Q&A: practice questions with answers

Concept: Work is fixed: number of pipes × time = constant (inverse proportion).

1. $39\times 19=741$ pipe-minutes.
2. Time with 23 pipes $=\frac{741}{23}=32.217\dots$
3. $\approx 32.22$ minutes.

Trick: Fewer pipes → more time, so the answer must exceed 19; $M_1{T}_1=M_2{T}_2$.

Answer: (B) $32.22$

Concept: Same exam → same maximum marks; find it from Kalpana's data.

1. Maximum marks $=\frac{656}{0.32}=2050$.
2. Sudha's % $=\frac{697}{2050}\times 100=34\%$.

Trick: Ratio method: $\frac{697}{656}\times 32=\frac{17}{16}\times 32=34\%$ (since $656=41\times 16$, $697=41\times 17$).

Answer: (C) $34\%$

Concept: Area $=\sqrt{s(s-a)(s-b)(s-c)}$ and altitude $h=\frac{2\times \text{Area}}{\text{base}}$.

1. $s=\frac{109+171+100}{2}=190$.
2. Area $=\sqrt{190\times 81\times 19\times 90}=\sqrt{26316900}=5130{\text{cm}}^2$.
3. $h=\frac{2\times 5130}{171}=60\text{cm}$.

Trick: Check with Pythagoras: foot divides 171 into 91 and 80; ${91}^2+{60}^2={109}^2$ and ${80}^2+{60}^2={100}^2$ ✓.

Answer: (C) $60$

Concept: Median = middle value after arranging data in order; for n odd it is the ${\left(\frac{n+1}{2}\right)}^{th}$ value.

1. Ascending: 15, 25, 33, 34, 36, 36, 37, 54, 70, 74, 78.
2. $n=11$ → position $\frac{11+1}{2}=6$.
3. 6th value = 36.

Trick: Always sort first — the unsorted middle value (36) can mislead; here it coincides.

Answer: (B) 36

Concept: Multiples of 13 form an AP; average of first n multiples of k $=k\times \frac{n+1}{2}$.

1. First 17 multiples: $13\times 1,13\times 2,\dots ,13\times 17$.
2. Average $=13\times \frac{1+2+\dots +17}{17}=13\times \frac{17+1}{2}$.
3. $=13\times 9=117$.

Trick: Odd count → average = middle term: 9th multiple $=13\times 9=117$.

Answer: (C) 117

Concept: The value of $a+b$ is given directly, so only squaring is needed.

1. $a+b=50$
2. $(a+b{)}^2={50}^2=2500$.
3. ($a-b=21$ is extra data; it would be needed for $ab$ or $a^2+b^2$.)

Trick: Read what is asked — sometimes extra data is a distraction.

Answer: (B) $2500$

Concept: Greatest number leaving remainders $r_1,r_2$ $=\text{HCF}(a-r_1,b-r_2)$.

1. $137-7=130$ and $198-3=195$.
2. $130=2\times 5\times 13$, $195=3\times 5\times 13$.
3. HCF $=5\times 13=65$. (Divisor 65 is larger than both remainders ✓.)

Trick: Subtract the remainders first, then take the HCF.

Answer: (C) 65

Concept: $\text{SP}=\text{MP}\times (1-\text{discount})$; $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. SP = 50% of ₹1,032 = ₹516.
2. Profit = ₹516 − ₹300 = ₹216.
3. Profit% $=\frac{216}{300}\times 100=72\%$.

Trick: 50% discount → SP is simply half the MP.

Answer: (D) $72\%$

Concept: If a, b, c, d are in proportion, $a:b=c:d$, i.e. $ad=bc$ (product of extremes = product of means).

1. $14:7=12:y$
2. $14y=7\times 12=84$
3. $y=\frac{84}{14}=6$.

Trick: $14:7=2:1$, so y is half of 12 = 6.

Answer: (B) $6$

Concept: Area of rectangle = length × breadth; use $(x+y)(x-y)=x^2-y^2$.

1. Length $=\frac{2a+2\sqrt{5}}{4}=\frac{a+\sqrt{5}}{2}$.
2. Area $=\frac{a+\sqrt{5}}{2}\times \frac{a-\sqrt{5}}{2}=\frac{a^2-(\sqrt{5}{)}^2}{4}$.
3. $=\frac{a^2-5}{4}{\text{m}}^2$.

Trick: Spot the conjugate pair $(a+\sqrt{5})(a-\sqrt{5})$ — the roots vanish.

Answer: (B) $\left(\frac{a^2-5}{4}\right){\text{m}}^2$

Concept: Write the numbers as $10k,15k,13k$ and find $k$ from the given condition.

1. $10\%$ of first $=66\Rightarrow$ first $=660$, so $10k=660$, $k=66$.
2. Second $=15\times 66=990$, third $=13\times 66=858$.
3. Difference $=990-858=132$; $50\%$ of $132=66$.

Trick: Difference of 3rd and 2nd $=(15-13)k=2k$; half of it $=k=66$ directly.

Answer: (B) 66

Concept: In $△ABC\cong △PQR$, corresponding parts are equal: $\angle B=\angle Q$, $\angle C=\angle R$, $AC=PR$.

1. $\angle PQR=\angle ABC$: $3x-4=77\Rightarrow x=27$.
2. $PR=AC$: $x+3y=48\Rightarrow 27+3y=48\Rightarrow y=7$.
3. $\angle QRP=\angle BCA=(x-y{)}^{\circ }=(27-7{)}^{\circ }={20}^{\circ }$.

Trick: Match letters in order: A↔P, B↔Q, C↔R, so $\angle QRP$ pairs with $\angle BCA$.

Answer: (D) ${20}^{\circ }$

Concept: Median $=l+\frac{\frac{N}{2}-cf}{f}\times h$, where the median class contains the $\frac{N}{2}$-th value.

1. $N=11+18+22+30+40+16+33=170$, $\frac{N}{2}=85$.
2. Cumulative frequencies: $11,29,51,81,121,\dots$ → median class $55$–$65$.
3. $l=55,cf=81,f=40,h=10$.
4. Median $=55+\frac{85-81}{40}\times 10=55+1=56$.

Trick: Stop the cumulative total as soon as it crosses $\frac{N}{2}$ – that class is the median class.

Answer: (C) 56 years

Concept: For 2 years: $CI-SI=P{\left(\frac{r}{100}\right)}^2$ and $S{I}_{1\text{ yr}}=\frac{Pr}{100}$.

1. SI for 2 years $=40\Rightarrow$ SI for 1 year $=20$.
2. $CI-SI=41-40=1$ = interest on first year’s interest: $20\times \frac{r}{100}=1\Rightarrow r=5\%$.
3. $P=\frac{20\times 100}{5}=400$.

Trick: $P=\frac{(S{I}_1{)}^2}{CI-SI}=\frac{{20}^2}{1}=400$.

Answer: (A) 400

Concept: $SI=\frac{P\times R\times T}{100}$ with $T$ in years; count days excluding the first day and including the last.

1. Days: February $28-15=13$, March $31$, April $29$ → $73$ days $=\frac{73}{365}=\frac{1}{5}$ year.
2. $SI=\frac{2000\times 5.75\times 1}{100\times 5}=23$.

Trick: $73$ days $=\frac{1}{5}$ year – a very common exam figure.

Answer: (D) 23

Concept: B’s one-day work $=\frac{1}{18}-\frac{1}{24}$.

1. $\frac{1}{18}-\frac{1}{24}=\frac{4-3}{72}=\frac{1}{72}$ → B alone takes $72$ days.
2. One-fourth work $=\frac{72}{4}=18$ days.

Trick: Total work $=\text{LCM}(18,24)=72$ units; A+B $=4$, A $=3$, so B $=1$ unit/day.

Answer: (A) 18

Concept: Downstream speed $=b+s$, upstream speed $=b-s$; time $=\frac{\text{distance}}{\text{speed}}$.

1. Downstream $=17+16=33\text{km/h}$; upstream $=17-16=1\text{km/h}$.
2. Time $=\frac{26.4}{33}+\frac{26.4}{1}=0.8+26.4=27.2$ hours.

Trick: When $b-s$ is tiny, the upstream trip dominates – the answer must exceed $26.4$.

Answer: (D) 27.2

Concept: Use ${\csc }^{2}A-1={\cot }^{2}A$ and ${\sec }^{2}A-1={\tan }^{2}A$.

1. $3\cot A=2\Rightarrow \cot A=\frac{2}{3}$, $\tan A=\frac{3}{2}$.
2. $\sqrt{\frac{3{\cot }^{2}A}{{\tan }^{2}A}}=\sqrt{3{\cot }^{4}A}=\sqrt{3}{\cot }^{2}A$.
3. $=\sqrt{3}\times \frac{4}{9}=\frac{4\sqrt{3}}{9}$.

Trick: $\frac{{\cot }^{2}A}{{\tan }^{2}A}={\cot }^{4}A$ – the root then just gives ${\cot }^{2}A$.

Answer: (C) $\frac{4\sqrt{3}}{9}$

Concept: Effective price $=MP\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})(1-\frac{d_3}{100})$.

1. $0.92\times 0.83\times 0.83=0.92\times 0.6889$.
2. $=0.633788\approx 63.38\%$ of marked price.

Trick: Discounts cannot simply be added ($42\%$ would give $58\%$) – always multiply the remaining fractions.

Answer: (C) $63.38\%$

Concept: Let a chair cost $c$ and a table $t$; solve the two equations.

1. $15c+3t=7800$ … (i), $14c+6t=9200$ … (ii).
2. $2\times$(i) $-$ (ii): $30c-14c=15600-9200\Rightarrow 16c=6400\Rightarrow c=400$.
3. $3t=7800-6000=1800\Rightarrow t=600$.
4. $17c+18t=6800+10800=17600$.

Trick: Doubling (i) makes the table terms equal ($6t$), so subtract directly.

Answer: (C) ₹17,600