Q&A: practice questions with answers

Concept: When equal quantities are mixed, make each mixture the same total (LCM of $3$ and $4$ parts = $12$ units) and add the parts.

1. Mixture A: $1+2=3$ parts → take $12$ units: copper $4$, tin $8$.
2. Mixture B: $1+3=4$ parts → $12$ units: copper $3$, tin $9$.
3. Mixture C: copper $4+3=7$, tin $8+9=17$.

Trick: Equal quantities → use fractions directly: copper $=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$, tin $=\frac{2}{3}+\frac{3}{4}=\frac{17}{12}$.

Answer: (C) $7:17$

Concept: $V=\frac{4}{3}\pi r^3$ and surface area $S=4\pi r^2$.

1. $\frac{4}{3}\times \frac{22}{7}\times r^3=38808$
2. $r^3=\frac{38808\times 21}{88}=9261$, so $r=21\text{cm}$
3. $S=4\times \frac{22}{7}\times {21}^2=88\times 63=5544{\text{cm}}^2$

Trick: Shortcut: $S=\frac{3V}{r}=\frac{3\times 38808}{21}=5544$.

Answer: (D) $5544{\text{cm}}^2$

Concept: Diagonals of a rhombus bisect each other at right angles; area $=\frac{1}{2}{d}_1{d}_2$.

1. Half of the given diagonal $=8\text{cm}$.
2. Half of the other diagonal $=\sqrt{{17}^2-8^2}=\sqrt{225}=15$, so $d_2=30\text{cm}$.
3. Area $=\frac{1}{2}\times 16\times 30=240{\text{cm}}^2$.

Trick: $(8,15,17)$ is a Pythagorean triplet – spot it instantly.

Answer: (C) $240{\text{cm}}^2$

Concept: Profit = SP − CP. Here profit on $42$ m = SP of $7$ m.

1. SP of $42$ m − CP of $42$ m = SP of $7$ m
2. So CP of $42$ m = SP of $35$ m.
3. Profit $\%=\frac{7}{35}\times 100=20\%$

Trick: Gain = SP of $x$ out of $n$ → Profit $\%=\frac{x}{n-x}\times 100=\frac{7}{35}\times 100$.

Answer: (C) $20\%$

Concept: Successive discounts multiply: SP = MP $\times (1-d_1)(1-d_2)$.

1. SP $=\text{MP}\times 0.75\times 0.85=0.6375\text{MP}$
2. $0.6375\text{MP}=12750$
3. MP $=\frac{12750}{0.6375}=20000$

Trick: Check options: $20000\times \frac{3}{4}=15000$, $15000\times 0.85=12750$ ✓

Answer: (A) ₹20,000

Concept: Leak rate = (filling rate of pipes) − (net rate with leak).

1. Both pipes: $\frac{1}{12}+\frac{1}{16}=\frac{7}{48}$, time $=\frac{48}{7}\text{h}$.
2. With leak: $\frac{48}{7}+\frac{3}{2}=\frac{117}{14}\text{h}$, rate $=\frac{14}{117}$.
3. Leak rate $=\frac{7}{48}-\frac{14}{117}=\frac{147}{5616}$
4. Leak time $=\frac{5616}{147}=38\frac{10}{49}\text{h}$

Trick: Leak time $=\frac{T_1{T}_2}{T_2-T_1}=\frac{\frac{48}{7}\times \frac{117}{14}}{\frac{3}{2}}$.

Answer: (B) $38\frac{10}{49}\text{h}$

Concept: Amounts of consecutive years differ by one year's interest on the previous amount.

1. Interest in 3rd year $=1260-1200=60$
2. Rate $=\frac{60}{1200}\times 100=5\%$

Trick: $r=\frac{A_3-A_2}{A_2}\times 100$.

Answer: (A) $5\%$

Concept: For $a{\csc }^2\theta +b{\sin }^2\theta$ (with $a\le b$), AM ≥ GM gives minimum $2\sqrt{ab}$.

1. $16{\csc }^2\theta +25{\sin }^2\theta \ge 2\sqrt{16\times 25}=40$
2. Equality when $25{\sin }^4\theta =16$, i.e. ${\sin }^2\theta =\frac{4}{5}\le 1$ – possible.

Trick: Min $=2\sqrt{ab}$ only if ${\sin }^2\theta =\sqrt{\frac{a}{b}}\le 1$.

Answer: (A) $40$

Concept: Growth $\%=\frac{\text{New}-\text{Old}}{\text{Old}}\times 100$.

1. Total 2015 $=650+450+1008=2108$
2. Total 2020 $=830+501+1330=2661$
3. Growth $=\frac{553}{2108}\times 100\approx 26.23\%$

Trick: $\frac{553}{2108}$ is a little above $\frac{1}{4}$ → only $26.23\%$ fits.

Answer: (C) $26.23$ percent

Concept: Draw a right triangle: $\cos A=\frac{\text{base}}{\text{hyp}}$, $\cot A=\frac{\text{base}}{\text{perp}}$.

1. Base $=1$, hypotenuse $=11$.
2. Perpendicular $=\sqrt{121-1}=\sqrt{120}=2\sqrt{30}$
3. $\cot A=\frac{1}{2\sqrt{30}}$

Trick: $\cot A=\frac{\cos A}{\sqrt{1-{\cos }^{2}A}}$.

Answer: (C) $\frac{1}{2\sqrt{30}}$

Concept: Average per year = total production of all brands ÷ number of years.

1. P: $40+20+40+20=120$
2. Q: $50+25+30+30=135$
3. R: $30+30+20+25=105$
4. Average $=\frac{360}{4}=90$ million tons

Trick: Add year-wise: $120,75,90,75$ → total $360$.

Answer: (A) $90$

Concept: ${\csc }^{2}A-{\cot }^{2}A=1\Rightarrow (\csc A+\cot A)(\csc A-\cot A)=1$.

1. $\csc A-\cot A=\frac{1}{3}$
2. Adding: $2\csc A=\frac{10}{3}$ → $\csc A=\frac{5}{3}$; subtracting: $\cot A=\frac{4}{3}$
3. $\cos A=\frac{\cot A}{\csc A}=\frac{4}{5}$

Trick: If $\csc A+\cot A=x$, then $\cos A=\frac{x^2-1}{x^2+1}=\frac{8}{10}$.

Answer: (D) $\frac{4}{5}$

Concept: $\frac{M_1{D}_1}{W_1}=\frac{M_2{D}_2}{W_2}$.

1. $\frac{36\times 48}{25}=\frac{45\times D}{60}$
2. $D=\frac{36\times 48\times 60}{25\times 45}=\frac{103680}{1125}=92.16$ days

Trick: Cancel first: $\frac{36}{45}=\frac{4}{5}$, so $D=\frac{4\times 48\times 60}{5\times 25}=92.16$.

Answer: (B) $92.16$

Concept: A number leaving the same remainder $r$ with $6,8,9$ is of the form $72k+r$ (LCM $=72$).

1. HCF of $6,8,9=1$, so $r=1$.
2. Need $N=72k+1$.
3. $433=72\times 6+1$ ✓; $291=72\times 4+3$, $575=72\times 7+71$, $506=72\times 7+2$ ✗

Trick: Subtract $1$ and test divisibility by $8$ and $9$: $432$ ✓.

Answer: (D) $433$

Concept: If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

1. $a=1+x$, $b=1-x$, $c=-2$: $a+b+c=0$.
2. Value $=3(1+x)(1-x)(-2)=-6(1-x^2)$

Trick: Check $x=0$: $1+1-8=-6$ ✓

Answer: (C) $-6(1-x^2)$

Concept: Convert mixed fractions; apply ÷ and × from left to right.

1. LHS $=\frac{4}{3}\times \frac{5}{2}\times \frac{x}{5}=\frac{2x}{3}$
2. RHS $=\frac{5}{4}\times \frac{2}{3}\times 6=5$
3. $\frac{2x}{3}=5\Rightarrow x=7.5$

Trick: Division by $\frac{1}{6}$ = multiply by $6$.

Answer: (B) $7.5$

Concept: New total − old total = weight of the new student.

1. Old total $=49\times 53=2597\text{kg}$
2. New total $=50\times 52.5=2625\text{kg}$
3. New student $=2625-2597=28\text{kg}$

Trick: New weight $=53-50\times 0.5=28$.

Answer: (C) $28\text{kg}$

Concept: Winner $75\%$, loser $25\%$; margin $=50\%$ of total votes.

1. $50\%$ of total $=2500$ → total $=5000$
2. Loser $=25\%$ of $5000=1250$

Trick: Loser's share = half of margin here ($25\%$ vs $50\%$): $\frac{2500}{2}=1250$.

Answer: (C) $1250$

Concept: Go step by step: enrolled → polled → valid → margin.

1. Polled $=70\%$ of $E$; valid $=80\%$ of that $=0.56E$
2. Margin $=(64\%-36\%)$ of valid $=0.28\times 0.56E=0.1568E$
3. $0.1568E=3136\Rightarrow E=20000$

Trick: Check option: $20000\to 14000\to 11200$; $28\%$ of $11200=3136$ ✓

Answer: (A) $20000$

Concept: Mean proportional of $a$ and $b$ $=\sqrt{ab}$.

1. $\sqrt{x^{3}y\times x{y}^3}=\sqrt{x^4{y}^4}$
2. $=x^2{y}^2$

Yaad rakhein: Mean proportional = geometric mean $\sqrt{ab}$.

Answer: (A) $x^2{y}^2$