Q&A: practice questions with answers

Concept: TSA of a solid hemisphere $=2\pi r^2+\pi r^2=3\pi r^2$.

1. $3\times \frac{22}{7}\times r^2=281$.
2. $r^2=\frac{281\times 7}{66}=\frac{1967}{66}$.
3. $r=\sqrt{\frac{1967}{66}}\text{cm}$.

Trick: Solid hemisphere → $3\pi r^2$ (with base); hollow bowl surface → $2\pi r^2$. Here numerator must be $281\times 7=1967$.

Answer: (B) $\sqrt{\frac{1967}{66}}$

Concept: Convert each pipe to "full tank per hour", then add fillers and subtract the emptier.

1. P fills $\frac{5}{9}$ in 15 h → full tank in $15\times \frac{9}{5}=27$ h. Q fills $\frac{7}{9}$ in 21 h → $21\times \frac{9}{7}=27$ h.
2. P + Q in 3 h $=3\times \frac{2}{27}=\frac{2}{9}$ of tank. R empties $\frac{2}{9}$ in 4 h → R empties full tank in 18 h.
3. Together: $\frac{1}{27}+\frac{1}{27}-\frac{1}{18}=\frac{4-3}{54}=\frac{1}{54}$ → 54 hours.

Trick: Take LCM $54$ units: P $=2$, Q $=2$, R $=-3$ per hour → net $1$ unit/h → $54$ h.

Answer: (D) 54 hours

Concept: B's one-day work $=$ (A+B)'s $-$ A's.

1. $\frac{1}{20}-\frac{1}{45}=\frac{9-4}{180}=\frac{5}{180}=\frac{1}{36}$ → B alone takes 36 days.
2. Three-fourth of the work: $\frac{3}{4}\times 36=27$ days.

Trick: LCM $180$ units: A+B $=9$/day, A $=4$/day, B $=5$/day → $\frac{135}{5}=27$ days.

Answer: (D) 27

Concept: Successive increases multiply: $P\times (1+r_1)(1+r_2)$.

1. Net factor $=1.15\times 1.20=1.38$ (same as $15+20+\frac{15\times 20}{100}=38\%$).
2. Population $=2,90,000\times 1.38=4,00,200$.

Trick: $a+b+\frac{ab}{100}$ for two successive changes – here $38\%$ of $2,90,000=1,10,200$; add it.

Answer: (B) 4,00,200

Concept: Downstream speed $=u+v$, upstream speed $=u-v$; total time $=\frac{d}{u+v}+\frac{d}{u-v}$.

1. Downstream $=8+2=10\text{km/h}$, upstream $=8-2=6\text{km/h}$; time $=3\text{h}12\text{min}=\frac{16}{5}$ h.
2. $\frac{d}{10}+\frac{d}{6}=\frac{16}{5}$ → $\frac{8d}{30}=\frac{16}{5}$.
3. $d=\frac{16}{5}\times \frac{30}{8}=12$ km.

Trick: $d=\frac{T(u^2-v^2)}{2u}=\frac{3.2\times 60}{16}=12$.

Answer: (D) 12 km

Concept: Take LCM and use ${\sin }^2\theta +{\cos }^2\theta =1$.

1. $\frac{\cos \theta }{1+\sin \theta }+\frac{1+\sin \theta }{\cos \theta }=\frac{{\cos }^2\theta +(1+\sin \theta {)}^2}{(1+\sin \theta )\cos \theta }$.
2. Numerator $={\cos }^2\theta +1+2\sin \theta +{\sin }^2\theta =2+2\sin \theta =2(1+\sin \theta )$.
3. Cancel $(1+\sin \theta )$: result $=\frac{2}{\cos \theta }=2\sec \theta$.

Trick: Put $\theta =0^{\circ }$: $\frac{1}{1}+\frac{1}{1}=2$; only $2\sec 0^{\circ }=2$ matches.

Answer: (C) $2\sec \theta$

Concept: To double, SI must equal the principal: $\frac{P\times R\times T}{100}=P$ → $R\times T=100$.

1. $T=20$ years → $R=\frac{100}{20}=5\%$ per annum.

Trick: Double → $RT=100$; triple → $RT=200$; $n$ times → $RT=100(n-1)$.

Answer: (B) 5

Concept: Set up the ratio of the fractional parts and simplify.

1. $\frac{a/5}{b/8}=\frac{3}{4}$ → $\frac{8a}{5b}=\frac{3}{4}$ → $32a=15b$.
2. $a:b=15:32$.

Trick: $a:b=(3\times 5):(4\times 8)=15:32$ – multiply each ratio term by its own denominator. The total 2820 is not needed.

Answer: (B) 15 : 32

Concept: Find the speed ratio from the first race and S's actual speed from the second.

1. 150 m race: R runs 150 while S runs 144 → speeds R : S $=25:24$.
2. 250 m race: when R finishes 250, S has run 240 and needs 2 s for the last 10 m → S's speed $=5\text{m/s}$; R's speed $=5\times \frac{25}{24}=\frac{125}{24}\text{m/s}$.
3. In 30 s: R $=156.25$ m, S $=150$ m → R beats S by $6.25$ m.

Trick: Lead in 30 s $=30\times (\text{R}-\text{S speed})=30\times \frac{5}{24}=6.25$.

Answer: (D) 6.25

Concept: Past population $=\frac{\text{present}}{(1+r{)}^n}$; $25\%$ growth means factor $\frac{5}{4}$ per year.

1. 2 years ago: $450000\times \frac{4}{5}\times \frac{4}{5}=288000$.
2. 3 years ago: $288000\times \frac{4}{5}=230400$.
3. Difference $=288000-230400=57600$.

Trick: Difference $=$ (population 3 yrs ago) $\times 25\%$ $=230400\times \frac{1}{4}=57600$.

Answer: (D) 57600

Concept: First find the maximum marks from Girish's score.

1. $92\%$ of max $=768+60=828$ → max $=\frac{828\times 100}{92}=900$.
2. Friend's percentage $=\frac{738}{900}\times 100=82\%$.

Trick: $828=92\times 9$ → max $=900$; then $738÷9=82$.

Answer: (A) $82\%$

Concept: For $a{x}^2+bx+c=0$, the discriminant $D=b^2-4ac$ decides the roots: $D>0$ real and distinct (rational if $D$ is a perfect square, else irrational), $D=0$ equal, $D<0$ no real roots.

1. Multiply $x-\frac{1}{x}=17$ by $x$: $x^2-17x-1=0$.
2. $D=(-17{)}^2-4(1)(-1)=289+4=293>0$.
3. $293$ is not a perfect square (${17}^2=289,{18}^2=324$), so the roots are real, distinct and irrational.

Trick: Whenever $c$ is negative with $a$ positive, $D=b^2-4ac$ is surely positive – roots are always real and distinct; just check perfect square for rational/irrational.

Answer: (C) Real and distinct (irrational)

Concept: Two angles are supplementary when their sum is ${180}^{\circ }$.

1. Let the smaller angle be $x$; the larger is $x+32$.
2. $x+(x+32)=180\Rightarrow 2x=148\Rightarrow x=74$.
3. Check: ${74}^{\circ }+{106}^{\circ }={180}^{\circ }$ ✓

Trick: Smaller $=\frac{180-\text{difference}}{2}=\frac{180-32}{2}=74$; larger $=\frac{180+32}{2}=106$.

Answer: (C) 74

Concept: A plane parallel to the base cuts off a smaller cone similar to the original. If the linear ratio is $k$, areas scale by $k^2$ and volumes by $k^3$.

1. CSA of the small cone $=\frac{1}{2}$ CSA of the original $\Rightarrow k^2=\frac{1}{2}\Rightarrow k=\frac{1}{\sqrt{2}}$.
2. Radius of the small cone $=\frac{6\sqrt{2}}{2}=3\sqrt{2}\text{cm}$.
3. Original radius $R=\frac{3\sqrt{2}}{k}=3\sqrt{2}\times \sqrt{2}=6\text{cm}$.
4. Check with $l=3R$: CSA $=\pi R·3R=3\pi R^2$; small cone $3\pi (3\sqrt{2}{)}^2=54\pi$, original $3\pi ·36=108\pi$ – exactly half ✓

Trick: For similar solids: length ratio $k$, area ratio $k^2$, volume ratio $k^3$. The "slant height = 3 × radius" data is only a distraction here.

Answer: (B) 6

Concept: Take CP of one article as ₹100. The customer pays for 14 articles but the trader gives away 16, so the cost is on 16 articles while the sale is on 14 at the discounted price.

1. CP of 16 articles $=16\times 100=1600$. Gain 33% $\Rightarrow$ SP $=1600\times 1.33=2128$.
2. SP per paid article $=\frac{2128}{14}=152$.
3. SP $=95\%$ of MP $\Rightarrow$ MP $=\frac{152}{0.95}=160$.
4. MP is $160-100=60\%$ above CP.

Trick: Chain formula: $\text{MP}=\text{CP}\times \frac{16}{14}\times \frac{133}{100}\times \frac{100}{95}=100\times \frac{16\times 133}{14\times 95}=160$.

Answer: (B) 60%

Concept: Profit% is calculated on the cost price of the same unit: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. CP of one article $=\frac{102}{12}=8.5$ rupees.
2. SP of one article $=68$ rupees. Profit $=68-8.5=59.5$.
3. Profit% $=\frac{59.5}{8.5}\times 100=7\times 100=700\%$.

Trick: Note $68=8\times 8.5$, i.e. SP is 8 times CP → profit is 7 times CP → $700\%$. The "32 dozens" is irrelevant.

Answer: (C) 700%

Concept: For $n$ observations arranged in order, the median is the ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value when $n$ is odd.

1. Arrange: 45, 49, 54, 57, 58, 60, 62, 63, 66, 68, 71.
2. $n=11\Rightarrow$ median $=\frac{11+1}{2}=6^{\text{th}}$ value $=60$.

Trick: Odd count → middle value directly; even count → average of the two middle values. Always sort first!

Answer: (A) 60

Concept: First simplify each expression using the laws of indices ($a^m·a^n=a^{m+n}$, $\sqrt{a}=a^{0.5}$), then find LCM and HCF.

1. First: $\frac{x^{2.5}}{y^{0.5}}\times \sqrt{xy}=x^{2.5}{y}^{-0.5}·x^{0.5}{y}^{0.5}=x^3$.
2. Second: $\frac{y^{1.5}}{\sqrt{x}}\times \sqrt{xy}=y^{1.5}{x}^{-0.5}·x^{0.5}{y}^{0.5}=y^2$.
3. $x$ and $y$ are co-prime, so $x^3$ and $y^2$ share no factor: HCF $=1$, LCM $=x^3{y}^2=x(xy{)}^2$.

Trick: For co-prime bases the HCF of pure powers is always 1 and the LCM is just their product.

Answer: (D) $x(xy{)}^2,1$

Concept: Eliminate one variable using the simplest equation, reduce to two equations in two unknowns, then back-substitute. Options can also be verified directly.

1. From (1): $w=2-u-v$. Put in (2): $6u+4v+5(2-u-v)=31\Rightarrow u-v=21$.
2. Put in (3): $5u+2v+2(2-u-v)=13\Rightarrow 3u=9\Rightarrow u=3$.
3. Then $v=u-21=-18$ and $w=2-3+18=17$.
4. Check (2): $18-72+85=31$ ✓

Trick: Equation (3) collapses fast because $2v+2w=2(v+w)=2(2-u)$. Look for such shortcuts before heavy elimination – or simply test options in equation (1) first (only B and D give 2).

Answer: (D) u = 3, v = −18, w = 17

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}=\frac{\sum n_i{\overline{x}}_i}{\sum n_i}$.

1. Totals: $16\times 447=7152$, $86\times 396=34056$, $21\times 158=3318$.
2. Sum $=7152+34056+3318=44526$; members $=16+86+21=123$.
3. Average $=\frac{44526}{123}=362$.

Trick: Use deviations from 400: $16(+47)+86(-4)+21(-242)=752-344-5082=-4674$; $\frac{-4674}{123}=-38$ → $400-38=362$.

Answer: (A) 362