Q&A: practice questions with answers

Concept: When one object chases another, time to catch up $=\frac{\text{gap}}{\text{relative speed}}$.

1. By 4 p.m. the first car is $50\times 1=50\text{km}$ ahead.
2. Relative speed $=75-50=25\text{km/h}$.
3. Time to catch up $=\frac{50}{25}=2\text{h}$, so they meet at $4+2=6$ p.m.

Trick: Head start ÷ speed difference gives the catch-up time; add it to the start time of the second car.

Answer: (A) 6 p.m.

Concept: Area of a trapezium $=\frac{1}{2}\times (a+b)\times h$.

1. $1785=\frac{1}{2}\times (42+b)\times 35$.
2. $42+b=\frac{1785\times 2}{35}=102$.
3. $b=102-42=60\text{ft}$.

Trick: Sum of parallel sides $=\frac{2\times \text{Area}}{h}$, then subtract the known side.

Answer: (A) $60$ feet

Concept: If numbers are $3x,7x,11x$ (pairwise co-prime ratio terms), their LCM $=3\times 7\times 11\times x$.

1. LCM $=231x=1386\Rightarrow x=6$.
2. Numbers: $18,42,66$.
3. Greatest + least $=66+18=84$.

Trick: Directly: $(3+11)\times 6=84$.

Answer: (C) $84$

Concept: Divide the equation by ${\sin }^2\theta$ and use ${\csc }^2\theta =1+{\cot }^2\theta$.

1. $\frac{1}{{\sin }^2\theta }+1-3\cot \theta =0\Rightarrow {\csc }^2\theta +1-3\cot \theta =0$.
2. $1+{\cot }^2\theta +1-3\cot \theta =0\Rightarrow {\cot }^2\theta -3\cot \theta +2=0$.
3. $(\cot \theta -1)(\cot \theta -2)=0\Rightarrow \cot \theta =1$ or $2$.
4. Only $2$ is among the options.

Trick: Check: $\cot \theta =2\Rightarrow {\sin }^2\theta =\frac{1}{5}$, $\sin \theta \cos \theta =\frac{2}{5}$; $1+\frac{1}{5}-\frac{6}{5}=0$ ✓.

Answer: (B) $2$

Concept: $a^2-b^2=(a+b)(a-b)$ and ${\cos }^2\alpha +{\sin }^2\alpha =1$.

1. ${\cos }^4\alpha -{\sin }^4\alpha =({\cos }^2\alpha +{\sin }^2\alpha )({\cos }^2\alpha -{\sin }^2\alpha )={\cos }^2\alpha -{\sin }^2\alpha$.
2. ${\cos }^2\alpha -{\sin }^2\alpha ={\cos }^2\alpha -(1-{\cos }^2\alpha )=2{\cos }^2\alpha -1$.
3. So $2{\cos }^2\alpha -1=\frac{5}{6}$.

Trick: ${\cos }^4\alpha -{\sin }^4\alpha =\cos 2\alpha =2{\cos }^2\alpha -1$ – the value is the same.

Answer: (B) $\frac{5}{6}$

Concept: Average $=\frac{\text{sum}}{\text{number of years}}$; values are in hundreds.

1. Units sold (S): $1.7+1.5+1.0+1.4+2.1=7.7$ hundred.
2. Average $=\frac{7.7}{5}=1.54$ hundred.
3. $1.54\times 100=154$ units.

Trick: Don't forget the unit – multiply the final average by $100$.

Answer: (C) $154$

Concept: A number is divisible by $99$ if the sum of its two-digit pairs (from the right) is divisible by $99$.

1. $55p1067q9$ → pairs from right: $q9,67,10,5p,5$.
2. Sum $=(10q+9)+67+10+(50+p)+5=141+10q+p$.
3. $141+10q+p$ must be $198\Rightarrow 10q+p=57\Rightarrow q=5,p=7$.
4. $pq=7\times 5=35$ (check: digit sum $=45$, divisible by $9$ ✓).

Trick: $99=9\times 11$: pair-sum method checks both at once.

Answer: (A) $35$

Concept: Perpendicular from the centre bisects a chord: $d=\sqrt{r^2-{\left(\frac{l}{2}\right)}^2}$.

1. Half chord $=4\text{m}$, radius $=5\text{m}$.
2. Distance of each chord from centre $=\sqrt{5^2-4^2}=3\text{m}$.
3. Equal parallel chords lie on opposite sides of the centre, so distance $=3+3=6\text{m}$.

Trick: $3$-$4$-$5$ triplet: half chord $4$, radius $5$ → distance $3$.

Answer: (B) $6\text{m}$

Concept: In BODMAS, 'of' is done before division and multiplication.

1. $5\text{ of }12=60$, $27\text{ of }3=81$.
2. $25÷60\times 2=\frac{5}{6}$; $9÷81\times 5=\frac{5}{9}$; $14÷28\times 4=2$.
3. $\frac{5}{6}-\frac{5}{9}-2=\frac{15-10-36}{18}=-\frac{31}{18}$.

Trick: Solve every 'of' first, then go left to right for ÷ and ×.

Answer: (D) $-\frac{31}{18}$

Concept: If $2\angle A=3\angle B=6\angle C=k$, the angles are $\frac{k}{2},\frac{k}{3},\frac{k}{6}$; their sum is ${180}^{\circ }$.

1. $\frac{k}{2}+\frac{k}{3}+\frac{k}{6}=k={180}^{\circ }$.
2. $\angle A={90}^{\circ }$, $\angle B={60}^{\circ }$, $\angle C={30}^{\circ }$.
3. Largest angle $={90}^{\circ }$.

Trick: Ratio $A:B:C=\frac{1}{2}:\frac{1}{3}:\frac{1}{6}=3:2:1$ → largest $=\frac{3}{6}\times {180}^{\circ }={90}^{\circ }$.

Answer: (B) ${90}^{\circ }$

Concept: Net change for $+a\%$ then $-b\%$ $=a-b-\frac{ab}{100}$.

1. $15-10-\frac{15\times 10}{100}=5-1.5=3.5\%$.
2. Check: $100\to 115\to 115\times 0.9=103.5$, i.e. $3.5\%$ increase.

Trick: Multiply factors: $1.15\times 0.90=1.035$.

Answer: (C) $3.5\%$

Concept: In an isosceles right triangle the legs are equal and hypotenuse $=a\sqrt{2}$.

1. $a^2+a^2={10}^2\Rightarrow 2a^2=100$.
2. $a^2=50\Rightarrow a=5\sqrt{2}\text{cm}$.

Trick: Leg $=\frac{\text{hypotenuse}}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}$.

Answer: (D) $5\sqrt{2}$ and $5\sqrt{2}$

Concept: Required $\%=\frac{\text{E in March}}{\text{Total in May}}\times 100$.

1. Total in May $=270+390+280+250+350=1540$ kg.
2. E in March $=308$ kg.
3. $\frac{308}{1540}\times 100=20\%$.

Trick: $1540÷5=308$, so it is exactly one-fifth $=20\%$.

Answer: (A) $20\%$

Concept: Compare $\frac{E}{\text{Total}}$ for each year.

1. 2007: $\frac{9}{54}=16.7\%$; 2008: $\frac{12}{57}=21.1\%$; 2009: $\frac{18}{84}=21.4\%$.
2. 2010: $\frac{12}{67}=17.9\%$; 2011: $\frac{14}{71}=19.7\%$; 2012: $\frac{15}{72}=20.8\%$.
3. Minimum share is in 2007.

Trick: $\frac{9}{54}=\frac{1}{6}$ is the only ratio below $\frac{1}{5.5}$ – spot it quickly.

Answer: (B) 2007

Concept: TSA of a cone $=\pi rl+\pi r^2$ and CSA $=\pi rl$; difference $=\pi r^2$ (base area).

1. Difference $=\pi r^2=\frac{22}{7}\times 35\times 35$.
2. $=22\times 5\times 35=3850{\text{cm}}^2$.

Trick: TSA − CSA of any cone = area of its circular base.

Answer: (A) $3850{\text{cm}}^2$

Concept: Take CP $=100$ for all goods; MP $=140$. Find total SP of the three parts.

1. $\frac{1}{5}$ at $10\%$ off: $\frac{1}{5}\times 126=25.2$.
2. $\frac{1}{2}$ at MP: $\frac{1}{2}\times 140=70$.
3. Rest $\frac{3}{10}$ at $20\%$ off: $\frac{3}{10}\times 112=33.6$.
4. Total SP $=128.8$ → profit $=28.8\%$.

Trick: With CP $=100$, total SP minus $100$ is directly the profit $\%$.

Answer: (C) $28.8\%$

Concept: Base area $=\pi r^2$, CSA $=\pi rl$, $h=\sqrt{l^2-r^2}$, $V=\frac{1}{3}\pi r^{2}h$.

1. $\pi r^2=154\Rightarrow r^2=49\Rightarrow r=7\text{cm}$.
2. $\pi rl=550\Rightarrow 22l=550\Rightarrow l=25\text{cm}$.
3. $h=\sqrt{{25}^2-7^2}=24\text{cm}$.
4. $V=\frac{1}{3}\times 154\times 24=1232{\text{cm}}^3$.

Trick: $7$-$24$-$25$ is a Pythagorean triplet.

Answer: (A) $1232{\text{cm}}^3$

Concept: Somya's 1-day work $=\text{(together)}-\text{(Ankita)}$.

1. $\frac{1}{12}-\frac{1}{20}=\frac{5-3}{60}=\frac{2}{60}=\frac{1}{30}$.
2. Somya alone takes $30$ days.

Trick: LCM method: work $=60$ units; together $5$/day, Ankita $3$/day → Somya $2$/day → $\frac{60}{2}=30$.

Answer: (D) $30$

Concept: Standard values: $\tan {45}^{\circ }=1$, $\sec {60}^{\circ }=\frac{1}{\cos {60}^{\circ }}=2$.

1. $x=\tan {45}^{\circ }+\sec {60}^{\circ }=1+2$.
2. $x=3$.

Trick: $\cos {60}^{\circ }=\frac{1}{2}$, so $\sec {60}^{\circ }=2$.

Answer: (C) $3$

Concept: Score $=5\times (\text{correct})-2\times (\text{incorrect})$.

1. Marks from correct answers $=4\times 5=20$.
2. $20-2x=-12\Rightarrow 2x=32$.
3. $x=16$ incorrect answers.

Trick: Deduction needed $=20-(-12)=32$; $\frac{32}{2}=16$.

Answer: (C) $16$