Q&A: practice questions with answers

Concept: Mode = the observation that occurs most frequently.

1. Frequencies: 3 → 4 times, 5 → 3 times, 4, 6, 8 → 2 times each, 2 and 7 → once.
2. Highest frequency is 4, for the value 3.

Trick: Tally only the repeating values – the one with the most tallies is the mode.

Answer: (D) 3

Concept: Area $=l\times b$, Perimeter $=2(l+b)$.

1. Let longer side $=L$, shorter $=L-18$.
2. $L(L-18)=6\times 2(2L-18)=24L-216$
3. $L^2-42L+216=0\Rightarrow (L-36)(L-6)=0$
4. $L=6$ gives a negative side, so $L=36$.

Trick: Option check: $L=36$, $b=18$: area $648$, perimeter $108$, $6\times 108=648$ ✓.

Answer: (D) 36

Concept: After successive discounts, SP $=\text{MP}\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})(1-\frac{d_3}{100})$.

1. $0.92\times 0.84=0.7728$
2. $0.7728\times 0.97=0.749616$
3. Effective price $\approx 74.96\%$ of MP

Trick: Never add discounts ($27\%$ would give $73\%$) – multiply the remaining fractions.

Answer: (B) $74.96\%$

Concept: Leak rate = pump rate − net rate (work per hour).

1. $7\frac{1}{5}=\frac{36}{5}$ hours, so net rate $=\frac{5}{36}$ per hour.
2. Pump rate $=\frac{1}{6}=\frac{6}{36}$
3. Leak rate $=\frac{6}{36}-\frac{5}{36}=\frac{1}{36}$ → leak empties the tank in 36 hours.

Trick: Leak time $=\frac{a\times b}{b-a}=\frac{6\times \frac{36}{5}}{\frac{36}{5}-6}=36$.

Answer: (C) 36 hours

Concept: In similar triangles, ratio of perimeters = ratio of corresponding sides.

1. BC corresponds to QR: $\frac{BC}{QR}=\frac{4}{8}=\frac{1}{2}$
2. $\frac{\text{Perimeter of }△ABC}{32}=\frac{1}{2}$
3. Perimeter of $△ABC=16\text{cm}$

Trick: Perimeters scale like sides; areas scale like the square of sides.

Answer: (C) 16 cm

Concept: Everything except rent forms the remaining $24\%$ of the salary.

1. Remaining part $=100\%-76\%=24\%$
2. $627+3947+712=5286$
3. $24\%$ of salary $=5286\Rightarrow$ salary $=\frac{5286\times 100}{24}=22025$

Trick: Unit digit check: $0.24\times 22025=5286$ exactly.

Answer: (B) ₹22,025

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

1. $18-8\sqrt{5}=2(9-4\sqrt{5})$
2. $ab=2(9+4\sqrt{5})(9-4\sqrt{5})=2(81-80)=2$
3. Mean proportional $=\sqrt{2}$

Trick: Spot the conjugate pair: $(9+4\sqrt{5})(9-4\sqrt{5})=1$.

Answer: (D) $\sqrt{2}$

Concept: Surface area of sphere $=4\pi r^2$, volume $=\frac{4}{3}\pi r^3$.

1. $4\pi r^2=25\pi \Rightarrow r^2=\frac{25}{4}\Rightarrow r=\frac{5}{2}$
2. $V=\frac{4}{3}\pi \times \frac{125}{8}$
3. $=\frac{125\pi }{6}{\text{cm}}^3$

Trick: $V=\frac{S^{3/2}}{6\sqrt{\pi }}$; with $S=25\pi$: $\frac{125\pi \sqrt{\pi }}{6\sqrt{\pi }}=\frac{125\pi }{6}$.

Answer: (D) $\frac{125\pi }{6}{\text{cm}}^3$

Concept: ${\sec }^2\theta -{\tan }^2\theta =1\Rightarrow (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$.

1. $\sec \theta -\tan \theta =\frac{1}{2}$
2. Adding: $2\sec \theta =2+\frac{1}{2}=\frac{5}{2}\Rightarrow \sec \theta =\frac{5}{4}$
3. $3\sec \theta +4=\frac{15}{4}+4=\frac{31}{4}$

Trick: If $\sec \theta +\tan \theta =x$, then $\sec \theta =\frac{x^2+1}{2x}=\frac{5}{4}$.

Answer: (A) $\frac{31}{4}$

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Let CP $=100$; MP $=152$
2. For $14\%$ gain, SP $=114$
3. Discount $=\frac{152-114}{152}\times 100=\frac{38}{152}\times 100=25\%$

Trick: $\frac{38}{152}=\frac{1}{4}$ → $25\%$.

Answer: (D) $25\%$

Concept: Write both situations in terms of the original cost price $x$.

1. Original SP $=0.78x$
2. New CP $=0.95x$, new SP $=1.15\times 0.95x=1.0925x$
3. $1.0925x-0.78x=0.3125x=935$
4. $x=\frac{935}{0.3125}=2992$

Trick: $0.3125=\frac{5}{16}$, so $x=\frac{935\times 16}{5}=2992$.

Answer: (A) ₹2,992

Concept: If $\frac{x}{a}=\frac{b}{x}$ then $x^2=ab$, i.e. $x=\sqrt{ab}$.

1. $\frac{x}{48}=\frac{243}{x}\Rightarrow x^2=48\times 243$.
2. $48\times 243=11664$.
3. $x=\sqrt{11664}=108$ (positive value).

Trick: $48\times 243=(16\times 3)\times (81\times 3)=16\times 81\times 9$, so $x=4\times 9\times 3=108$.

Answer: (D) 108

Concept: For a fixed distance, speed $\propto \frac{1}{\text{time}}$.

1. Distance $=92\times 6=552\text{km}$.
2. New speed $=\frac{552}{3}=184\text{km/h}$.
3. Increase $=184-92=92\text{km/h}$.

Trick: Time halves → speed doubles → increase equals the original speed, 92.

Answer: (C) 92

Concept: Let Sujatha = $S$, Vanita = $V$ (present ages) and form two equations.

1. $6S=3V+6\Rightarrow V=2S-2$.
2. After 7 years: $4(V+7)=6(S+7)-6\Rightarrow 4V=6S+8\Rightarrow 2V=3S+4$.
3. Substitute: $2(2S-2)=3S+4\Rightarrow S=8$, $V=14$.
4. $k=V-S=14-8=6$.

Trick: Check: $6\times 8=48=3\times 14+6$ ✓ and $4\times 21=84=6\times 15-6$ ✓.

Answer: (B) 6

Concept: One-day work of B = (A + B)'s one-day work − A's one-day work.

1. $\frac{1}{B}=\frac{1}{6}-\frac{1}{7}=\frac{7-6}{42}=\frac{1}{42}$.
2. So B alone takes 42 days.

Trick: B's time $=\frac{6\times 7}{7-6}=42$ days.

Answer: (B) 42

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

1. $\sqrt{0.09\times 16}=\sqrt{1.44}$.
2. $\sqrt{1.44}=1.2$.

Trick: $\sqrt{0.09}\times \sqrt{16}=0.3\times 4=1.2$.

Answer: (C) 1.2

Concept: If every number is increased by the same amount, the average increases by that amount.

1. Old average = 18.
2. Each number + 5 → new average $=18+5=23$.
3. Check: sum $=6\times 18=108$, new sum $=108+30=138$, $\frac{138}{6}=23$.

Trick: Add / subtract / multiply every item by $k$ → do the same to the average.

Answer: (C) 23

Concept: Upstream speed $=b-s$, downstream speed $=b+s$.

1. Upstream speed $=\frac{59.4}{18/60}=198\text{km/h}$.
2. $b:s=3:2$ → upstream $=3-2=1$ part $=198$, downstream $=3+2=5$ parts $=990\text{km/h}$.
3. Time $=\frac{29.3}{198}+\frac{51.5}{990}=\frac{146.5+51.5}{990}=\frac{198}{990}=0.2\text{h}$.

Trick: Write both times over 990: upstream distance × 5 + downstream distance $=146.5+51.5=198$.

Answer: (D) 0.2

Concept: Supplementary angles add up to ${180}^{\circ }$.

1. Let smaller angle $=x$, larger $=x+{60}^{\circ }$.
2. $x+x+{60}^{\circ }={180}^{\circ }\Rightarrow 2x={120}^{\circ }$.
3. $x={60}^{\circ }$ (larger $={120}^{\circ }$).

Trick: Smaller $=\frac{{180}^{\circ }-{60}^{\circ }}{2}={60}^{\circ }$.

Answer: (C) ${60}^{\circ }$

Concept: For $a{x}^3+b{x}^2+cx+d=0$, sum of roots $=-\frac{b}{a}$.

1. Here $a=2$, $b=7$.
2. Sum of roots $=-\frac{7}{2}$.

Trick: Sum $=-\frac{b}{a}$, product $=-\frac{d}{a}$ for a cubic; options $\frac{9}{2}$ or $-\frac{2}{7}$ come from using wrong coefficients.

Answer: (C) $\frac{-7}{2}$