Q&A: practice questions with answers

Concept: Volume (capacity) of a cone $=\frac{1}{3}\pi r^{2}h$.

1. Here $r=7\text{ft}$, $h=9\text{ft}$, $\pi =3.14$.
2. $V=\frac{1}{3}\times 3.14\times 7^2\times 9=3.14\times 49\times 3$.
3. $V=3.14\times 147=461.58{\text{ft}}^3$.

Trick: Cancel the $\frac{1}{3}$ with $h=9$ first: $V=3.14\times 49\times 3$ – one quick multiplication.

Answer: (D) $461.58$ cubic feet

Concept: Value of a sector $=\frac{\text{its percentage}}{100}\times \text{total}$.

1. Village D has $17\%$ of the total population.
2. Population of D $=\frac{17}{100}\times 50000=17\times 500$.
3. $=8500$.

Trick: $1\%$ of $50000$ is $500$, so just multiply the percentage by $500$: $17\times 500=8500$.

Answer: (C) $8500$

Concept: Add the required entries from the table and compare them as a ratio.

1. First employee, first three days (Mon–Wed): $200+300+200=700\text{km}$.
2. Second employee, last three days (Thu–Sat): $250+350+100=700\text{km}$.
3. Ratio $=700:700=1:1$.

Trick: Read the right column for the right rows – a common slip is to take the first employee's last three days ($150+250+350=750$).

Answer: (B) $1:1$

Concept: Third proportional to $a,b$ is $\frac{b^2}{a}$; mean proportional between $a,b$ is $\sqrt{ab}$.

1. Third proportional to $16$ and $40$: $\frac{{40}^2}{16}=\frac{1600}{16}=100$.
2. Mean proportional between $10$ and $40$: $\sqrt{10\times 40}=\sqrt{400}=20$.
3. Required ratio $=100:20=5:1$.

Yaad rakhein: $a:b=b:c\Rightarrow c=\frac{b^2}{a}$ (third); $a:x=x:b\Rightarrow x=\sqrt{ab}$ (mean).

Answer: (A) $5:1$

Concept: From $\cos A={\sin }^{2}A$ we get ${\cos }^{2}A={\sin }^{4}A$, so ${\sin }^{4}A+{\sin }^{2}A={\cos }^{2}A+{\sin }^{2}A=1$.

1. Cube both sides: $({\sin }^{4}A+{\sin }^{2}A{)}^3=1$.
2. Using $(x+y{)}^3=x^3+3x^{2}y+3x{y}^2+y^3$: ${\sin }^{12}A+3{\sin }^{10}A+3{\sin }^{8}A+{\sin }^{6}A=1$.
3. Compare with the given expression: $a=1,b=3,c=3$.
4. $a+b+c=1+3+3=7$.

Trick: Powers $12,10,8,6$ with last coefficient $1$ signal a cube expansion – coefficients $1,3,3,1$.

Answer: (A) $7$

Concept: For equal distances at speeds $x$ and $y$, average speed $=\frac{2xy}{x+y}$.

1. $x=12\text{km/h}$, $y=18\text{km/h}$.
2. Average speed $=\frac{2\times 12\times 18}{12+18}=\frac{432}{30}$.
3. $=14.4=14\frac{2}{5}\text{km/h}$.

Trick: The average speed is always a little less than the simple mean ($15$), so $14\frac{2}{5}$ is the only sensible choice.

Answer: (C) $14\frac{2}{5}\text{km/h}$

Concept: With a false weight, compare the cost of the quantity actually given with the money received.

1. For $900\text{g}$ given, cost $=\frac{900}{1000}\times 16=$ ₹$14.40$.
2. He receives the price of $1\text{kg}$ = ₹$18$.
3. Profit $=18-14.40=3.60$; Profit$\%=\frac{3.60}{14.40}\times 100=25\%$.

Trick: $(1+\text{gain})=\frac{18}{16}\times \frac{1000}{900}=\frac{9}{8}\times \frac{10}{9}=\frac{5}{4}$, i.e. $25\%$.

Answer: (D) $25\%$

Concept: Longest rod in a room = diagonal of the cuboid $=\sqrt{l^2+b^2+h^2}$.

1. $l=12,b=9,h=8$ (in m).
2. $d=\sqrt{144+81+64}=\sqrt{289}$.
3. $d=17\text{m}$.

Trick: $12,9\to 15$ (a $3\text{-}4\text{-}5$ triple), then $15,8\to 17$ (an $8\text{-}15\text{-}17$ triple).

Answer: (B) $17\text{m}$

Concept: For the same distance, time is inversely proportional to speed.

1. Speeds $A:B=3:5\Rightarrow$ times $A:B=5:3$.
2. Difference $=5-3=2$ units $=30$ minutes, so $1$ unit $=15$ minutes.
3. Time of A $=5\times 15=75$ minutes $=1$ hour $15$ minutes.

Trick: Flip the speed ratio to get the time ratio, then use the difference to find one unit.

Answer: (A) $1$ hour $15$ minutes

Concept: Compare the cost of the quantity actually given with the price received.

1. He gives $800\text{g}$, whose cost $=0.8\times 23=$ ₹$18.40$.
2. He receives ₹$20$ (price of $1\text{kg}$).
3. Profit $=20-18.40=1.60$; Profit$\%=\frac{1.60}{18.40}\times 100=\frac{200}{23}\%$.
4. $\frac{200}{23}=8\frac{16}{23}\%$.

Trick: $\frac{\text{SP}\times 1000}{\text{CP}\times 800}=\frac{20\times 1000}{23\times 800}=\frac{25}{23}$, gain $=\frac{2}{23}=8\frac{16}{23}\%$.

Answer: (B) $8\frac{16}{23}\%$

Concept: Read the two 2020 bars and simplify the ratio.

1. Economics in 2020 $=75$ students.
2. Physical Education in 2020 $=125$ students.
3. Ratio $=75:125=3:5$ (divide by $25$).

Trick: Use the blue (2020) bars only – the orange bars ($85,110$) are for 2021.

Answer: (B) $3:5$

Concept: Both values come from the same total, so compare the percentages directly.

1. Company A $=31\%$, company D $=20\%$.
2. Required $\%=\frac{31}{20}\times 100=155\%$.
3. (Check: $A=16120$, $D=10400$; $\frac{16120}{10400}\times 100=155$.)

Trick: The total $52000$ is not needed – it cancels out.

Answer: (D) $155$

Concept: $88=8\times 11$: last three digits divisible by $8$, and (sum of odd-place digits − sum of even-place digits) divisible by $11$.

1. $2y8$ divisible by $8$: $208,248,288\Rightarrow y=0,4,8$. Smallest natural $y=4$.
2. Number $97x456248$: odd places $9+x+5+2+8=24+x$; even places $7+4+6+4=21$.
3. $(24+x)-21=x+3$ must be $0$ or $11\Rightarrow x=8$.
4. $x^2+y^2=64+16=80$.

Trick: $y=0$ is rejected because $y$ must be a natural number.

Answer: (D) $80$

Concept: Find each average, then (first average ÷ second average) $\times 100$.

1. Rugby, hockey, badminton, squash: $\frac{30+60+20+10}{4}=\frac{120}{4}=30$.
2. Tennis and football: $\frac{20+80}{2}=50$.
3. Required $\%=\frac{30}{50}\times 100=60\%$.

Trick: Divide the totals by the right counts ($4$ and $2$) – comparing totals $120$ and $100$ gives the wrong $120\%$.

Answer: (A) $60$ Percent

Concept: Area to paint = area of 4 walls + ceiling $=2(l+b)h+lb$ (floor excluded).

1. 4 walls $=2(10+5)\times 6=180{\text{m}}^2$.
2. Ceiling $=10\times 5=50{\text{m}}^2$.
3. Total $=180+50=230{\text{m}}^2$.

Trick: Total surface area is $280{\text{m}}^2$; subtract the floor ($50$) to get $230$.

Answer: (D) $230{\text{m}}^2$

Concept: $\cos (x-y)\cos (x+y)={\cos }^{2}x-{\sin }^{2}y$, and $\cos {54}^{\circ }=\sin {36}^{\circ }$.

1. First product $={\cos }^2{36}^{\circ }-{\sin }^{2}A$.
2. Second product $={\cos }^2{54}^{\circ }-{\sin }^{2}A={\sin }^2{36}^{\circ }-{\sin }^{2}A$.
3. Sum $=({\cos }^2{36}^{\circ }+{\sin }^2{36}^{\circ })-2{\sin }^{2}A=1-2{\sin }^{2}A$.
4. $1-2{\sin }^{2}A=\cos 2A$.

Trick: Put $A=0^{\circ }$: value $={\cos }^2{36}^{\circ }+{\cos }^2{54}^{\circ }=1$; only $\cos 2A$ gives $1$ (and not $\cos A$, check $A={90}^{\circ }$: value $-1$).

Answer: (C) $\cos 2A$

Concept: Same total, so compare percentages: $\frac{\text{Cooking}\%}{\text{Dancing}\%}\times 100$.

1. From the legend: Cooking $=22\%$, Dancing $=21\%$.
2. Required $\%=\frac{22}{21}\times 100=104.7619\dots$
3. Rounded off: $104.76\%$.

Trick: Cooking is slightly more than dancing, so the answer is just above $100\%$ – this rules out $92.65\%$ and $125.25\%$ at once.

Answer: (D) $104.76\%$

Concept: Total = average × number; then split the total in the given ratio.

1. Total height $=56\times 3=168\text{cm}$.
2. Ratio $7:8:9$ has $24$ parts; $1$ part $=\frac{168}{24}=7$.
3. Heights $=7\times 7,8\times 7,9\times 7=49,56,63\text{cm}$.

Trick: The middle term of $7:8:9$ equals the average, so $8$ parts $=56$; only option (D) has $56$ in the middle.

Answer: (D) $49\text{cm},56\text{cm},63\text{cm}$

Concept: Add/subtract one-day works (efficiencies).

1. $P+Q=\frac{1}{14}$, $Q+R=\frac{1}{21}$.
2. Subtract: $P-R=\frac{1}{14}-\frac{1}{21}=\frac{1}{42}$. Since $P=2R$: $R=\frac{1}{42}$, $P=\frac{2}{42}$.
3. $Q=\frac{1}{14}-\frac{2}{42}=\frac{3-2}{42}=\frac{1}{42}$.
4. Q alone takes $42$ days.

Trick: Total work $=42$ (LCM). $P+Q=3$, $Q+R=2$ per day $\Rightarrow P-R=1$; with $P=2R$: $R=1,P=2,Q=1\Rightarrow \frac{42}{1}=42$ days.

Answer: (A) $42$ days

Concept: The extra discount given (in % of marked price) equals the loss in profit.

1. Planned discount $5\%$, actual discount $8\%$.
2. Extra discount $=8\%-5\%=3\%$ of MP $=$ ₹$45$.
3. MP $=\frac{45\times 100}{3}=$ ₹$1500$.

Trick: Cost price is the same in both cases, so the profit falls exactly by the difference in selling prices: $3\%\text{ of MP}=45$.

Answer: (A) Rs.$1500$