Q&A: practice questions with answers

Concept: 24 places → try equal blocks; a 6-letter block PCTPRS repeats.

1. PCTPRS | PCTPRS | PCTPRS | PCTPRS
2. Blanks in order: C R P P T S C P → CRPPTSCP.

Trick: Count total letters + blanks, then try block sizes that divide it (4, 6, 8) – the visible letters confirm the block.

Answer: (D) CRPPTSCP

Concept: Letter part and number part follow separate rules.

1. Letters: P, Q, R, S, T, U → next V.
2. Numbers: 0, 1, 4, 9, 16, 25 = $0^2,1^2,\dots$ → next $6^2=36$.
3. Term = V36.

Trick: All options start with V, so only the number matters: next square after 25 is 36.

Answer: (D) V36

Concept: Numbers use $\times 1,+0.5,\times 2,+0.5,\times 3,+0.5$ …

1. $1\times 1=1$ → $+0.5=1.5$ → $\times 2=3$ → $+0.5=3.5$ → $\times 3=10.5$.
2. Next step is +0.5 → 11; letter after L = M.
3. Term = M11.

Trick: If numbers jump then barely move, suspect two alternating operations.

Answer: (C) M11

Concept: Swap every sign first (+ → −, − → ×, × → +, / → /), then solve by BODMAS.

1. $96/(12+4)\times (6-2)+10$ → $96÷(12-4)+(6\times 2)-10$
2. $=96÷8+12-10$
3. $=12+12-10=$ 14

Trick: Rewrite the whole expression with new signs on paper before calculating – mixing old and new signs is the usual mistake.

Answer: (C) 14

Concept: Replace signs as given (+ → ×, − → +, × → +, / → −), then simplify.

1. $[(6\times 3)+(18+6)]-(20+5)$
2. $=[18+24]-25$
3. $=42-25=$ 17

Trick: Brackets stay where they are – only the signs change.

Answer: (C) 17

Concept: Gap in the word must equal gap in the alphabet (either direction).

1. Alphabet positions: D4 A1 U21 G7 H8 T20 E5 R18.
2. Forward: D–G (3 = 3), D–H (4 = 4), G–H (1 = 1).
3. Backward: T–R (word gap 2, alphabet R←T gap 2).
4. Total = 4.

Trick: Write the alphabet rank under each letter; a pair works when |position difference| = |rank difference|.

Answer: (A) Four

Concept: A word can be formed only if every letter (with count) is in the given word.

1. Letters: P, A×3, R×2, L×3, E, O, G, M.
2. ROLE, LARGE, LEAP – all letters available.
3. ALERT needs T – not present ✗

Trick: First scan options for a letter missing altogether (here T) – the fastest elimination.

Answer: (D) ALERT

Concept: Count pairs whose gap in the word equals their gap in the alphabet.

1. Ranks: C3 L12 U21 S19 T20 E5 R18 S19.
2. S–T (adjacent, 19→20) ✓; R–S (adjacent, 18→19) ✓.
3. T–R: word gap 2, alphabet gap 2 (T→S→R backward) ✓
4. Total = 3.

Trick: Start with adjacent letters that are also alphabet neighbours, then check gaps of 2, 3 … – most pairs fail fast.

Answer: (D) Three

Concept: Fast growth → check ratios.

1. $7\times 2=14,\times 3=42,\times 4=168$.
2. Next $\times 5$ → $168\times 5=$ 840.

Trick: These are $7\times 1!,7\times 2!,7\times 3!,7\times 4!$, so next = $7\times 5!=7\times 120=840$.

Answer: (A) 840

Concept: All terms are primes; check which primes are skipped.

1. Skipped primes: 3, 7, 13, 19, 29, 37 – each lies between two terms.
2. Every alternate prime is taken → after 41 skip 43 → 47.

Trick: If differences look random, test whether every term is prime – a very common SSC pattern.

Answer: (B) 47

Concept: Real numbers = rational numbers ∪ irrational numbers (no overlap).

1. So every real number is either rational or irrational ✓
2. (B) false – rationals can be written as $\frac{p}{q}$, irrationals cannot.
3. (C) false – $\frac{1}{2}$, 5 are real and rational; (D) false – every integer $n=\frac{n}{1}$ is rational.

Yaad rakhein: Natural ⊂ Whole ⊂ Integer ⊂ Rational ⊂ Real; irrational numbers ($\sqrt{2}$, $\pi$) sit outside the rationals but inside the reals.

Answer: (A) All real numbers are either rational or irrational

Concept: Remaining volume = prism volume − cavity volume.

1. Hexagon area = $\frac{3\sqrt{3}}{2}\times 4^2=24\sqrt{3}=24\times 1.732=41.568{\text{cm}}^2$.
2. Prism volume = $41.568\times 15=623.52{\text{cm}}^3$.
3. Cavity = $2\times 2\times 15=60{\text{cm}}^3$.
4. Remaining = $623.52-60=$ $563.52\approx 563.5{\text{cm}}^3$.

Trick: Since the cavity runs the full height: $(\text{base area}-\text{hole area})\times h=(41.568-4)\times 15=563.5$ in one step.

Answer: (A) $563.5{\mathrm{cm}}^3$

Concept: $\text{Profit}=\text{volume}\times (\text{SP}-{\text{CP per cm}}^3)\times \text{number of pieces}$.

1. Volume of one prism = $12\times 12\times 30=4320{\text{cm}}^3$.
2. Profit per cm³ = $0.80-0.50=$ ₹0.30.
3. Total = $4320\times 0.30\times 20=4320\times 6=$ ₹25920.

Trick: Combine the small numbers first: $0.30\times 20=6$, then $4320\times 6=25920$.

Answer: (D) ₹25920

Concept: Profit ratio = capital × months each capital stayed.

1. A: $40000\times 8=3,20,000$.
2. B: $30000\times 12=3,60,000$.
3. Ratio = 32 : 36 = 8 : 9.

Trick: Cancel common zeros first: $4\times 8:3\times 12=32:36=8:9$.

Answer: (D) 8 : 9

Concept: Multiply each capital by the months it was invested.

1. A: $50000\times 12=6,00,000$.
2. B joins after 6 months → $20000\times 6=1,20,000$.
3. Ratio = 6 : 1.2 = 5 : 1.

Trick: In thousands: $50\times 12:20\times 6=600:120=5:1$.

Answer: (A) $5:1$

Concept: Height of a sphere = its diameter = 2r, so the cone also has $h=2r$.

1. Sphere = $\frac{4}{3}\pi r^3$.
2. Cone = $\frac{1}{3}\pi r^2\times 2r=\frac{2}{3}\pi r^3$.
3. Ratio = $\frac{4}{3}:\frac{2}{3}=$ 2 : 1.

Trick: Same r and $h=2r$: cone : sphere : cylinder = 1 : 2 : 3 (Archimedes' ratio).

Answer: (C) 2 : 1

Concept: Perpendicular bisectors meet at the point equidistant from all three vertices.

1. That point is the centre of the circle through A, B, C → circumcentre.
2. Incentre – angle bisectors; centroid – medians; orthocentre – altitudes.

Trick: Pairs: Bisector of Angle → Incentre; Perpendicular bisector → Circumcentre; Median → Centroid; Altitude → Orthocentre.

Answer: (B) Circumcenter

Concept: A sphere fitting exactly in a cylinder gives cylinder radius r and height 2r.

1. Sphere = $\frac{4}{3}\pi r^3$.
2. Cylinder = $\pi r^2\times 2r=2\pi r^3$.
3. Ratio = $\frac{4}{3}:2=$ 2 : 3.

Trick: Use the 1 : 2 : 3 rule (cone : sphere : cylinder) → sphere : cylinder = 2 : 3 instantly.

Answer: (A) $2:3$

Concept: $(a\cos A+b\sin A{)}^2+(a\sin A-b\cos A{)}^2=a^2+b^2$.

1. Cross terms $\pm 2·7·24\sin A\cos A$ cancel on adding.
2. $x^2+y^2=7^2+{24}^2=49+576=625$.
3. $2(x^2+y^2)=2\times 625=$ 1250.

Trick: Put $A=0^{\circ }$: $x=7,y=-24$ → $x^2+y^2=49+576=625$ – answer in seconds.

Answer: (A) 1250

Concept: Angle sum of a triangle = 180°.

1. Two angles = x and 3x, sum = 4x.
2. Third = $4x+50\%\text{ of }4x=6x$.
3. $10x={180}^{\circ }\to x={18}^{\circ }$; angles ${18}^{\circ },{54}^{\circ },{108}^{\circ }$.
4. Largest = 108°.

Trick: Third angle is $1.5\times (\text{sum of other two})$, so it takes $\frac{1.5}{2.5}=\frac{3}{5}\text{ of }{180}^{\circ }={108}^{\circ }$.

Answer: (A) 108°