Q&A: practice questions with answers

Concept: $(a,b,c)$ is a Pythagorean triplet if $a^2+b^2=c^2$.

1. $(8,15,17)$: $64+225=289={17}^2$ ✔.
2. Others fail, e.g. $36+81=117\ne 121$; $4+9=13\ne 25$; $25+49=74\ne 81$.

Yaad rakhein: Standard triplets: $(3,4,5)$, $(5,12,13)$, $(8,15,17)$, $(7,24,25)$.

Answer: (B) (8, 15, 17)

Concept: Angle sum of a triangle is ${180}^{\circ }$; with $QR\parallel PS$, alternate angles are equal; equal sides face equal angles.

1. In $△PQR$: $\angle QRP={180}^{\circ }-{96}^{\circ }-{36}^{\circ }={48}^{\circ }$.
2. $QR\parallel PS$ (transversal $PR$): $\angle RPS=\angle QRP={48}^{\circ }$.
3. $PR=RS\Rightarrow \angle RSP=\angle RPS={48}^{\circ }$.
4. $\angle PRS={180}^{\circ }-{48}^{\circ }-{48}^{\circ }={84}^{\circ }$.

Trick: In fact $\angle PRS={180}^{\circ }-2\angle QRP={180}^{\circ }-2({180}^{\circ }-{96}^{\circ }-{36}^{\circ })$.

Answer: (C) ${84}^{\circ }$

Concept: Rodney’s gain or loss depends only on his own cost price and selling price.

1. Rodney’s CP $=10000\times 1.10=11000$.
2. Rodney’s SP $=11000\times 0.90=9900$.
3. Loss $=11000-9900=1100$.

Trick: Loss $=10\%$ of his CP $=10\%$ of ₹11,000 $=$ ₹1,100.

Answer: (D) loss of ₹1,100

Concept: Sum of observations $=\text{mean}\times \text{number of observations}$.

1. Total $=9\times 57=513$.
2. Known sum $=50+63+62+64+48+51+55+48=441$.
3. $x=513-441=72$.

Trick: Deviation method: deviations from 57 of the known values add to $-15$, so $x=57+15=72$.

Answer: (B) 72

Concept: $a\%$ of $b\%=\frac{a}{100}\times \frac{b}{100}$.

1. $\frac{25}{100}\times \frac{25}{100}=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$.
2. $\frac{1}{16}=0.0625$.

Trick: $25\%=0.25$; $0.25\times 0.25=0.0625$ (four decimal places).

Answer: (D) 0.0625

Concept: $V=\frac{4}{3}\pi r^3$, so volume varies as $r^3$.

1. New radius $=1.2r$.
2. New volume $=(1.2{)}^{3}V=1.728V$.
3. $1.728V=172.8\%$ of original.

Trick: Volume ratio $={\left(\frac{6}{5}\right)}^3=\frac{216}{125}=1.728$; note the increase is $72.8\%$, the new volume is $172.8\%$.

Answer: (C) 172.8%

Concept: ${\sec }^{2}A-{\tan }^{2}A=1$ and ${\text{cosec}}^{2}A-{\cot }^{2}A=1$.

1. Group: $2({\sec }^{2}A-{\tan }^{2}A)+6({\text{cosec}}^{2}A-{\cot }^{2}A)$.
2. $=2(1)+6(1)=8$.

Trick: Put $A={45}^{\circ }$: $2(2)+6(2)-2(1)-6(1)=4+12-2-6=8$.

Answer: (D) 8

Concept: Check the gaps between consecutive letters using alphabet positions.

1. QOL: $17,15,12$ → $-2,-3$; MKH: $13,11,8$ → $-2,-3$; YWT: $25,23,20$ → $-2,-3$.
2. VTR: $22,20,18$ → $-2,-2$ (different).

Trick: Write each letter’s position number once; the odd gap pattern stands out immediately.

Answer: (B) VTR

Concept: Compare the position gaps between letters in each cluster.

1. NPK: $14,16,11$ → $+2,-5$; JLG: $10,12,7$ → $+2,-5$; FHC: $6,8,3$ → $+2,-5$.
2. HJF: $8,10,6$ → $+2,-4$ (different).

Trick: Here the first and third letters differ by $3$ in the group (N–K, J–G, F–C), but H–F differ by $2$.

Answer: (C) HJF

Concept: Vowels move one step forward (+1); consonants move one step back (−1).

1. F→E, L→K, O→P, W→V, I→J, N→M, G→F.
2. New group: EKPVJMF. Vowels: only E → $1$.

Trick: Vowels +1 never give a vowel (A→B, E→F, I→J, O→P, U→V); only a consonant that turns into a vowel (here F→E) adds one.

Answer: (D) 1

Concept: Look for three symbols in a row: symbol – symbol – symbol; the middle one is counted.

1. Scan the series for runs of consecutive symbols: (& %), (* £), (+ & @).
2. Only “+ & @” has three symbols together, so only & qualifies.
3. (& %) and (* £) have just two symbols – no middle symbol.

Trick: Mark every symbol, then circle only runs of length 3 or more; count the inner ones.

Answer: (A) One

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

1. EVIL → 6824 and BONE → 8753.
2. Common letter: E; common digit: $8$. So E = $8$.

Trick: Common letter ↔ common digit – order of digits does not matter.

Answer: (A) 8

Concept: Find the operation linking the three numbers of each set.

1. $(4,12,24)$: $4\times 3=12$, $12\times 2=24$; $(7,21,42)$: $7\times 3=21$, $21\times 2=42$.
2. $(9,27,54)$: $9\times 3=27$, $27\times 2=54$ ✔.

Trick: Pattern $(a,3a,6a)$: third $=6\times$ first. Only $(9,27,54)$ fits.

Answer: (B) (9, 27, 54)

Concept: Draw the least-overlap Venn diagram; a conclusion follows only if it is true in every possible diagram.

1. Some tables are knives and all knives are paper → the table–knife overlap lies inside paper, so some papers are tables: (II) follows.
2. Blue only touches paper somewhere; it need not touch tables → (I) does not follow.

Yaad rakhein: “Some + All” gives a definite “Some”; a link through “Some + Some” gives no conclusion.

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

Concept: Swap the signs first, then apply BODMAS.

1. After swapping: $63÷9-6+13\times 4$.
2. $=7-6+52=53$.

Trick: Rewrite the whole expression with new signs before calculating – never solve mentally while swapping.

Answer: (B) 53

Concept: Second term = first two letters swapped; third term = first term written in reverse.

1. WEST → EWST (swap W, E) → TSEW (reverse of WEST). POLE → OPLE → ELOP likewise.
2. VIEW → IVEW (swap V, I) → WEIV (reverse of VIEW) ✔.
3. FANG’s third term should be GNAF, GROW’s second should be RGOW, ROAD’s third should be DAOR – all fail.

Trick: Check the reverse (third term) first – it eliminates options fastest.

Answer: (A) VIEW - IVEW - WEIV

Concept: Track position on a grid (East = +x, North = +y) after each turn.

1. A$(0,0)$ → 4 km East → $(4,0)$. Left (North) 2 km → $(4,2)$; left (West) 2 km → $(2,2)$; left (South) 3 km → $(2,-1)$.
2. Right (West) 1 km → $(1,-1)$; right (North) 1 km → B$(1,0)$.
3. B is 1 km East of A, so she must drive 1 km West.

Trick: Net East $=4-2-1=1$; net North $=2-3+1=0$ → B is 1 km East of A.

Answer: (D) 1 km West

Concept: Decode each symbol one link at a time and build the family tree.

1. A – B: A is the son of B. B × C: B is the wife of C (so C is A’s father).
2. C ÷ D: C is the brother of D. D + E: D is the father of E.
3. So C is the brother of E’s father.

Trick: Decode from the person asked about (C) to the target (E): C → brother of D → D is E’s father.

Answer: (A) Father's brother

Concept: Replace each sign by its meaning, then apply BODMAS.

1. $65\times 13-7÷24+9$ becomes $65÷13\times 7+24-9$.
2. $=5\times 7+24-9=35+15=50$.

Trick: Write the mapping (+→−, −→×, ×→÷, ÷→+) above the expression before substituting.

Answer: (C) 50

Concept: Track each letter position separately across the terms.

1. 1st letters: G, I, K, M, O ($+2$).
2. 2nd letters: J, L, N, P, R ($+2$); 3rd letters: E, G, I, K, M ($+2$).
3. Next term: ORM.

Trick: Every letter moves $+2$, so just add 2 to each letter of MPK.

Answer: (B) ORM