Q&A: practice questions with answers

Concept: Internal bisectors of $\angle B$ and $\angle C$ meet at O: $\angle BOC={90}^{\circ }+\frac{\angle A}{2}$.

1. ${90}^{\circ }+\frac{\angle A}{2}={134}^{\circ }$
2. $\frac{\angle A}{2}={44}^{\circ }\Rightarrow \angle A={88}^{\circ }$

Yaad rakhein: Internal bisectors → ${90}^{\circ }+\frac{A}{2}$; external bisectors → ${90}^{\circ }-\frac{A}{2}$.

Answer: (D) ${88}^{\circ }$

Concept: Doubling the speed halves the time for the same distance.

1. Let B take $t$ h; A takes $t+2.5$ h
2. With double speed A takes $\frac{t+2.5}{2}=t-1$
3. $t+2.5=2t-2\Rightarrow t=4.5$ h; A $=7$ h
4. Average $=\frac{7+4.5}{2}=5.75$ h $=5$ h 45 min

Trick: Check: A's half time 3.5 h is 1 h less than 4.5 h ✓.

Answer: (B) 5 hours 45 minutes

Concept: Increase in percentage points × value of one point.

1. 2005: strawberry + butterscotch $=18.9+10.8=29.7\%$
2. 2015: $20.5+16.6=37.1\%$
3. Shift $=37.1-29.7=7.4$ points
4. Increase $=7.4\times 5000=$ ₹37000

Trick: Add the individual changes: $1.6+5.8=7.4$.

Answer: (B) ₹37,000

Concept: "% less than" $=\frac{\text{difference}}{\text{base}}\times 100$, base = value after "than".

1. 165–170: 14 students; 150–155: 15 students
2. $\frac{15-14}{15}\times 100=\frac{100}{15}=6.67\%$
3. Correct to one decimal place: $6.7\%$

Trick: $\frac{1}{15}=6.67\%$ – worth remembering.

Answer: (B) $6.7\%$

Concept: Form an equation for $x$, then use % more $=\frac{\text{difference}}{\text{base}}\times 100$.

1. $110(1-\frac{x}{100})=50(1+\frac{x}{100})\Rightarrow 60=1.6x\Rightarrow x=37.5$
2. $37.5\%$ of 650 $=243.75$; $27.5\%$ of 780 $=214.5$
3. $\frac{243.75-214.5}{214.5}\times 100=\frac{29.25}{214.5}\times 100\approx 13.6\%\approx 14\%$

Trick: Decrease of 110 = increase of 50 → $1.6x=60$.

Answer: (C) $14\%$

Concept: Find the value of $1\%$ from the known sector, then scale.

1. Chocolate $25.8\%=$ ₹5160 → $1\%=\frac{5160}{25.8}=$ ₹200
2. Vanilla $16.5\%=16.5\times 200=$ ₹3300

Trick: Vanilla : Chocolate $=16.5:25.8$, so ₹$5160\times \frac{16.5}{25.8}=3300$.

Answer: (C) ₹3,300

Concept: Change from 16% loss to 8% gain is $24\%$ of CP.

1. $24\%$ of CP $=$ ₹660 → CP $=\frac{660\times 100}{24}=$ ₹2750
2. Sold at ₹3080: profit $=3080-2750=$ ₹330
3. Profit % $=\frac{330}{2750}\times 100=12\%$

Trick: $1\%$ of CP $=\frac{660}{24}=27.5$; $\frac{330}{27.5}=12$.

Answer: (D) $12\%$

Concept: The three sector angles are the triangle's angles, which add to ${180}^{\circ }$ – together they form a semicircle.

1. Total angle $={180}^{\circ }$ → area $=\frac{180}{360}\pi r^2=\frac{1}{2}\pi r^2$
2. $=\frac{1}{2}\times \frac{22}{7}\times 4.2\times 4.2=\frac{1}{2}\times 55.44=27.72{\text{cm}}^2$

Trick: Side lengths 24, 26, 10 are not needed at all – only $r$ matters.

Answer: (D) $27.72$

Concept: The difference of ages never changes; take ages 7 years ago as $4k$ and $5k$.

1. 8 years hence (15 years after that): $\frac{4k+15}{5k+15}=\frac{9}{10}$
2. $40k+150=45k+135\Rightarrow k=3$
3. Difference $=5k-4k=k=3$ years

Trick: Ratio $4:5$ and $9:10$ both have difference 1 part; check $12,15\to 27,30$ ✓.

Answer: (A) $3$

Concept: Remainder on division by 10 = unit digit of the product.

1. Unit digits: 3, 2 and 5
2. $3\times 2\times 5=30$ → unit digit 0
3. Remainder $=0$

Trick: An even number × a number ending in 5 always ends in 0.

Answer: (C) $0$

Concept: Area $\propto$ side${}^2$; for a change of $a\%$ in each side, area changes by $(2a+\frac{a^2}{100})\%$ (with sign).

1. New side $=0.83$ of old
2. New area $=(0.83{)}^2=0.6889$ of old
3. Decrease $=1-0.6889=0.3111=31.11\%$

Trick: $-17-17+\frac{17\times 17}{100}=-34+2.89=-31.11\%$.

Answer: (C) $31.11\%$

Concept: For every year find $|A-B|$ and pick the largest.

1. 2008: 90, 2009: 20, 2010: 100, 2011: 50
2. 2012: $900-500=400$; 2013: $750-480=270$
3. Maximum difference is in 2012

Trick: Look for the pair of bars with the biggest visual gap – 2012 stands out.

Answer: (A) 2012

Concept: If A is $r\%$ more than B, then B is $\frac{r}{100+r}\times 100\%$ less than A.

1. Let B $=100$, then A $=130$
2. B is less than A by $\frac{30}{130}\times 100=23.08\%$
3. Correct to one decimal: $23.1\%$

Trick: $\frac{3}{13}\approx 0.2308$.

Answer: (B) $23.1\%$

Concept: $72=8\times 9$: last three digits divisible by 8 and digit sum divisible by 9.

1. $78y$ divisible by 8: $784=8\times 98\Rightarrow y=4$
2. Digit sum $=7+8+5+x+3+6+7+8+4=48+x$
3. $48+x$ divisible by 9 $\Rightarrow x=6$
4. $x+y=6+4=10$

Trick: Always fix the 8-rule digit first ($y$), then use the 9-rule for $x$.

Answer: (C) $10$

Concept: Replace ${\sin }^2\theta =1-{\cos }^2\theta$ to get a quadratic in $\cos \theta$.

1. $4-2(1-{\cos }^2\theta )-5\cos \theta =0\Rightarrow 2{\cos }^2\theta -5\cos \theta +2=0$
2. $(2\cos \theta -1)(\cos \theta -2)=0\Rightarrow \cos \theta =\frac{1}{2}$ (since $\cos \theta \le 1$), $\theta ={60}^{\circ }$
3. $\cos \theta -\tan \theta =\frac{1}{2}-\sqrt{3}=\frac{1-2\sqrt{3}}{2}$

Trick: $\tan {60}^{\circ }>\cos {60}^{\circ }$, so the answer must be negative – only (D).

Answer: (D) $\frac{1-2\sqrt{3}}{2}$

Concept: Take tank capacity = LCM of times and work with units per hour.

1. Capacity $=\text{LCM}(16,24,40)=240$ units; A $=15$, B $=10$, C $=-6$ per hour
2. First 10 hours: $(15+10-6)\times 10=190$ units
3. Remaining $=50$ units by A + B at 25 per hour $=\frac{50}{25}=2$ hours

Trick: The LCM method avoids fractions completely.

Answer: (B) $2$

Concept: Step down using ${(x^n+\frac{1}{x^n})}^2=x^{2n}+\frac{1}{x^{2n}}+2$.

1. ${(x^2+\frac{1}{x^2})}^2=194+2=196\Rightarrow x^2+\frac{1}{x^2}=14$
2. ${(x+\frac{1}{x})}^2=14+2=16\Rightarrow x+\frac{1}{x}=4$ ($x>0$)
3. $x+\frac{1}{x}+2=4+2=6$

Trick: Chain $194\to 14\to 4$: each time add 2 and take the square root.

Answer: (C) $6$

Concept: BODMAS – brackets first, then "of", then division/multiplication, then addition/subtraction.

1. $3÷6\text{ of }2=3÷12=\frac{1}{4}$
2. $(4÷4\text{ of }\frac{1}{4})÷8=(4÷1)÷8=\frac{1}{2}$
3. $(4\times 8÷\frac{1}{4})\times \frac{1}{8}=128\times \frac{1}{8}=16$
4. $1-\frac{1}{4}+\frac{1}{2}+16=\frac{69}{4}$

Trick: "of" always beats $÷$ – compute it first.

Answer: (A) $\frac{69}{4}$

Concept: $\frac{a^3+b^3}{a^2-ab+b^2}=a+b$.

1. Let $a=5.35$, $b=3.65$; the denominator $=100(a^2+b^2-ab)$ because $53.5=10a$, $36.5=10b$
2. Value $=\frac{a^3+b^3}{100(a^2-ab+b^2)}=\frac{a+b}{100}$
3. $=\frac{5.35+3.65}{100}=\frac{9}{100}=0.09$

Trick: Watch the decimal shift – the denominator numbers are 10 times bigger, so divide by 100.

Answer: (A) $0.09$

Concept: Diagonals of a rhombus bisect each other at right angles; area $=\frac{1}{2}{d}_1{d}_2$.

1. Half of the given diagonal $=12$; side $=13$
2. Half of the other diagonal $=\sqrt{{13}^2-{12}^2}=5$ → $d_2=10$
3. Area $=\frac{1}{2}\times 24\times 10=120{\text{cm}}^2$

Trick: 5-12-13 is a Pythagorean triplet.

Answer: (A) $120$