Q&A: practice questions with answers

Concept: If every observation is multiplied by $k$, the average is also multiplied by $k$.

1. Sum of 13 numbers $=13\times 8=104$.
2. New sum $=104\times 7=728$.
3. New average $=\frac{728}{13}=56$.

Trick: Skip the sum: new average $=8\times 7=56$ directly. (Adding $k$ to each number adds $k$ to the average too.)

Answer: (C) 56

Concept: Write $y-3$ as a perfect square of the form $(\sqrt{a}+\sqrt{b}{)}^2=a+b+2\sqrt{ab}$, then use rationalisation.

1. $y=12+4\sqrt{5}$ ⇒ $y-3=9+4\sqrt{5}=5+4+2\times 2\sqrt{5}=(\sqrt{5}+2{)}^2$.
2. $\sqrt{y-3}=\sqrt{5}+2$ and $\frac{1}{\sqrt{y-3}}=\frac{1}{\sqrt{5}+2}=\sqrt{5}-2$.
3. Difference $=(\sqrt{5}+2)-(\sqrt{5}-2)=4$.

Trick: If $x=\sqrt{a}+\sqrt{b}$ with $a-b=1$, then $\frac{1}{x}=\sqrt{a}-\sqrt{b}$ and $x-\frac{1}{x}=2\sqrt{b}$. Here $b=4$ ⇒ $2\times 2=4$.

Answer: (B) 4

Concept: First find the maximum marks from Girish's data, then convert the friend's marks to a percentage.

1. $28\%$ of max $=572+30=602$ ⇒ max $=\frac{602\times 100}{28}=2150$.
2. Friend's percentage $=\frac{473}{2150}\times 100=22\%$.

Trick: $602÷28=21.5$, so $1\%=21.5$ marks; $473÷21.5=22$. Working in "marks per 1%" avoids big numbers.

Answer: (A) $22\%$

Concept: Work per hour adds up: filling taps positive, emptying tap negative.

1. Net rate $=\frac{1}{4}+\frac{1}{38}-\frac{1}{38}=\frac{1}{4}$ of the cistern per hour.
2. Full cistern in $4$ hours ⇒ one-fourth in $4\times \frac{1}{4}=1$ hour.

Trick: A filling and an emptying tap with the same time cancel completely – just ignore both.

Answer: (A) 1

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$, i.e. $\text{Mode}-\text{Median}=2(\text{Median}-\text{Mean})$.

1. Given $\text{Mode}-\text{Median}=61.54$.
2. So $\text{Median}-\text{Mean}=\frac{61.54}{2}=30.77$.

Trick: Mean, Median, Mode lie in the ratio of gaps $1:2$ – Median is twice as far from Mode as from Mean.

Answer: (C) 30.77

Concept: Opposite directions ⇒ relative speed = sum of speeds; distance covered while crossing = sum of lengths.

1. Relative speed $=60+84=144\text{km/h}=144\times \frac{5}{18}=40\text{m/s}$.
2. Combined length $=40\times 30=1200$ m.

Trick: km/h → m/s: multiply by $\frac{5}{18}$. $144$ is $18\times 8$, so $144\to 8\times 5=40$ instantly.

Answer: (C) 1200

Concept: Same time ⇒ distances are in the ratio of speeds: $\frac{\text{downstream}}{\text{upstream}}=\frac{x+3}{x-3}$.

1. $\frac{x+3}{x-3}=\frac{36}{24}=\frac{3}{2}$.
2. $2x+6=3x-9$ ⇒ $x=15$.

Trick: Speed ratio $3:2$ ⇒ boat speed $=\frac{3+2}{3-2}\times$ stream $=5\times 3=15\text{km/h}$.

Answer: (A) $15\text{km/h}$

Concept: In the right triangle, $\tan \theta =\frac{\text{height}}{\text{distance}}$.

1. $\tan {45}^{\circ }=\frac{h}{28.6}$.
2. $\tan {45}^{\circ }=1$ ⇒ $h=28.6$ m.

Trick: Angle of elevation ${45}^{\circ }$ ⇒ height = distance from the foot, always.

Answer: (B) 28.6 m

Concept: Mean $=\frac{\sum f_i{x}_i}{\sum f_i}$.

1. $\sum f_i{x}_i=81\times 15+35\times 4+73\times 3+56\times 16=1215+140+219+896=2470$.
2. $\sum f_i=15+4+3+16=38$.
3. Mean $=\frac{2470}{38}=65$.

Trick: Options are far apart – 31 of the 38 students scored 81 or 56, so the mean must be well above 60 but below 70; only 65 fits.

Answer: (C) 65

Concept: A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

1. Last two digits of 816A0408 are 08, and $8÷4=2$.
2. So the number is divisible by 4 for every digit A ($0$ to $9$); the maximum value is $9$.

Trick: Divisibility by 4 (last 2 digits) and by 8 (last 3 digits) never depends on digits further left – do not waste time adding digits.

Answer: (C) 9

Concept: BODMAS – do division and multiplication before addition and subtraction; watch the signs.

1. $(-56)÷(-14)=4$ and $(-4)\times 7=-28$.
2. Expression $=-9-4+(-28)=-9-4-28=-41$.

Trick: Negative ÷ negative = positive, but here it is being subtracted, so it becomes $-4$. Sign slips cause the wrong option $-43$ or $-40$.

Answer: (B) $-41$

Concept: Profit is shared in the ratio of (capital × time).

1. Let capitals be $4k,2k,5k$. X: $4k\times 6+2k\times 6=36k$; Y: $2k\times 12=24k$; Z: $5k\times 12=60k$.
2. Profit ratio X : Y : Z $=36:24:60=3:2:5$ (total 10 parts).
3. Difference X − Y $=1$ part $=\frac{37000}{10}=$ ₹$3700$.

Trick: Z's ₹23,000 is a distractor – ratios alone decide the profit split. X withdrawing half for half the year makes his effective capital $4\times \frac{6+3}{12}\times 12=36$, i.e. $\frac{3}{4}$ of a full year at 4.

Answer: (A) ₹3700

Concept: Express all present ages in one variable, then subtract 3 from each for "3 years ago".

1. Son $=s$, Sharad $=3s$, daughter $=s-3$.
2. 3 years ago: $(3s-3)+(s-3)+(s-6)=63$ ⇒ $5s-12=63$ ⇒ $s=15$.
3. Sharad $=3\times 15=45$ years.

Trick: Sum of present ages $=63+3\times 3=72$; $3s+s+(s-3)=72$ ⇒ $5s=75$ ⇒ $s=15$. Adding $3\times$(number of people) first is faster.

Answer: (C) 45

Concept: A regular hexagon is six equilateral triangles: Area $=6\times \frac{\sqrt{3}}{4}{a}^2=\frac{3\sqrt{3}}{2}{a}^2$.

1. $a=14\sqrt{3}$ ⇒ $a^2=196\times 3=588$.
2. Area $=\frac{3\sqrt{3}}{2}\times 588=3\sqrt{3}\times 294=882\sqrt{3}{\text{cm}}^2$.

Trick: With side $k\sqrt{3}$, area $=\frac{3\sqrt{3}}{2}\times 3k^2=\frac{9\sqrt{3}}{2}{k}^2$; here $k=14$ ⇒ $\frac{9\times 196}{2}\sqrt{3}=882\sqrt{3}$.

Answer: (C) $882\sqrt{3}{\text{cm}}^2$

Concept: Express the actual selling price in terms of the cost price using the loss condition, then apply the new fraction.

1. $\frac{6}{15}\times SP=72\%$ of $CP$ ⇒ $SP=\frac{0.72\times 15}{6}CP=1.8CP$.
2. New price $=89\%$ of $SP=0.89\times 1.8CP=1.602CP$.
3. Profit $=60.2\%$.

Trick: Take $CP=100$: $\frac{2}{5}SP=72$ ⇒ $SP=180$; $89\%$ of $180=160.2$ ⇒ profit $60.2\%$.

Answer: (C) $60.2\%$

Concept: If two quadratics have both roots common, their coefficients are proportional: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.

1. First: $(8A+4)x^2+Bx+(4A-8)=0$. Second: $(6A+3)x^2+Kx+(3A-6)=0$.
2. $\frac{8A+4}{6A+3}=\frac{4(2A+1)}{3(2A+1)}=\frac{4}{3}$ and $\frac{4A-8}{3A-6}=\frac{4}{3}$ ⇒ so $\frac{B}{K}=\frac{4}{3}$ ⇒ $3B=4K$.
3. $9B^2=16K^2$ ⇒ $9B^2-16K^2=0$.

Trick: The second equation is just $\frac{3}{4}$ of the first in $x^2$ and constant terms – so $K$ must be $\frac{3}{4}B$; square both sides and the answer is 0.

Answer: (D) 0

Concept: When only a ratio is given, put the simplest values ($a=2,b=3$) directly into the expression.

1. $7a^3-2b^3=7\times 8-2\times 27=56-54=2$
2. $b^3-a^3=27-8=19$
3. Required ratio $=2:19$

Trick: Ratio questions never need variables — plug the ratio numbers and compute.

Answer: (A) $2:19$

Concept: For in-centre O of triangle PQR, $\angle QOR={90}^{\circ }+\frac{\angle P}{2}$.

1. ${130}^{\circ }={90}^{\circ }+\frac{\angle P}{2}$
2. $\frac{\angle P}{2}={40}^{\circ }\Rightarrow \angle QPR={80}^{\circ }$

Yaad rakhein: In-centre: ${90}^{\circ }+\frac{A}{2}$; Circumcentre: $2A$; Orthocentre: ${180}^{\circ }-A$.

Answer: (B) ${80}^{\circ }$

Concept: Take total work $=\text{LCM}(8,16,24)=48$ units; P, Q, R do $6,3,2$ units/hour.

1. One P-Q-R round (3 h) $=6+3+2=11$ units; 9 h = 3 rounds $=33$ units.
2. Remaining $=48-33=15$ units.
3. Now Q and R alternate: every 2 h they do $3+2=5$ units $\Rightarrow 15$ units in $\frac{15}{5}\times 2=6$ h (Q–R–Q–R–Q–R exactly).

Trick: In alternate-work problems, first find work per full cycle, then handle the leftover cycle by cycle.

Answer: (B) 6

Concept: Cylinders are cut whole, so count how many fit along each edge; then TSA of cylinder $=2\pi r(r+h)$.

1. Base $8\times 8$ m, diameter 1 m $\Rightarrow 8\times 8=64$ cylinders per layer; height $8÷4=2$ layers $\Rightarrow 128$ cylinders.
2. TSA of one cylinder $=2\times \frac{22}{7}\times \frac{1}{2}\times (\frac{1}{2}+4)=\frac{22}{7}\times \frac{9}{2}=\frac{99}{7}{\text{m}}^2$
3. Total area $=128\times \frac{99}{7}$; cost $=128\times \frac{99}{7}\times 14=128\times 198=25344$

Trick: Keep $\frac{99}{7}$ unsimplified — the rate ₹14 cancels the 7 neatly.

Answer: (C) ₹25,344