Q&A: practice questions with answers

Concept: In any triangle, the straight side between two vertices is the shortest path, so going around through the third vertex is always longer.

1. For sides $a,b,c$: $a+b>c$, $b+c>a$, $c+a>b$.
2. Equality $a+b=c$ would flatten the triangle into a straight line, so it is not allowed.
3. Hence the sum of any two sides is always greater than the third side.

Trick: Sum of two sides $>$ third side $>$ difference of the two sides.

Answer: (B) greater than the third side

Concept: Marks in a subject $=\frac{\text{percentage}}{100}\times \text{maximum marks}$; add over the five subjects.

1. Hindi: $82\%\times 150=123$; English: $64\%\times 100=64$
2. Geography: $56\%\times 175=98$; Mathematics: $70\%\times 120=84$
3. Physics: $80\%\times 75=60$
4. Aggregate $=123+64+98+84+60=429$

Trick: Use friendly forms: $56\%$ of $175=0.56\times 175=98$, $80\%$ of $75=60$ — no long multiplication needed.

Answer: (B) 429

Concept: When a solid is melted and recast, volume is conserved. Flat surfaces: cylinder has two circles, each cone has one circular base.

1. Cylinder volume $=\pi \times 7^2\times 12=588\pi$; its flat area $=2\pi \times 7^2=98\pi$.
2. Cone 1: $\frac{1}{3}\pi r_1^2\times 12=\frac{1}{3}\times 588\pi \Rightarrow r_1^2=49$.
3. Cone 2: $\frac{1}{3}\pi r_2^2\times 12=\frac{2}{3}\times 588\pi \Rightarrow r_2^2=98$.
4. Flat area of cones $=\pi (49+98)=147\pi$.
5. Change $=\frac{147\pi -98\pi }{98\pi }\times 100=50\%$ (increase).

Trick: With the same height, a cone needs 3 times the base area of a cylinder of equal volume: $3\times 49\pi =147\pi$ vs $98\pi$, i.e. $\frac{3}{2}$ times.

Answer: (C) 50%

Concept: Perimeter of a triangle = sum of its three sides.

1. Sides: 13 cm, 12 cm, 5 cm.
2. Perimeter $=13+12+5=30\text{cm}$.
3. (Since $5^2+{12}^2={13}^2$, it is a right triangle — a useful check.)

Yaad rakhein: 5–12–13 is a Pythagorean triplet; its perimeter is always 30 units.

Answer: (C) 30 cm

Concept: BODMAS: brackets first; inside, 'of' acts as multiplication; then ÷ and × from left to right, then + and −.

1. $(4.2\times 2.5÷3.5)=10.5÷3.5=3$
2. $(3.6÷2.4+2.5\times 3÷7.5)=1.5+1=2.5$
3. $(6-3\times 2+4\times 3÷6)=6-6+2=2$
4. Total $=3-2.5+2=2.5$

Trick: Look for neat pairs: $10.5÷3.5=3$, $7.5÷7.5=1$ — decimals cancel quickly.

Answer: (D) 2.5

Concept: Area of trapezium $=\frac{1}{2}\times (\text{sum of parallel sides})\times \text{height}$.

1. $1188=\frac{1}{2}\times (a+b)\times 18$
2. $1188=9(a+b)$
3. $a+b=\frac{1188}{9}=132\text{cm}$

Trick: Divide area by half the height directly: $1188÷9=132$.

Answer: (C) 132 cm

Concept: Use known values: $\sin {90}^{\circ }=1$ and $\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$.

1. $\sin (x+y)=1\Rightarrow x+y={90}^{\circ }$
2. $\cos (x-y)=\frac{\sqrt{3}}{2}\Rightarrow x-y={30}^{\circ }$
3. Subtracting: $2y={60}^{\circ }\Rightarrow y={30}^{\circ }$ (and $x={60}^{\circ }$).

Trick: Sum and difference given → $y=\frac{\text{sum}-\text{difference}}{2}=\frac{{90}^{\circ }-{30}^{\circ }}{2}$.

Answer: (B) ${30}^{\circ }$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$ (not the average of the speeds).

1. Distance 1 $=8\times 4=32\text{km}$; distance 2 $=4\times 8=32\text{km}$
2. Total distance $=64\text{km}$, total time $=8+4=12$ h
3. Average speed $=\frac{64}{12}=\frac{16}{3}=5\frac{1}{3}\text{km/h}$

Trick: When times are given, average speed $=\frac{t_1{s}_1+t_2{s}_2}{t_1+t_2}=\frac{32+32}{12}$.

Answer: (B) $5\frac{1}{3}$

Concept: Required $\%=\frac{\text{Pune (2018)}}{\text{total of all towns (2018)}}\times 100$.

1. Total 2018 $=874+850+179+857+820=3580$
2. Pune 2018 $=179$
3. $\frac{179}{3580}\times 100=5\%$

Trick: Check with options: $5\%$ of $3580=179$ exactly.

Answer: (D) 5%

Concept: Remainder theorem: remainder of $f(x)$ divided by $(x-2)$ is $f(2)$.

1. $f(2)=2(8)+4a+6-5=17+4a$
2. $g(2)=8+4-4+a=8+a$
3. Equal remainders: $17+4a=8+a\Rightarrow a=-3$
4. Remainder $=8+a=8-3=5$

Trick: Put $x=2$ in both, equate, solve for $a$, then substitute back into the simpler one.

Answer: (D) 5

Concept: Simple interest $=\frac{P\times R\times T}{100}$; the difference of 8 years and 4 years is the SI of 4 years.

1. Extra years $=8-4=4$
2. $\frac{P\times 14\times 4}{100}=6160$
3. $0.56P=6160\Rightarrow P=11000$

Trick: 4 years at $14\%$ = $56\%$ of sum = ₹6,160 → sum $=\frac{6160}{56}\times 100$.

Answer: (C) ₹11,000

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

1. $ab=a\times \frac{4}{a}=4$
2. Mean proportional $=\sqrt{4}=2$ (positive since $a>0$).

Yaad rakhein: If $b=\frac{k}{a}$, the mean proportional is always $\sqrt{k}$.

Answer: (C) 2

Concept: Use BODMAS separately in numerator and denominator; 'of' is done before ÷.

1. Numerator: $7-9=-2$; $6÷(-2)=-3$; $-3\times 8=-24$
2. $\{6-(-24)+9\}=39$; $13÷39=\frac{1}{3}$
3. Denominator: $5\text{ of }5=25$; $5\times 5÷25=1$; $5+1=6$
4. Value $=\frac{1/3}{6}=\frac{1}{18}$

Trick: 'of' binds tighter than ÷: $5\times 5÷5\text{ of }5=25÷25$, not $25÷5\times 5$.

Answer: (A) $\frac{1}{18}$

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

1. $\frac{a}{b}+\frac{b}{a}=1\Rightarrow a^2+b^2=ab$
2. So $a^2-ab+b^2=0$
3. $a^3+b^3=(a+b)\times 0=0$
4. Hence $\frac{a^3+b^3}{3ab}=0$

Yaad rakhein: $\frac{a}{b}+\frac{b}{a}=1$ always gives $a^3+b^3=0$ (i.e. $a^3=-b^3$).

Answer: (D) 0

Concept: Solve innermost brackets first, then ÷, ×, +, − (BODMAS).

1. $4÷\frac{1}{4}=16$; $16-4=12$; $2\times 12=24$
2. $81÷9=9$ (twice)
3. Inside $[]$: $9+3-24+9+9+3-9=0$
4. Value $=0+9=9$

Trick: Dividing by $\frac{1}{4}$ = multiplying by 4.

Answer: (B) 9

Concept: Take total votes polled = 100 and track each share.

1. Invalid $=\frac{1}{10}\times 100=10$
2. R and S together $=50$, so R $=25$, S $=25$ (P gets $40$)
3. Fault: R gets $\frac{1}{2}$ of S $=12.5$ and $\frac{1}{2}$ of invalid $=5$
4. R's votes $=25+12.5+5=42.5$, i.e. $42.5\%$ of votes polled

Trick: Assume total = 100 — the final count is directly the percentage.

Answer: (A) 42.5%

Concept: Write $12=x+1$ (since $x=11$) so the terms telescope.

1. $x^5-(x+1)x^4+(x+1)x^3-(x+1)x^2+(x+1)x-1$
2. $=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-1$
3. $=x-1=11-1=10$

Trick: When a coefficient is $x+1$, rewrite it — almost every term cancels.

Answer: (D) 10

Concept: Joining the centres gives an equilateral triangle; the enclosed region = triangle − three ${60}^{\circ }$ sectors.

1. Side $=5+5=10$ cm; area $=\frac{\sqrt{3}}{4}\times {10}^2=25\sqrt{3}$
2. Three sectors of ${60}^{\circ }$ = half a circle $=\frac{1}{2}\pi \times 5^2=\frac{25\pi }{2}$
3. Required area $=25\sqrt{3}-\frac{25\pi }{2}=25(\sqrt{3}-\frac{\pi }{2})$

Yaad rakhein: Three ${60}^{\circ }$ sectors always add to ${180}^{\circ }$ = half circle.

Answer: (D) $25(\sqrt{3}-\frac{\pi }{2})$

Concept: Use $2{\cos }^2\theta -1=\cos 2\theta$, $1-2{\sin }^2\theta =\cos 2\theta$ and $1+{\cot }^2\theta ={\text{cosec}}^2\theta$.

1. Numerator $=\cos \theta (2{\cos }^2\theta -1)=\cos \theta \cos 2\theta$
2. Denominator $=\sin \theta (1-2{\sin }^2\theta )=\sin \theta \cos 2\theta$
3. Ratio $=\cot \theta$ ($\cos 2\theta \ne 0$ as $\theta \ne {45}^{\circ }$)
4. ${\cot }^2\theta +1={\text{cosec}}^2\theta$

Trick: Put $\theta ={30}^{\circ }$: ratio $=\cot {30}^{\circ }=\sqrt{3}$, so $3+1=4={\text{cosec}}^2{30}^{\circ }$.

Answer: (C) ${\text{cosec}}^2\theta$

Concept: If price rises by $r\%$, consumption must fall by $\frac{r}{100+r}\times 100\%$ to keep expenditure same.

1. $r=65$
2. Reduction $=\frac{65}{165}\times 100$
3. $=\frac{13}{33}\times 100=39.39\%$

Trick: Expenditure = price × consumption; $1.65\times \frac{1}{1.65}=1$, so cut $=1-\frac{1}{1.65}$.

Answer: (C) 39.39%