Q&A: practice questions with answers

Concept: A change from loss to gain = change in SP as % of CP.

1. From $-10\%$ to $+5\%$ is a change of $15\%$ of CP.
2. $15\%$ of CP $=120$ ⇒ CP $=800$.
3. SP $=90\%$ of 800 $=720$.

Trick: Rupee difference ÷ % difference gives 1% of CP: $\frac{120}{15}=8$.

Answer: (B) ₹720

Concept: $SI=\frac{P\times R\times T}{100}$.

1. $SI=\frac{55000\times 5.5\times 5.5}{100}$.
2. $=550\times 30.25=16637.5$.

Trick: $5.5\times 5.5=30.25$ – multiply this by $\frac{P}{100}=550$.

Answer: (A) 16,637.5

Concept: Convert the fraction to a decimal: $\frac{3}{8}=0.375$.

1. Expression $=0.375+0.375-0.125$.
2. $=0.75-0.125=0.625$.

Yaad rakhein: Eighths: $\frac{1}{8}=0.125$, $\frac{3}{8}=0.375$, $\frac{5}{8}=0.625$.

Answer: (B) 0.625

Concept: $A=P{(1+\frac{R}{100})}^T$.

1. $9075=P\times (1.1{)}^2=1.21P$.
2. $P=\frac{9075}{1.21}=7500$.

Trick: $10\%$ for 2 years ⇒ $P:A=100:121$; $\frac{9075}{121}=75$, so $P=7500$.

Answer: (C) 7,500

Concept: Angle of depression = angle of elevation from the lower point (alternate angles).

1. Bottom (60°): $d=\frac{20}{\tan {60}^{\circ }}=\frac{20}{\sqrt{3}}$.
2. Top (30°): drop $=d\tan {30}^{\circ }=\frac{20}{\sqrt{3}}\times \frac{1}{\sqrt{3}}=\frac{20}{3}$.
3. Tower $h=20-\frac{20}{3}=\frac{40}{3}\approx 13.33$ m.

Trick: For 30° & 60° depressions, tower $=\frac{2}{3}$ of building height.

Answer: (A) 13.33 m

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$.

1. Take AB $=120$ km. Onward: $\frac{48}{15}+\frac{72}{20}=3.2+3.6=6.8$ h.
2. Return: $\frac{30}{16}+\frac{90}{24}=1.875+3.75=5.625$ h.
3. Average $=\frac{240}{12.425}\approx 19.32\text{km/h}$.

Trick: Choose AB as LCM of 5 and 4 multiples (120) to keep pieces whole.

Answer: (A) 19.32 km/hr

Concept: $\overline{x}=\frac{\sum fx}{\sum f}$.

1. $\sum f=12+12+14+13+14+11+14=90$.
2. $\sum fx=10800+12000+15400+15600+18200+15400+21000=108400$.
3. $\overline{x}=\frac{108400}{90}\approx 1204.44$.

Trick: Assumed mean 1200: deviations sum $=400$ ⇒ $1200+\frac{400}{90}=1204.44$.

Answer: (A) ₹1204.44

Concept: Form two equations from past and future conditions.

1. $G-8=4(R-8)$ ⇒ $G=4R-24$.
2. $G+6=2(R+6)$ ⇒ $4R-18=2R+12$ ⇒ $R=15$.
3. $G=60-24=36$ years.

Trick: Check options: $G=36$ ⇒ 8 years ago 28, Rohit 7 ✓; after 6 years $42=2\times 21$ ✓.

Answer: (B) 36 years

Concept: With the two resulting numbers $a,b$ in continued proportion $a:b::b:c$, the last term is $c=\frac{b^2}{a}$.

1. Numbers: $13-x$ and $25-x$; required $\frac{(25-x{)}^2}{13-x}=49$.
2. Try $x=4$: $\frac{{21}^2}{9}=\frac{441}{9}=49$ ✓.
3. $x=2,3$ give $\frac{529}{11}$, $48.4$ – not 49; so smallest $x=4$.

Trick: $49=7^2$, so look for $25-x=7k$ and $13-x=k^2$: $k=3$ gives $x=4$.

Answer: (A) 4

Concept: Let the numbers be $3k,5k,7k$.

1. $9k^2+25k^2+49k^2=83k^2=2075$.
2. $k^2=25$ ⇒ $k=5$.
3. Middle number $=5k=25$.

Trick: $2075÷83=25$ – a perfect square confirms the ratio method.

Answer: (B) 25

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

1. All vans are trailers → Vans circle lies inside Trailers.
2. No trailer is a scooter → Scooters circle is completely separate from Trailers.
3. Since vans are inside trailers, no van can touch scooters → Conclusion I follows.
4. Scooters and trailers never meet, so ‘Some scooters are trailers’ is false → II does not follow.

Trick: All A are B + No B is C ⇒ No A is C (always).

Answer: (C) Only conclusion (I) follows

Concept: Decode each symbol from left to right and build the family tree.

1. P £ D → P is the daughter of D (P is female).
2. D ← M → D is the wife of M; so P is the daughter of M too.
3. M @ L → M is the father of L; L ≥ U → L is the brother of U, so U is also M's child.
4. P, L and U are children of M and D → P (female) is U's sister.

Trick: Children of the same parents: a female is ‘sister’, a male is ‘brother’.

Answer: (B) Sister

Concept: Apply the rule digit by digit, then count repeated digits.

1. Even digits ($-2$): $8\to 6,2\to 0,4\to 2,6\to 4$.
2. Odd digits ($+3$): $5\to 8,3\to 6,1\to 4$.
3. 8254316 → 6 0 8 2 6 4 4.
4. 6 appears twice and 4 appears twice → two digits repeat.

Trick: Write the new number under the old one in a table – mistakes vanish.

Answer: (A) Two

Concept: Check the differences between consecutive terms.

1. Known differences: $98-89=9,110-98=12,125-110=15$ → they rise by 3.
2. So the differences are 3, 6, 9, 12, 15.
3. $80+3=83$ and $83+6=89$ ✓.

Trick: Differences in steps of 3 → multiples of 3 series.

Answer: (C) 83

Concept: Mark the vowels first, then look only at the letters sitting between two vowels.

1. Vowels in the series: E (2nd), I (4th), I (10th), A (19th).
2. Only one consonant has a vowel on both sides: E S I (S is the 3rd letter).
3. Around the other vowels the neighbours are consonants (e.g. R I Y, S A P), so no other case.

Trick: Search for the pattern ‘vowel–consonant–vowel’ – it needs two vowels just two places apart.

Answer: (D) One

Concept: Find each vowel and check its left and right neighbours.

1. Vowels: I (3rd), E (6th), U (13th), A (19th), U (20th).
2. R I D ✓, P E Q ✓, Q U Y ✓ – consonants on both sides.
3. H A U ✗ (followed by a vowel) and A U Z ✗ (preceded by a vowel).
4. Count = 3.

Trick: Two vowels side by side (AU) can never count – skip them quickly.

Answer: (B) Three

Concept: Compare the shift from each letter of the first cluster to the matching letter of the second (A = 1 … Z = 26).

1. DP – ZL: $D-4=Z$ (looping back), $P-4=L$.
2. KS – GO: $K-4=G,S-4=O$. YH – UD: $Y-4=U,H-4=D$.
3. RM – IN: $R-9=I$ but $M+1=N$ → different rule.

Trick: Learn reverse jumps: D−4 = Z (wrap round the alphabet).

Answer: (B) RM – IN

Concept: Number the letters of the first word 1 2 3 4 and track their positions.

1. FORT → RTFO = letters 3 4 1 2; third word TROF = FORT written backwards (4 3 2 1).
2. SIGN → GNSI (3 4 1 2) → NGIS (reverse) ✓.
3. RAFT → FTRA (3 4 1 2) → TFAR (reverse) ✓.
4. DIRT would give RTDI; DOVE gives VEDO; FILE gives LEFI – but its reverse is ELIF, not ELFI.

Trick: Middle = swap the two halves; last = full reverse of the first word.

Answer: (A) RAFT - FTRA - TFAR

Concept: Find the gaps between consecutive letters using their positions.

1. KMP: $K\stackrel{+2}{\to }M\stackrel{+3}{\to }P$; QSV: $+2,+3$; FHK: $+2,+3$.
2. DFJ: $D\stackrel{+2}{\to }F\stackrel{+4}{\to }J$ → the second gap is 4, not 3.

Trick: Write the gaps as a pair (+2, +3) – the different pair is the odd one.

Answer: (D) DFJ

Concept: Number the letters 1 2 3 4 and see where each moves.

1. WORD → RWOD = letters 3 1 2 4; DROW = WORD reversed (4 3 2 1).
2. JAMS → MJAS (3 1 2 4) → SMAJ (reverse) ✓.
3. YEAR → AYER (3 1 2 4) → RAEY (reverse) ✓.
4. AGES should give EAGS; VAST gives SVAT but its reverse is TSAV; POLE gives LPOE.

Trick: Step 1: 3rd letter jumps to the front; Step 2: write the original word backwards.

Answer: (C) YEAR - AYER - RAEY