Q&A: practice questions with answers

Concept: $\sin \theta =\cos ({90}^{\circ }-\theta )$.

1. $({50}^{\circ }+A)+({40}^{\circ }-A)={90}^{\circ }$ – the angles are complementary.
2. So $\sin ({50}^{\circ }+A)=\cos ({90}^{\circ }-{50}^{\circ }-A)=\cos ({40}^{\circ }-A)$.
3. Difference $=0$

Trick: Put $A=0$: $\sin {50}^{\circ }-\cos {40}^{\circ }=0$.

Answer: (B) 0

Concept: In simple interest, interest is the same every year; the difference of amounts of consecutive years = one year's interest.

1. 1 year SI $=8120-7656=$ ₹464.
2. 4 years SI $=4\times 464=1856$ ⇒ Principal $=7656-1856=$ ₹5,800.
3. $R=\frac{464}{5800}\times 100=8\%$

Trick: $R=\frac{\text{one-year SI}\times 100}{P}$; check: 8% of 5800 = 464.

Answer: (D) 8%

Concept: ${(a+\frac{1}{a})}^2={(a-\frac{1}{a})}^2+4$.

1. ${(a+\frac{1}{a})}^2=4^2+4=20$
2. $a+\frac{1}{a}=\sqrt{20}=2\sqrt{5}$

Trick: "minus" known ⇒ add 4 after squaring; "plus" known ⇒ subtract 4.

Answer: (C) $2\sqrt{5}$

Concept: Add one-day work of all persons; time $=\frac{1}{\text{combined rate}}$.

1. One-day work $=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
2. $=\frac{3+2+1}{6}=1$
3. Time $=1$ day

Trick: Total work = LCM(2, 3, 6) = 6 units; rates 3 + 2 + 1 = 6 units/day ⇒ 1 day.

Answer: (B) 1

Concept: Equivalent single discount $=a+b-\frac{ab}{100}$.

1. $30+20-\frac{30\times 20}{100}$
2. $=50-6=44\%$

Trick: On ₹100: after 30% → 70; after 20% → 56; total discount $100-56=44\%$.

Answer: (A) 44%

Concept: Factorise each polynomial; LCM = product of all factors with highest powers.

1. $x^2-8x+15=(x-3)(x-5)$
2. $x^2-5x+6=(x-2)(x-3)$
3. $\text{LCM}=(x-5)(x-2)(x-3)$

Trick: Common factor $(x-3)$ is taken only once – so option A with $(x-3{)}^2$ is wrong.

Answer: (D) $(x-5)(x-2)(x-3)$

Concept: Correct average = old average $+\frac{\text{correct}-\text{wrong}}{n}$.

1. Error $=65-56=9$ (sum was 9 less).
2. Change in average $=\frac{9}{45}=0.2$
3. Correct average $=39+0.2=39.2$

Trick: Sum was under-counted, so the average must go UP – only 39.2 is more than 39.

Answer: (A) 39.2

Concept: Time and efficiency are inversely proportional.

1. Let Robert take $6t$ hours ⇒ Tom $3t$, George $2t$.
2. Efficiencies (per hour) Robert : Tom : George $=1:2:3$, total 6 units/hr.
3. Work $=6\times 23=138$ units; Robert alone $=138$ hours $=6t$ ⇒ $t=23$.
4. Tom $=3t=69$ hours

Trick: Tom does 2 of the 6 units each hour, i.e. $\frac{1}{3}$ of the team ⇒ Tom alone takes $3\times 23=69$ hours.

Answer: (B) 69 hours

Concept: Volume of a sphere $\propto r^3$ (or $d^3$).

1. $d_P=4d_Q\Rightarrow r_P:r_Q=4:1$
2. $V_P:V_Q=4^3:1^3=64:1$

Trick: Length ratio $k$ ⇒ area ratio $k^2$ ⇒ volume ratio $k^3$; 16 : 1 would be the surface-area ratio.

Answer: (D) 64 : 1

Concept: Speed = distance ÷ time; m/s × $\frac{18}{5}$ = km/h.

1. Speed $=\frac{750}{54}=\frac{125}{9}\text{m/s}$
2. $\frac{125}{9}\times \frac{18}{5}=50\text{km/h}$

Trick: $\frac{750}{54}\times 3.6=\frac{2700}{54}=50$.

Answer: (D) 50

Concept: The radius to the point of contact is perpendicular to the tangent, and the perpendicular from the centre bisects a chord.

1. OM ⊥ PQ, so $PM=MQ=5\text{cm}$.
2. In right $△OMQ$: $OQ=6$ (outer radius).
3. $r=OM=\sqrt{6^2-5^2}=\sqrt{11}\text{cm}$

Trick: $R^2=r^2+{\left(\frac{\text{chord}}{2}\right)}^2$ for a chord of the outer circle touching the inner one.

Answer: (D) $\sqrt{11}$

Concept: Order: Brackets → "of" → Division → Multiplication → Addition → Subtraction.

1. Bracket: $6-7=-1$; "of": $3\text{ of }24=72$, $3\text{ of }5=15$.
2. $9\times 2÷72=\frac{18}{72}=\frac{1}{4}$
3. $3÷2\times (-1)\times 2=-3$; $3÷15=\frac{1}{5}$
4. $\frac{1}{4}-(-3)+\frac{1}{5}=3+\frac{9}{20}=\frac{69}{20}$

Trick: "of" is solved before ÷ – treat "3 of 24" as one number, 72.

Answer: (D) $\frac{69}{20}$

Concept: Area of a quadrilateral $=\frac{1}{2}\times \text{diagonal}\times (h_1+h_2)$, where $h_1,h_2$ are the perpendiculars from the opposite vertices on that diagonal.

1. Diagonal $=20\text{m}$, $h_1=12$, $h_2=18$.
2. Area $=\frac{1}{2}\times 20\times (12+18)=300{\text{m}}^2$

Trick: The diagonal splits it into two triangles: $\frac{1}{2}\times 20\times 12+\frac{1}{2}\times 20\times 18=120+180$.

Answer: (D) 300 m²

Concept: People owning a car = "Car" only + "Both"; the rest do not own a car.

1. Own a car $=(25+10)+(15+5)=55$
2. Do not own a car $=200-55=145$
3. $\frac{145}{200}\times 100=72.5\%$

Trick: Do not forget the "Both" row – it also owns a car. Ignoring it gives 82.5% (not an option) and counting only "no vehicle" gives 47.5%.

Answer: (D) 72.5%

Concept: Required % $=\frac{\text{part}}{\text{row total}}\times 100$.

1. School U total $=150+160+162+160+161+140=933$
2. Class VII of U $=162$
3. $\frac{162}{933}\times 100\approx 17.36\%$

Trick: 162 is a little more than one-sixth of 933 (155.5), so the answer must be a little above 16.67% – only 17.36% fits.

Answer: (D) 17.36%

Concept: Scan the whole table for the smallest entry.

1. Options: A-2010 = 560000, B-2015 = 550000, B-2010 = 540000, A-2011 = 580000.
2. The smallest value in the entire table is 540000 (B in 2010).

Trick: Compare only the first three digits: 540 is the least.

Answer: (C) B, 2010

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

1. $(\sin \theta +\cos \theta {)}^2=\frac{11}{9}$
2. $(\sin \theta -\cos \theta {)}^2=2-\frac{11}{9}=\frac{7}{9}$
3. $\sin \theta -\cos \theta =\frac{\sqrt{7}}{3}$

Trick: Numerators of the two squares add to $2\times 9=18$: $11+7=18$ ⇒ $\frac{\sqrt{7}}{3}$.

Answer: (B) $\frac{\sqrt{7}}{3}$

Concept: Add the row of the required student.

1. Ayush: $88+80+90+82+85$
2. $=168+172+85=425$

Trick: Take 85 as base: $3-5+5-3+0=0$ ⇒ total $=5\times 85=425$.

Answer: (D) 425

Concept: Expenditure = Income − Savings; split the expenditure step by step.

1. Savings $=10\%$ of 75000 = ₹7,500 ⇒ expenditure = ₹67,500.
2. Education $=30\%$ of 67500 = ₹20,250.
3. Food $=67500-20250-30375=$ ₹16,875.
4. $\frac{16875}{75000}\times 100=22.5\%$

Trick: Education is 30% of expenditure, not of income – using income gives a wrong value.

Answer: (B) 22.5%

Concept: Compare the column totals of the organisations given in the options.

1. A $=1050+1017+1382+1542+786+48=5825$
2. B = 5703, D = 5613, E = 5625.
3. Among the options, A has the maximum (5825).

Trick: Read the "Total" row directly – 5825 is the largest among A, B, D, E.

Answer: (A) A