Q&A: practice questions with answers

Concept: Write cosec and sec in terms of sin and cos to get $\tan A$.

1. $\frac{\sqrt{2}}{\sin A}=\frac{2}{\cos A}\Rightarrow \tan A=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$.
2. Triangle: opposite $1$, adjacent $\sqrt{2}$, hypotenuse $\sqrt{1+2}=\sqrt{3}$, so $\sec A=\frac{\sqrt{3}}{\sqrt{2}}$.
3. $\tan A\sec A=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}}{2}$.

Trick: ${\sec }^{2}A=1+{\tan }^{2}A=\frac{3}{2}$ gives sec without drawing.

Answer: (A) $\frac{\sqrt{3}}{2}$

Concept: Find total investment from the profit percentage, then share it in the given ratio.

1. $28\%$ of total $=84700\Rightarrow \text{Total}=\frac{84700\times 100}{28}=302500$.
2. Ratio sum $=21+9+20=50$.
3. Lalita $=\frac{9}{50}\times 302500=54450$.

Trick: Ratio sum 50 ⇒ $\frac{9}{50}=18\%$ of total: $0.18\times 302500$.

Answer: (D) ₹$54,450$

Concept: Net filling rate = filling rate − emptying rate.

1. Net rate $=\frac{1}{8}-\frac{1}{16}=\frac{1}{16}$ tank per hour (full in $16$ h).
2. Time for $\frac{3}{4}$ tank $=\frac{3}{4}\times 16=12$ hours.

Trick: LCM method: capacity 16 units; filler $+2$, emptier $-1$, net $1$/h; $\frac{3}{4}$ of 16 $=12$ units ⇒ $12$ h.

Answer: (D) $12$

Concept: Let the numbers be $x$ and $y$; equate and find $\frac{x}{y}$.

1. $\frac{22}{100}x=\frac{4}{9}y$
2. $\frac{x}{y}=\frac{4}{9}\times \frac{100}{22}=\frac{400}{198}=\frac{200}{99}$

Trick: $x:y$ = (fraction of 2nd) : (fraction of 1st) $=\frac{4}{9}:\frac{22}{100}$.

Answer: (D) $200:99$

Concept: Use $(a\pm b{)}^2=a^2+b^2\pm 2ab$.

1. $(a-b{)}^2=179-28=151$, $(a+b{)}^2=179+28=207$.
2. Since $a>b$ and $ab>0$: $\frac{a-b}{a+b}=\sqrt{\frac{151}{207}}$.

Trick: Ratio of squares first, then take one square root.

Answer: (C) $\sqrt{\frac{151}{207}}$

Concept: For 2 years, CI − SI = interest on the first year's interest.

1. SI for 1 year $=\frac{40}{2}=20$.
2. $\text{CI}-\text{SI}=42-40=2=20\times \frac{r}{100}\Rightarrow r=10\%$.
3. $P=\frac{20\times 100}{10}=200$.

Trick: $P=\frac{(\text{SI}/2{)}^2}{\text{CI}-\text{SI}}=\frac{{20}^2}{2}=200$.

Answer: (A) $200$

Concept: $\text{Average}=\frac{\text{Sum}}{\text{Number of values}}$.

1. Sum $=28+30+32+31+29=150$.
2. Average $=\frac{150}{5}={30}^{\circ }\text{C}$.

Trick: Deviations from 30: $-2,0,+2,+1,-1$ sum to $0$ ⇒ average $30$.

Answer: (B) ${30}^{\circ }\text{C}$

Concept: $\text{SP}=\text{CP}\times \frac{100\pm \text{gain/loss}\%}{100}$.

1. At $31\%$ loss: $\text{CP}=\frac{483\times 100}{69}=700$.
2. For $34\%$ profit: $\text{SP}=700\times \frac{134}{100}=938$.

Trick: Direct: $483\times \frac{134}{69}=7\times 134=938$.

Answer: (D) $938$

Concept: If bisectors of two adjacent angles R and S of a quadrilateral meet at T, then $\angle STR=\frac{\angle P+\angle Q}{2}$.

1. In $△STR$: $\angle STR={180}^{\circ }-\frac{\angle R+\angle S}{2}$.
2. $\angle R+\angle S={360}^{\circ }-(\angle P+\angle Q)$, so $\angle STR=\frac{\angle P+\angle Q}{2}$.
3. $\frac{\angle P+\angle Q}{2}=8^{\circ }\Rightarrow \angle P+\angle Q={16}^{\circ }$.
4. Ratio $7:9$ → 16 units total: $\angle P=7^{\circ }$, $\angle Q=9^{\circ }$; difference $=2^{\circ }$.

Trick: Bisector angle = half the sum of the other two angles, so the sum is $2\times 8^{\circ }={16}^{\circ }$; difference $=\frac{9-7}{16}\times {16}^{\circ }=2^{\circ }$.

Answer: (C) $2^{\circ }$

Concept: $\text{Required }\%=\frac{\text{first}}{\text{second}}\times 100$.

1. $\frac{1}{20}÷\frac{1}{30}=\frac{1}{20}\times 30=\frac{3}{2}$.
2. $\frac{3}{2}\times 100=150\%$.

Trick: For $\frac{1}{a}$ as % of $\frac{1}{b}$, just do $\frac{b}{a}\times 100=\frac{30}{20}\times 100=150\%$.

Answer: (A) $150\%$

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

1. 400% more than ${360}^{\circ }$ means $5\times {360}^{\circ }={1800}^{\circ }$.
2. $(n-2)\times 180=1800\Rightarrow n=12$.
3. Each interior angle $=\frac{{1800}^{\circ }}{12}={150}^{\circ }$.

Trick: Interior : exterior = $5:1$, so each exterior $=\frac{{180}^{\circ }}{6}={30}^{\circ }$ and interior $={150}^{\circ }$.

Answer: (C) 150

Concept: $a^3+b^3=(a+b{)}^3-3ab(a+b)$.

1. $a=0.54$, $b=0.46$, $a+b=1$.
2. $ab=0.54\times 0.46=0.2484$.
3. $a^3+b^3=1-3\times 0.2484=1-0.7452=0.2548$.

Trick: When $a+b=1$: $a^3+b^3=1-3ab$.

Answer: (D) $0.2548$

Concept: A number is divisible by 12 if it is divisible by both 3 and 4.

1. 2800: digit sum 10 – not divisible by 3.
2. 2349: odd – not divisible by 4.
3. 1264: digit sum 13 – not divisible by 3.
4. 2184: last two digits 84 (divisible by 4), digit sum 15 (divisible by 3) ✓; $2184=12\times 182$.

Trick: Test 3 first (digit sum) – it removes three options at once.

Answer: (D) 2184

Concept: For a fixed distance, time is inversely proportional to speed.

1. Normal time = $T$; at $\frac{1}{4}$ speed time $=4T$.
2. Extra time $=4T-T=3T=39$ min.
3. $T=13$ min.

Trick: Speed becomes $\frac{1}{n}$ → late by $(n-1)T$; here $3T=39$.

Answer: (C) 13

Concept: Combined average $=\frac{n_1{a}_1+n_2{a}_2}{n_1+n_2}$.

1. Total of 14 boys $=14\times 40.25=563.5$ kg.
2. Total of 4 boys $=4\times 38=152$ kg.
3. Average $=\frac{563.5+152}{18}=\frac{715.5}{18}=39.75$ kg.

Trick: Deviation method: $\frac{4\times (38-40.25)}{18}=-0.5$, so $40.25-0.5=39.75$.

Answer: (C) $39.75$

Concept: Take tank capacity as the LCM of the times; each tap's rate = capacity ÷ time.

1. Capacity $=\text{LCM}(20,50,60)=300$ units.
2. A + B + C $=\frac{300}{20}=15$, A $=\frac{300}{50}=6$, B $=\frac{300}{60}=5$ units/min.
3. C $=15-6-5=4$ units/min.
4. Time for C $=\frac{300}{4}=75$ min.

Trick: $\frac{1}{C}=\frac{1}{20}-\frac{1}{50}-\frac{1}{60}=\frac{15-6-5}{300}=\frac{1}{75}$.

Answer: (D) 75

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

1. $\frac{1}{3}\pi r^2\times 11=297\pi$.
2. $r^2=\frac{297\times 3}{11}=81\Rightarrow r=9$ cm.
3. Diameter $=2r=18$ cm.

Trick: $297=11\times 27$, so $r^2=27\times 3=81$ instantly.

Answer: (D) 18 cm

Concept: Required number $=\text{LCM}+7$.

1. $16=2^4$, $20=2^2\times 5$, $10=2\times 5$, $27=3^3$.
2. $\text{LCM}=2^4\times 5\times 3^3=2160$.
3. Number $=2160+7=2167$.

Trick: Check options: subtract 7 and test divisibility by 27 and 16 – only $2160$ works.

Answer: (A) 2167

Concept: Curved surface area of a hemisphere $=2\pi r^2$; inside + outside $=4\pi r^2$.

1. Area $=4\times \frac{22}{7}\times 3.5\times 3.5=154{\text{cm}}^2$.
2. Rate ₹17 per $10{\text{cm}}^2$ = ₹1.7 per ${\text{cm}}^2$.
3. Cost $=154\times 1.7=$ ₹261.8.

Trick: $4\pi r^2$ with $r=3.5$ is $154$ (same as the area of a circle of radius 7).

Answer: (B) ₹261.8

Concept: If A is $r\%$ more than B, B is less than A by $\frac{r}{100+r}\times 100\%$.

1. $r=51$.
2. $\frac{51}{151}\times 100=33.77\%$ (approx).

Trick: Let B = 100, A = 151; difference 51 out of 151.

Answer: (B) $33.77\%$