Q&A: practice questions with answers

Concept: Take total work = LCM of the days; add the three pair-rates to get twice the rate of $(A+B+C)$.

1. Total work $=\text{LCM}(36,60,45)=180$ units.
2. Rates: $A+B=5$, $B+C=3$, $A+C=4$ units/day.
3. $2(A+B+C)=12\Rightarrow A+B+C=6$ units/day.
4. $B=6-(A+C)=6-4=2$ units/day, so B alone takes $\frac{180}{2}=90$ days.

Trick: B alone $=(A+B+C)-(A+C)$; always remove the pair that does not contain B.

Answer: (C) 90 days

Concept: Express every share in terms of one share and use the total.

1. Let B = $x$. Then A $=x+500$, C $=x+800$.
2. $x+(x+500)+(x+800)=8200\Rightarrow 3x=6900\Rightarrow x=2300$.
3. C $=2300+800=3100$.

Trick: Check: $2300+2800+3100=8200$ ✔.

Answer: (A) ₹3,100

Concept: Sum of ratio terms $=3+4+7=14$; highest share $=\frac{7}{14}$ of the amount being divided.

1. If ₹14,014 is divided in $3:4:7$: one part $=\frac{14014}{14}=1001$.
2. Highest price $=7\times 1001=7007$.
3. (Taking ₹14,014 strictly as the average, the total is $3\times 14014=42042$, one part $=3003$ and the highest price would be $7\times 3003=21021$, which is not among the choices; the choices are built on dividing ₹14,014 in $3:4:7$.)

Trick: $14014=14\times 1001$, so each ratio unit is 1001 — highest share is simply $7\times 1001$.

Answer: (C) ₹7,007

Concept: Opposite angles of a cyclic quadrilateral add to ${180}^{\circ }$; the angle in a semicircle is ${90}^{\circ }$.

1. $\angle ABC={180}^{\circ }-\angle ADC={180}^{\circ }-{118}^{\circ }={62}^{\circ }$.
2. AB is a diameter, so $\angle ACB={90}^{\circ }$.
3. In $△ABC$: $\angle BAC={180}^{\circ }-{90}^{\circ }-{62}^{\circ }={28}^{\circ }$.

Trick: Directly: $\angle BAC=\angle ADC-{90}^{\circ }={118}^{\circ }-{90}^{\circ }={28}^{\circ }$.

Answer: (A) ${28}^{\circ }$

Concept: Area of trapezium $=\frac{1}{2}\times (\text{sum of parallel sides})\times h$; $1{\text{m}}^2=10000{\text{cm}}^2$.

1. Area $=\frac{1}{2}\times (17+15)\times 6=96{\text{cm}}^2$.
2. In ${\text{m}}^2$: $\frac{96}{10000}=0.0096{\text{m}}^2$.

Trick: ${\text{cm}}^2\to {\text{m}}^2$: divide by ${10}^4$ (shift the decimal 4 places left).

Answer: (D) 0.0096

Concept: Average students per school $=\frac{\text{total boys + total girls}}{5}$.

1. Boys: $600+450+750+700+800=3300$.
2. Girls: $500+550+650+600+650=2950$.
3. Average $=\frac{3300+2950}{5}=\frac{6250}{5}=1250$.

Trick: School totals 1100, 1000, 1400, 1300, 1450 — their average is $1250$.

Answer: (A) 1250

Concept: Add the class frequencies below 60 and above 70, then simplify.

1. Less than 60: $50+70=120$.
2. More than 70: $90+60+45=195$.
3. $120:195=8:13$ (divide by 15).

Trick: Both numbers are multiples of 15 — divide by it at once.

Answer: (A) 8 : 13

Concept: $\tan \theta =\frac{\text{height}}{\text{shadow}}$; from $\sec \theta =\frac{13}{12}$ use the triplet $5,12,13$.

1. $\sec \theta =\frac{13}{12}\Rightarrow \tan \theta =\frac{\sqrt{{13}^2-{12}^2}}{12}=\frac{5}{12}$.
2. Height $h=36\times \frac{5}{12}=15$.
3. So the pole is 15 units tall (15 m with the shadow in metres), matching the option 15.

Trick: $5$-$12$-$13$ triangle: shadow 36 $=12\times 3$, so height $=5\times 3=15$.

Answer: (B) 15

Concept: For the same distance, speed is inversely proportional to time.

1. Time of bus $=90$ min, car $=45$ min.
2. Speed car : bus $=90:45=2:1$.
3. (Speeds: car $80\text{km/h}$, bus $40\text{km/h}$.)

Trick: Same distance ⇒ invert the time ratio.

Answer: (D) 2 : 1

Concept: Write both selling prices in terms of CP and equate the difference to ₹1,000.

1. Let CP $=100x$; SP $=110x$.
2. New CP $=80x$; new SP $=1.4\times 80x=112x$.
3. $112x-110x=1000\Rightarrow x=500$.
4. Earlier SP $=110\times 500=55000$.

Trick: Difference in SP $=2\%$ of CP $=1000$ ⇒ CP $=50000$, SP $=55000$.

Answer: (A) ₹55,000

Concept: Radius ⟂ tangent, so OAPB has $\angle A=\angle B={90}^{\circ }$ and $\angle AOB={180}^{\circ }-\angle APB$.

1. $\angle AOB={180}^{\circ }-{130}^{\circ }={50}^{\circ }$.
2. $OA=OB$, so $△OAB$ is isosceles.
3. $\angle OAB=\frac{{180}^{\circ }-{50}^{\circ }}{2}={65}^{\circ }$.

Trick: $\angle OAB=\frac{\angle APB}{2}$ always: $\frac{{130}^{\circ }}{2}={65}^{\circ }$.

Answer: (D) ${65}^{\circ }$

Concept: Time $=\frac{\text{distance}}{\text{speed}}$; form the time-difference equation.

1. Let outward speed $=v$. $\frac{300}{v}-\frac{300}{v+10}=5$.
2. $\frac{3000}{v(v+10)}=5\Rightarrow v(v+10)=600$.
3. $v^2+10v-600=0\Rightarrow (v+30)(v-20)=0\Rightarrow v=20$.

Trick: Option test: $\frac{300}{20}-\frac{300}{30}=15-10=5$ ✔.

Answer: (B) $20\text{km/h}$

Concept: Central angle $=\frac{\text{percentage}}{100}\times {360}^{\circ }$, i.e. $1\%={3.6}^{\circ }$.

1. Department E $=20\%$.
2. Angle $=20\times {3.6}^{\circ }={72}^{\circ }$.

Trick: $1\%$ of a pie $={3.6}^{\circ }$.

Answer: (C) ${72}^{\circ }$

Concept: Volume is conserved on melting; surface area of a cube $=6a^2$.

1. Volume $=9^3+5\times 13\times 31=729+2015=2744{\text{cm}}^3$.
2. New side $a=\sqrt[3]{2744}=14\text{cm}$.
3. Surface area $=6\times {14}^2=1176{\text{cm}}^2$.
4. Cost $=1176\times 10=11760$.

Yaad rakhein: $2744={14}^3$ — remember cubes up to ${20}^3$.

Answer: (D) ₹11,760

Concept: Reduction is $12\%$ of the bill; amount paid $=88\%$ of the bill.

1. Bill $=\frac{2100}{12}\times 100=17500$.
2. Amount paid $=17500-2100=15400$.

Trick: Paid $=\frac{88}{12}\times 2100=15400$ directly. (17,500 is the bill, not the amount paid.)

Answer: (B) ₹15,400

Concept: Spot $a=0.325$, $b=0.175$ and use $(a+b{)}^2$ and $a^2-b^2$.

1. $25\times 0.00455=0.11375=2\times 0.325\times 0.175=2ab$, so numerator $=(a+b{)}^2={0.5}^2$.
2. $5\times 0.0065\times 3.25=0.105625=a^2$ and $7\times 0.175\times 0.025=0.030625=b^2$, so denominator $=a^2-b^2=(a+b)(a-b)$.
3. Fraction $=\frac{a+b}{a-b}=\frac{0.5}{0.15}=\frac{10}{3}$.
4. Total $=\frac{10}{3}+\frac{1}{3}=\frac{11}{3}$.

Trick: $\frac{(a+b{)}^2}{a^2-b^2}=\frac{a+b}{a-b}$.

Answer: (D) $\frac{11}{3}$

Concept: Total area = rectangle + two semicircles on the long sides + two on the short sides (two equal semicircles = one full circle).

1. Rectangle $=24\times 12=288{\text{m}}^2$.
2. Semicircles on 24 m sides (radius 12): $\pi \times {12}^2=144\pi$.
3. Semicircles on 12 m sides (radius 6): $\pi \times 6^2=36\pi$.
4. Total $=288+180\times 3.14=288+565.2=853.2{\text{m}}^2$.

Trick: Pair of semicircles on opposite sides = one full circle.

Answer: (A) $853.2{\text{m}}^2$

Concept: Percentage less $=\frac{\text{difference}}{\text{base (larger value)}}\times 100$.

1. (60–70): 85 cars; (70–80): 90 cars.
2. $\frac{90-85}{90}\times 100=\frac{500}{90}=5.56\%\approx 5.6\%$.

Trick: “Less than X” ⇒ X is the base. (5.9% comes from wrongly dividing by 85.)

Answer: (B) $5.6\%$

Concept: Divide by $x$ to get $x+\frac{1}{x}$; divide numerator and denominator by $x$ to get cubes.

1. $x^2+1=3x\Rightarrow x+\frac{1}{x}=3$.
2. $\frac{x^4+\frac{1}{x^2}}{x^2+1}=\frac{x^3+\frac{1}{x^3}}{x+\frac{1}{x}}$ (divide top and bottom by $x$).
3. $x^3+\frac{1}{x^3}=3^3-3\times 3=18$.
4. Value $=\frac{18}{3}=6$.

Trick: $x+\frac{1}{x}=k\Rightarrow x^3+\frac{1}{x^3}=k^3-3k$.

Answer: (B) 6

Concept: For 2 years: $\text{CI}-\text{SI}=P{\left(\frac{r}{100}\right)}^2$.

1. $P\times {0.17}^2=433.5\Rightarrow P=\frac{433.5}{0.0289}=15000$.
2. $\text{SI}=15000\times 0.34=5100$.
3. $\text{CI}=5100+433.5=5533.5$.

Trick: CI = SI + difference; don’t stop at 5,100 (that is SI).

Answer: (D) ₹5,533.50