Q&A: practice questions with answers

Concept: Total time $=$ sum of times of each stretch. In the 30-minute slow stretch the car covers $30\times \frac{1}{2}=15$ km.

1. Let distance $=d$ km. Onward time $=\frac{d}{60}$.
2. Return: $\frac{1}{2}$ h at 30 km/h (15 km) $+$ $\frac{d-15}{80}$ h at 80 km/h.
3. $\frac{d}{60}+\frac{1}{2}+\frac{d-15}{80}=7\Rightarrow \frac{4d+3d}{240}=6.5+\frac{15}{80}=\frac{107}{16}$.
4. $7d=\frac{107\times 240}{16}=1605\Rightarrow d=\frac{1605}{7}=229\frac{2}{7}$ km.

Trick: Convert the slow stretch into a fixed distance (15 km) first; the rest is a plain two-speed equation.

Answer: (A) $229\frac{2}{7}$ km

Concept: Sum $=$ mean $\times$ count; the missing value is total minus the known parts.

1. Total of 10 $=10\times 42=420$.
2. Known three: $30+38+50=118$; six others: $6\times 45=270$.
3. Missing $=420-118-270=32$.

Trick: Work with totals, never with averages directly.

Answer: (C) 32

Concept: $A=P{(1+\frac{r}{100})}^2$.

1. ${(1+\frac{r}{100})}^2=\frac{21218}{20000}=1.0609$.
2. ${1.03}^2=1.0609$ → $r=3\%$.

Trick: 2-year CI $\%=2r+\frac{r^2}{100}$; for $r=3$: $6+0.09=6.09\%$ of 20000 $=1218$ ✓.

Answer: (A) $3\%$

Concept: Convert mixed fractions to decimals and add step by step.

1. $3\frac{1}{4}=3.25$, $5\frac{1}{8}=5.125$.
2. $3.46+3.25=6.71$; $6.71-5.125=1.585$.
3. $1.585+4.25+0.165=6.000$.

Trick: Pair decimals that complete each other: $3.46+4.25+0.165=7.875$ and $3.25-5.125=-1.875$ → $6$.

Answer: (C) 6

Concept: $SI=\frac{P\times R\times T}{100}$.

1. $SI=\frac{8000\times 5\times 3}{100}=80\times 15=1200$.

Trick: $5\%$ for 3 years $=15\%$ of principal; $15\%$ of 8000 $=1200$.

Answer: (A) ₹1,200

Concept: $\tan \theta =\frac{\text{height}}{\text{distance}}$.

1. $\tan {45}^{\circ }=\frac{h}{40}$.
2. $\tan {45}^{\circ }=1\Rightarrow h=40$ m.

Yaad rakhein: At ${45}^{\circ }$ elevation, height = distance (isosceles right triangle).

Answer: (B) 40 m

Concept: Exterior angle $={180}^{\circ }-$ interior angle, and $n=\frac{{360}^{\circ }}{\text{exterior angle}}$.

1. Exterior angle $={180}^{\circ }-{150}^{\circ }={30}^{\circ }$.
2. $n=\frac{{360}^{\circ }}{{30}^{\circ }}=12$.

Yaad rakhein: ${108}^{\circ }\to 5$, ${120}^{\circ }\to 6$, ${135}^{\circ }\to 8$, ${144}^{\circ }\to 10$, ${150}^{\circ }\to 12$ sides.

Answer: (D) 12 sides

Concept: Simplify nested radicals from the innermost bracket outward.

1. $419+157=576$, $\sqrt{576}=24$.
2. $3225+24=3249={57}^2$, $\sqrt{3249}=57$.
3. $1624+57=1681={41}^2$, $\sqrt{1681}=41$.

Yaad rakhein: ${41}^2=1681$, ${57}^2=3249$, ${24}^2=576$ — memorise squares up to 60.

Answer: (D) 41

Concept: Total surface area of a closed cylinder $=2\pi r(r+h)$; red part $=80\%$ of it.

1. TSA $=2\times \frac{22}{7}\times 14\times (14+10)=88\times 24=2112{\text{cm}}^2$.
2. Red $=80\%$ of $2112=0.8\times 2112=1689.6{\text{cm}}^2$.

Trick: With $r=14$, $2\pi r=88$ exactly — then just multiply by $(r+h)$.

Answer: (C) $1689.6{\text{cm}}^2$

Concept: Average salary $=\frac{\text{total salary}}{\text{total staff}}$.

1. Total $=10\times 45000+30\times 25000=450000+750000=1200000$.
2. Average $=\frac{1200000}{40}=30000$.

Trick: Ratio 1:3 → average lies $\frac{1}{4}$ of the gap (20000) above 25000 → $25000+5000=30000$.

Answer: (B) ₹30,000

Concept: Required number $=$ (largest 4-digit multiple of LCM) $+3$.

1. $\text{LCM}(12,13,7)=1092$.
2. $9999÷1092=9$ rem $171$ → largest multiple $=9999-171=9828$.
3. Number $=9828+3=9831$.

Trick: Options check: $9831-3=9828=9\times 1092$ ✓ (divisible by 7, 12 and 13).

Answer: (D) 9831

Concept: Speed $=\frac{\text{distance}}{\text{time}}$; m/s to km/h multiply by $\frac{18}{5}$.

1. Speed $=\frac{40}{8}=5$ m/s.
2. $5\times \frac{18}{5}=18$ km/h.

Yaad rakhein: $5$ m/s $=18$ km/h, $10$ m/s $=36$ km/h.

Answer: (A) 18 km/hr

Concept: For two numbers, $\text{HCF}\times \text{LCM}=a\times b$.

1. $b=\frac{36\times 7560}{540}$.
2. $\frac{7560}{540}=14$ → $b=36\times 14=504$.

Trick: The other number must be a multiple of the HCF 36 — only 396, 432 and 504 are; check $504=36\times 14$ with LCM $7560$ ✓.

Answer: (D) 504

Concept: Area of a rhombus $=\frac{1}{2}{d}_1{d}_2$ (diagonals bisect each other at right angles).

1. Area $=\frac{1}{2}\times 19\times 29$.
2. $19\times 29=551$ → area $=275.5{\text{cm}}^2$.

Trick: $19\times 29=(24-5)(24+5)=576-25=551$.

Answer: (C) $275.5{\text{cm}}^2$

Concept: $a:c=\frac{a}{b}\times \frac{b}{c}$.

1. $\frac{a}{c}=\frac{12}{13}\times \frac{4}{5}=\frac{48}{65}$.
2. $a:b:c=48:52:65$ → $a:c=48:65$.

Trick: Multiply the first terms and the last terms: $a:c=(12\times 4):(13\times 5)$.

Answer: (D) $48:65$

Concept: Apply the multipliers one after another.

1. $600\times 1.20=720$.
2. $720\times 0.75=540$.

Trick: Net change $=20-25-\frac{20\times 25}{100}=-10\%$ → $600\times 0.9=540$.

Answer: (D) ₹540

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

1. $24+(-5)+(-19)=0$.
2. Sum $=3\times 24\times (-5)\times (-19)=3\times 24\times 95=6840$.

Trick: Whenever three cubes are asked, first check if the bases add to zero.

Answer: (A) 6840

Concept: The buyer pays for 1 kg at the discounted marked price but receives only 900 g; profit $=$ amount received $-$ cost of 900 g.

1. Marked price of 1 kg $=60\times 1.2=72$; after $10\%$ discount $=64.80$.
2. Cost of 900 g actually given $=0.9\times 60=54$.
3. Profit $=64.80-54=10.80$.

Trick: Profit $\%=\frac{1.2\times 0.9}{0.9}-1=20\%$ on cost 54 → $10.80$.

Answer: (A) ₹10.80

Concept: Form a linear equation.

1. Sister $=x$, Mary $=x+3$; $x+x+3=27\Rightarrow x=12$.
2. Mary $=15$ years.

Trick: Sum 27, difference 3 → larger $=\frac{27+3}{2}=15$.

Answer: (A) 15

Concept: Let the first pipe take $x$ h; then second $=x-5$, third $=x-9$ and $\frac{1}{x}+\frac{1}{x-5}=\frac{1}{x-9}$.

1. $(2x-5)(x-9)=x(x-5)\Rightarrow x^2-18x+45=0\Rightarrow x=15$ (reject 3).
2. Times: $15,10,6$ h.
3. Second + third: $\frac{1}{10}+\frac{1}{6}=\frac{4}{15}$ → $\frac{15}{4}=3\frac{3}{4}$ h.

Trick: The classic triple $(15,10,6)$: $\frac{1}{15}+\frac{1}{10}=\frac{1}{6}$ — remember it.

Answer: (B) $3\frac{3}{4}$ hours