Q&A: practice questions with answers

Concept: When the options give the values, substitute each option into all three equations – the one satisfying all is the answer.

1. Try $(a,b,c)=(1,4,2)$: $3(1)-4(4)+2(2)=3-16+4=-9$ ✓
2. $-4(1)+4(4)+10(2)=-4+16+20=32$ ✓
3. $-(1)+2(4)-7(2)=-1+8-14=-7$ ✓ – all three hold, so $(1,4,2)$ is the solution.

Trick: Check the shortest equation first ($-a+2b-7c=-7$): (3,2,2) gives $-13$, (2,1,4) gives $-28$, (5,1,1) gives $-10$ – only (1,4,2) survives; then confirm with another equation.

Answer: (A) (1, 4, 2)

Concept: A number is divisible by 3 when the sum of its digits is divisible by 3.

1. Digit sum of 713A8845 $=7+1+3+A+8+8+4+5=36+A$.
2. $36$ is already a multiple of 3, so $A$ must be a multiple of 3: $A=0,3,6$ or $9$.
3. The maximum single digit is $A=9$ (sum $=45$, divisible by 3).

Trick: If the known digits already sum to a multiple of 3, the missing digit can be 0, 3, 6 or 9 – take 9 for 'maximum'.

Answer: (A) 9

Concept: Volume of a cone $=\frac{1}{3}\pi r^{2}h$; then account for the spillage ratio.

1. $V=\frac{1}{3}\times \frac{22}{7}\times 6^2\times 7=\frac{1}{3}\times 22\times 36=264\text{cc}$.
2. For every $180\text{cc}$ that goes in, $20\text{cc}$ spills, so $200\text{cc}$ must be poured to retain $180\text{cc}$.
3. Water required $=264\times \frac{200}{180}=264\times \frac{10}{9}=293.33\text{cc}$.

Trick: Spillage of 20 per 180 means pour $\frac{10}{9}$ of the volume: $264\times \frac{10}{9}=293.33$.

Answer: (D) 293.33

Concept: Write every base as the same fraction: ${\left(\frac{10}{13}\right)}^n={\left(\frac{13}{10}\right)}^{-n}$, then equate the powers.

1. LHS $={\left(\frac{13}{10}\right)}^4\times {\left(\frac{13}{10}\right)}^{-16}\times {\left(\frac{13}{10}\right)}^{17}={\left(\frac{13}{10}\right)}^{4-16+17}={\left(\frac{13}{10}\right)}^5$.
2. RHS $={\left(\frac{10}{13}\right)}^{3m+18}={\left(\frac{13}{10}\right)}^{-(3m+18)}$.
3. So $-(3m+18)=5\Rightarrow 3m=-23\Rightarrow m=-\frac{23}{3}$.

Trick: Flipping a fraction changes the sign of the power; collect all powers on one base before equating.

Answer: (D) $-\frac{23}{3}$

Concept: Use $1+{\cot }^{2}A={\csc }^{2}A$ and $(1-\sec A)(1+\sec A)=1-{\sec }^{2}A=-{\tan }^{2}A$.

1. $\csc A=\frac{5}{3}\Rightarrow \sin A=\frac{3}{5},\cos A=\frac{4}{5},\tan A=\frac{3}{4}$.
2. Numerator $=9{\csc }^{2}A=9\times \frac{25}{9}=25$.
3. Denominator $=80\times (1-{\sec }^{2}A)=80\times (-{\tan }^{2}A)=80\times (-\frac{9}{16})=-45$.
4. Value $=\frac{25}{-45}=-\frac{5}{9}$.

Trick: The 3-4-5 triangle does it all: $\csc =\frac{5}{3}$ means opposite 3, hypotenuse 5, adjacent 4. $1-{\sec }^2$ is always negative, so look for a negative option.

Answer: (D) $\frac{-5}{9}$

Concept: Equal perimeters give the hexagon's side; area of a regular hexagon $=\frac{3\sqrt{3}}{2}{a}^2$.

1. Side of square $=\sqrt{8649}=93\text{cm}$; perimeter $=4\times 93=372\text{cm}$.
2. Hexagon side $a=\frac{372}{6}=62\text{cm}$.
3. Area $=\frac{3\sqrt{3}}{2}\times {62}^2=\frac{3\sqrt{3}}{2}\times 3844=5766\sqrt{3}{\text{cm}}^2$.

Trick: Hexagon side $=\frac{4}{6}=\frac{2}{3}$ of the square's side; a regular hexagon is 6 equilateral triangles, so area $=6\times \frac{\sqrt{3}}{4}{a}^2$.

Answer: (D) $5766\sqrt{3}$

Concept: Income − Savings = Expenditure; equate the expenditure ratio.

1. Let incomes be $6x$ and $7x$. Expenditures: $6x-69000$ and $7x-69000$.
2. $\frac{6x-69000}{7x-69000}=\frac{2}{3}\Rightarrow 18x-207000=14x-138000\Rightarrow 4x=69000\Rightarrow x=17250$.
3. Mohan's income $=6x=6\times 17250=$ ₹1,03,500.

Trick: Equal savings → ${\text{income}}_1-{\text{income}}_2={\text{exp}}_1-{\text{exp}}_2$: $x=\text{(one part of expenditure)}$; here $7x-6x=3k-2k$ gives $x=k$, so $6k-69000=2k\Rightarrow k=17250$.

Answer: (A) 1,03,500

Concept: For numbers $a$ and $b$: mean proportional $=\sqrt{ab}$, third proportional $=\frac{b^2}{a}$.

1. $\sqrt{ab}=14\Rightarrow ab=196$.
2. $\frac{b^2}{a}=112\Rightarrow b^2=112a$. Multiply by $b$: $b^3=112ab=112\times 196=21952={28}^3\Rightarrow b=28$.
3. $a=\frac{196}{28}=7$. Numbers: 7 and 28.

Trick: Test options: product must be 196 → only (14,14) and (7,28); third proportional of 7, 28 $=\frac{{28}^2}{7}=112$ ✓.

Answer: (B) 7, 28

Concept: Circumference $=\pi d=2\pi r$.

1. $d=14\text{cm}$, so circumference $=\frac{22}{7}\times 14$.
2. $=22\times 2=44\text{cm}$.

Trick: With $\pi =\frac{22}{7}$, a diameter that is a multiple of 7 cancels neatly: $14\to 44$, $7\to 22$, $21\to 66$.

Answer: (B) 44 cm

Concept: Do division and multiplication before addition and subtraction; mind the signs.

1. $(-56)÷(-14)=4$ and $(-2)\times 9=-18$.
2. Expression $=(-9)-4+(-18)=-9-4-18=-31$.

Trick: Negative ÷ negative = positive; then '− (positive)' subtracts. Keep a running total: $-9\to -13\to -31$.

Answer: (B) −31

Concept: Angle sum of a quadrilateral $={360}^{\circ }$; if a pair of co-interior (adjacent) angles adds to ${180}^{\circ }$, the sides containing them are parallel.

1. $4x+3x+7x+6x=20x={360}^{\circ }\Rightarrow x={18}^{\circ }$. Angles: $P={72}^{\circ }$, $Q={54}^{\circ }$, $R={126}^{\circ }$, $S={108}^{\circ }$.
2. $Q+R={54}^{\circ }+{126}^{\circ }={180}^{\circ }$ (and $P+S={180}^{\circ }$) → $PQ\parallel SR$. But $P+Q={126}^{\circ }\ne {180}^{\circ }$, so $PS$ is not parallel to $QR$.
3. Exactly one pair of opposite sides parallel, and opposite angles are unequal → a trapezium (not a parallelogram, rhombus or kite).

Trick: Opposite angles equal → parallelogram/rhombus; only one pair of adjacent angles supplementary → trapezium. Here 4 : 3 : 7 : 6 has unequal opposite terms.

Answer: (D) Trapezium

Concept: Same direction → relative speed = difference of speeds; distance to overtake = sum of lengths.

1. Relative speed $=72-54=18\text{km/h}=18\times \frac{5}{18}=5\text{m/s}$.
2. Distance $=150+120=270\text{m}$.
3. Time $=\frac{270}{5}=54\text{s}$.

Trick: km/h → m/s: multiply by $\frac{5}{18}$; 18 km/h is exactly 5 m/s – a value worth memorising.

Answer: (C) 54 seconds

Concept: A sum doubles under simple interest when the interest equals the principal, i.e. $\frac{P\times R\times T}{100}=P\Rightarrow R\times T=100$.

1. Here $T=18$ years, so $R=\frac{100}{18}$.
2. $\frac{100}{18}=5.555\dots \approx 5.56\%$ per year.

Trick: Doubling → $R\times T=100$; tripling → $R\times T=200$; $n$ times → $R\times T=100(n-1)$.

Answer: (A) $5.56$

Concept: Net work per hour = sum of filling rates − emptying rate.

1. Net rate $=\frac{1}{4}+\frac{1}{28}-\frac{1}{28}=\frac{1}{4}$ of the cistern per hour (the second tap and the emptying tap cancel).
2. Full cistern takes 4 hours, so one-fourth takes $4\times \frac{1}{4}=1$ hour.

Trick: Equal filling and emptying times cancel each other – spot this before doing LCM work.

Answer: (D) 1

Concept: Time is directly proportional to the number of pages.

1. Time per page $=\frac{40}{2}=20$ minutes.
2. For 89 pages: $89\times 20=1780$ minutes.

Trick: $89\times 20=(90-1)\times 20=1800-20=1780$.

Answer: (B) 1780

Concept: Apply the multipliers in sequence: $\text{Final price}=\text{CP}\times 1.5\times 0.68$.

1. Let P's cost price be $x$. Q pays $1.5x$; R pays $1.5x\times 0.68=1.02x$.
2. $1.02x=5406\Rightarrow x=\frac{5406}{1.02}=5300$.

Trick: Net change for +50% then −32% is $50-32-\frac{50\times 32}{100}=+2\%$, so $5406=102\%$ of CP → CP = 5300.

Answer: (A) 5300

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$.

1. Total $=53\times 424+86\times 337+57\times 122=22472+28982+6954=58408$.
2. Members $=53+86+57=196$.
3. Average $=\frac{58408}{196}=298$.

Trick: Use deviations from 300: $53(124)+86(37)+57(-178)=6572+3182-10146=-392$; $\frac{-392}{196}=-2$ → $300-2=298$.

Answer: (B) 298

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$.

1. $\sqrt{425\times 17}=\sqrt{25\times 17\times 17}$.
2. $=5\times 17=85$.

Trick: Factor instead of multiplying: $425=25\times 17$, so the root is $5\times 17$ directly.

Answer: (A) 85

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$, which gives $\text{Mode}-\text{Median}=\frac{2}{3}(\text{Mode}-\text{Mean})$.

1. Given $\text{Mode}-\text{Mean}=26.7$.
2. $\text{Mode}-\text{Median}=\frac{2}{3}\times 26.7=17.8$.

Trick: Mean, Median, Mode divide the gap in ratio 1 : 2 – Median is one-third of the way from Mean to Mode, so Mode − Median is two-thirds of Mode − Mean.

Answer: (C) 17.8

Concept: In $a:b::c:d$, $d=\frac{b\times c}{a}$.

1. $x=\frac{63\times 6.5}{7}=9\times 6.5=58.5$.
2. $y=\frac{42\times 4}{7}=6\times 4=24$.
3. $x:y=58.5:24=117:48=39:16$.

Trick: $7:63=1:9$ and $7:42=1:6$, so $x=9\times 6.5$, $y=6\times 4$ – no cross-multiplication needed.

Answer: (C) $39:16$