Q&A: practice questions with answers

Concept: Covered distance $d$ is $\frac{7}{9}$ of the remaining $54-d$, so the 54 km splits in the ratio $7:9$.

1. $d:(54-d)=7:9$ → total $16$ parts $=54\text{km}$.
2. $d=\frac{7}{16}\times 54=23.625\text{km}$.
3. Time $=1\text{ h }15\text{ min}=1.25\text{h}$.
4. Speed $=\frac{23.625}{1.25}=18.9\text{km/h}$.

Trick: "$\frac{a}{b}$ of the remaining" means the parts are $a:b$ – covered share is $\frac{a}{a+b}$ of the total.

Answer: (B) 18.9

Concept: The largest container that exactly empties both quantities is their HCF.

1. $168=2^3\times 3\times 7$.
2. $234=2\times 3^2\times 13$.
3. Common factors: $2\times 3=6$.
4. Maximum container volume $=6$ litres.

Trick: "Maximum size that divides both exactly" = HCF; "minimum that both divide" = LCM.

Answer: (B) 6

Concept: $\text{TSA}=2(lb+bh+hl)$.

1. Let $l=10k$, $b=17k$, $h=14k$.
2. $lb+bh+hl=170k^2+238k^2+140k^2=548k^2$.
3. $2\times 548k^2=1096k^2=1096\Rightarrow k=1$.
4. Length $=10k=10\text{cm}$.

Trick: When TSA comes out equal to the coefficient ($1096$), $k=1$ immediately.

Answer: (B) 10

Concept: If $x$ grows for $n$ years and $y$ for $n+1$ years at rate $r$ to equal amounts, then $x=y(1+\frac{r}{100})$.

1. Shan gets $x$, Piyush $1612-x$.
2. $x(1.08{)}^{19}=(1612-x)(1.08{)}^{20}\Rightarrow x=1.08(1612-x)$.
3. $x:(1612-x)=1.08:1=27:25$.
4. $x=\frac{27}{52}\times 1612=837$.

Trick: One extra year → ratio $(100+r):100=108:100=27:25$; split 1612 in 52 parts of 31.

Answer: (D) ₹837

Concept: For a pure recurring part, $0.\overline{ab}=\frac{ab}{99}$.

1. $8.2727\dots =8+0.\overline{27}$.
2. $0.\overline{27}=\frac{27}{99}=\frac{3}{11}$.
3. $8+\frac{3}{11}=\frac{88+3}{11}=\frac{91}{11}$.

Trick: Check: $91÷11=8.2727\dots$ ✓ – dividing the options is the fastest check.

Answer: (A) $\frac{91}{11}$

Concept: $\text{Expenditure}=\text{Income}-\text{Savings}$.

1. Incomes $7k$ and $8k$.
2. $\frac{7k-15000}{8k-9000}=\frac{11}{16}\Rightarrow 112k-240000=88k-99000$.
3. $24k=141000\Rightarrow k=5875$.
4. Expenditures: $41125-15000=26125$ and $47000-9000=38000$; total $=64125$.

Trick: Total expenditure = total income − total savings $=15k-24000=88125-24000$.

Answer: (C) 64,125

Concept: Difference of ratio parts × value of one part = actual difference.

1. Ratio A : B : C : D $=6:3:7:5$, total $21$ parts.
2. $D-B=5-3=2$ parts $=400$ → 1 part $=200$.
3. Total $=21\times 200=4200$.

Trick: Total $=\frac{400}{2}\times 21$ in one line.

Answer: (C) 4200

Concept: Number of pipes and time are inversely proportional: $n_1{t}_1=n_2{t}_2$.

1. Total work $=23\times 29=667$ pipe-minutes.
2. With 47 pipes: $t=\frac{667}{47}$.
3. $t=14.191\dots \approx 14.19$ minutes.

Trick: More pipes → less time; so the answer must be below 29 and about half of it ($47\approx 2\times 23$).

Answer: (C) 14.19

Concept: Follow BODMAS: division before addition and subtraction.

1. $\frac{76}{2}=38$.
2. $81+38+42-86$.
3. $=161-86=75$.

Trick: Add all positives first ($81+38+42=161$), then subtract once.

Answer: (C) 75

Concept: Net change for successive changes $a\%$ and $b\%$: $a+b+\frac{ab}{100}$.

1. $a=-16$, $b=+11$.
2. Net $=-16+11+\frac{(-16)(11)}{100}=-5-1.76=-6.76\%$.
3. So the final salary is $6.76\%$ lower.

Trick: Check: $0.84\times 1.11=0.9324$ → $100-93.24=6.76\%$.

Answer: (A) $6.76\%$

Concept: For similar triangles, ratio of areas = square of the ratio of corresponding sides.

1. Side ratio $=15:14$.
2. Area ratio $={15}^2:{14}^2$.
3. $=225:196$.

Trick: Areas go as $(\text{side}{)}^2$, volumes as $(\text{side}{)}^3$.

Answer: (C) $225:196$

Concept: Sum of exterior angles of any polygon $={360}^{\circ }$; sum of interior angles $=(n-2)\times {180}^{\circ }$.

1. 250% more means interior sum $=3.5\times {360}^{\circ }={1260}^{\circ }$.
2. $(n-2)\times 180=1260\Rightarrow n-2=7\Rightarrow n=9$.
3. Each interior angle $x=\frac{1260}{9}={140}^{\circ }$.

Trick: Each exterior angle $=\frac{360}{9}={40}^{\circ }$, so interior $=180-40={140}^{\circ }$.

Answer: (A) $140$

Concept: Required total SP = CP + desired gain; the remainder must cover the shortfall.

1. Total CP ₹450; for 20% gain total SP $=450\times 1.2=540$.
2. $\frac{1}{3}$ goods: CP $=150$, sold at 10% loss → SP $=135$.
3. Remainder: CP $=300$, needed SP $=540-135=405$.
4. Gain $\%=\frac{105}{300}\times 100=35\%$.

Trick: Weighted: $\frac{1}{3}(-10)+\frac{2}{3}g=20\Rightarrow g=35$.

Answer: (D) $35\%$

Concept: Use ${\sec }^2\theta {\text{cosec}}^2\theta ={\sec }^2\theta +{\text{cosec}}^2\theta$ (since $\frac{1}{{\sin }^2\theta {\cos }^2\theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }$).

1. ${\sec }^2\theta +{\text{cosec}}^2\theta =(1+{\tan }^2\theta )+(1+{\cot }^2\theta )=2+{\tan }^2\theta +{\cot }^2\theta$.
2. Expression $=3+{\tan }^2\theta +{\cot }^2\theta -2-{\tan }^2\theta -{\cot }^2\theta$.
3. $=1$.

Trick: Put $\theta ={45}^{\circ }$: $3+1+1-2\times 2=1$.

Answer: (A) $1$

Concept: Solve $5m^2+18m+16=0$ by splitting the middle term.

1. $5\times 16=80$; $10+8=18$, $10\times 8=80$.
2. $5m^2+10m+8m+16=5m(m+2)+8(m+2)=(5m+8)(m+2)$.
3. $m=-\frac{8}{5}$ or $m=-2$.

Trick: All coefficients positive ⇒ both roots negative; sum $=-\frac{18}{5}$, product $=\frac{16}{5}$.

Answer: (D) $-\frac{8}{5}$ and $-2$

Concept: Square step by step: ${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$ and ${(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2$.

1. $x^2+\frac{1}{x^2}=3^2+2=11$.
2. $x^4+\frac{1}{x^4}={11}^2-2=121-2=119$.

Trick: Minus-form: add 2 first, then subtract 2: $(9+2{)}^2-2$.

Answer: (D) $119$

Concept: Combine the two variations: $X\propto A^2$ and $A\propto \frac{1}{Y}$ give $X\propto \frac{1}{Y^2}$, i.e. $X{Y}^2=k$.

1. $k=97\times 6^2=97\times 36$.
2. When $Y=2$: $X=\frac{97\times 36}{4}=97\times 9$.
3. $X=873$.

Trick: Y becomes $\frac{1}{3}$, so X becomes $3^2=9$ times.

Answer: (A) $873$

Concept: Expenditure = Income − Savings.

1. Savings 40% ⇒ expenditure $=60\%$ of income.
2. $0.6\times I=960\Rightarrow I=\frac{960}{0.6}=1600$.

Trick: $60\%=960\Rightarrow 10\%=160\Rightarrow 100\%=1600$.

Answer: (C) ₹1,600

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$.

1. Onward time $=\frac{d}{11}+\frac{2d}{44}=\frac{3d}{22}$.
2. Return 3d at v: time $=\frac{3d}{v}$. Total distance $=6d$.
3. $\frac{6d}{\frac{3d}{22}+\frac{3d}{v}}=34\Rightarrow \frac{1}{22}+\frac{1}{v}=\frac{1}{17}$.
4. $\frac{1}{v}=\frac{22-17}{374}=\frac{5}{374}\Rightarrow v=74.8\text{km/h}$.

Trick: Onward average $=\frac{3d}{3d/22}=22$; equal legs ⇒ $\frac{2\times 22v}{22+v}=34$.

Answer: (C) $74.8$

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

1. $\frac{1}{3}\pi r^2\times 30=160\pi$.
2. $10r^2=160\Rightarrow r^2=16\Rightarrow r=4\text{cm}$.
3. Diameter $=2r=8\text{cm}$.

Trick: Cancel π first: $r^2=\frac{3\times 160}{30}=16$.

Answer: (C) 8 cm