, 1, 4, 5, 6, 9; a number ending in 3 or 7 gives a square ending in 9.

Answer: (B) 169

Concept: A square ending in 25 has a root ending in 5.

1. Unit digit of root = 5.
2. Remaining part 30 lies between $5^2=25$ and $6^2=36$ → tens digit 5.
3. $\sqrt{3025}=$ 55 (check: $55\times 55=3025$).

Trick: For $n{5}^2$: $n\times (n+1)$ followed by 25 → $5\times 6=30$ → 3025, so root = 55.

Answer: (C) 55

Concept: Cross-section of a pipe = $\pi (R^2-r^2)=\pi (R+r)(R-r)$.

1. $R=\frac{54}{2}=27\text{cm}$, $r=\frac{38}{2}=19\text{cm}$.
2. $R^2-r^2=(27+19)(27-19)=46\times 8=368$.
3. Area = $\frac{22}{7}\times 368=\frac{8096}{7}\approx$ 1156.57 cm².

Trick: Use $(R+r)(R-r)$ – no need to square 27 and 19.

Answer: (D) $1156.57{\mathrm{cm}}^2$

Concept: $a^3-b^3=(a-b)(a^2+ab+b^2)$.

1. $216x^3-343y^3=(6x{)}^3-(7y{)}^3$ → quotient = $36x^2+42xy+49y^2$.
2. Match with $A{x}^2-B{y}^2+Cxy$: $A=36$, $-B=49\Rightarrow B=-49$, $C=42$.
3. $2A+B-C=72-49-42=$ −19.

Trick: Watch the sign: the "−B" in the given form makes B negative.

Answer: (A) -19

Concept: If a sum of real squares is 0, each square is 0.

1. $x^2-8x+y^2-10y+z^2-6z+50=0$.
2. $50=16+25+9\to (x-4{)}^2+(y-5{)}^2+(z-3{)}^2=0\Rightarrow x=4,y=5,z=3$.
3. $2(4+5)-3\times 3=18-9=$ 9.

Trick: Values = half the coefficients inside the bracket: $2(4x+5y+3z)$ → $x=4$, $y=5$, $z=3$ directly.

Answer: (D) 9

Concept: CSA of cone = $\pi rl=\frac{1}{2}\times (\text{base circumference})\times l$.

1. $2\pi r=22\to \pi r=11$.
2. CSA = $\pi r\times l=11\times 7=$ 77 cm².

Trick: No need to find r: CSA = $\frac{\text{circumference}\times \text{slant height}}{2}=\frac{22\times 7}{2}=77$.

Answer: (B) $77{\mathrm{cm}}^2$

Concept: ${\csc }^{2}A-{\cot }^{2}A=1$ for every valid angle.

1. From $1+{\cot }^{2}A={\csc }^{2}A$ → ${\csc }^{2}A-{\cot }^{2}A=1$.
2. Check: $\sin A=\frac{5}{13}\to \csc A=\frac{13}{5}$, $\cot A=\frac{12}{5}\to \frac{169}{25}-\frac{144}{25}=\frac{25}{25}=1$.
3. Value = 1.

Trick: The given value of $\sin A$ is a distractor – the identity is always 1.

Answer: (D) 1